Modern Physics with Modern Computational Methods 3rd Edition By John Morrison SOLUTIONS MANUAL

1 The Wave-Particle Duality - Solutions

1. The energy of photons in terms of the wavelength of light is given by Eq. (1.5). Following Example 1.1 and substituting λ = 200 eV gives: Ephoton =

1240 eV · nm hc = = 6.2 eV λ 200 nm

2. The energy of the beam each second is: Etotal =

power 100 W = = 100 J time 1s

The number of photons comes from the total energy divided by the energy of each photon (see Problem 1). The photon’s energy must be converted to Joules using the constant 1.602 × 10−19 J/eV , see Example 1.5. The result is: Nphotons =

100 J Etotal = = 1.01 × 1020 Ephoton 9.93 × 10−19

for the number of photons striking the surface each second.

3. We are given the power of the laser in milliwatts, where 1 mW = 10−3 W .

The power may be expressed as: 1 W = 1 J/s. Following Example 1.1, the energy of a single photon is: Ephoton =

1240 eV · nm hc = = 1.960 eV λ 632.8 nm

We now convert to SI units (see Example 1.5): 1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J Following the same procedure as Problem 2: Rate of emission =

1 × 10−3 J/s photons = 3.19 × 1015 3.14 × 10−19 J/photon s

2

4. The maximum kinetic energy of photoelectrons is found using Eq. (1.6) and the work functions, W, of the metals are given in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV . For part (a), Na has W = 2.28 eV : (KE)max = 6.20 eV − 2.28 eV = 3.92 eV Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV .

5. This problem again concerns the photoelectric effect. As in Problem 4, we use Eq. (1.6): (KE)max =

hc −W λ

where W is the work function of the material and the term hc/λ describes the energy of the incoming photons. Solving for the latter: hc = (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV λ Solving Eq. (1.5) for the wavelength: λ=

1240 eV · nm = 387.5 nm 3.2 eV

6. A potential energy of 0.72 eV is needed to stop the flow of electrons. Hence, (KE)max of the photoelectrons can be no more than 0.72 eV. Solving Eq. (1.6) for the work function: W =

1240 eV · nm hc − (KE)max = − 0.72 eV = 1.98 eV λ 460 nm

7. Reversing the procedure from Problem 6, we start with Eq. (1.6): (KE)max =

hc 1240 eV · nm −W = − 1.98 eV = 3.19 eV λ 240 nm

Hence, a stopping potential of 3.19 eV prohibits the electrons from reaching the anode.

8. Just at threshold, the kinetic energy of the electron is zero. Setting (KE)max = 0 in Eq. (1.6), W =

1240 eV · nm hc = = 3.44 eV λ0 360 nm

9. A frequency of 1200 THz is equal to 1200 × 1012 Hz. Using Eq. (1.10), Ephoton = hf = 4.136 × 10−15 eV · s × 1.2 × 1015 Hz = 4.96 eV

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Next, using the work function for sodium (Na) metal and Eq. (1.6), (KE)max = Ephoton − W = 4.96 ev − 2.28 eV = 2.68 eV 10. We start from Eq. (1.8) for the case of m = 2:

1 1 1 =R − 2 λ 22 n Now invert the equation and plug in for the Rydberg constant, R: λ=

1 1.0971 × 105 cm−1

1 1 − 2 4 n

−1

Subtract the fractions by getting a common denominator: λ=

1 cm 1.0971 × 105

n2 − 4 4n2

−1

Invert the term in the parenthesis and factor out the common factor of 4

4 cm n2 λ= 1.0971 × 105 n2 − 4 Doing the division, we get Eq. (1.7) for the Balmer formula:

n2 λ = (3645.6 × 10−8 cm) n2 − 4 11. Following Example 1.2,

13.6 eV 13.6 eV ∆E = − = 2.86 eV − − 52 22 Using Eq. (1.12): λ=

1240 eV · nm hc = = 434 nm ∆E 2.86 eV

12. Since the initial state has m = 2, we can use Eq. (1.7) with n = 4: 2 4 = 486.1 nm λ = (364.56 nm) 2 4 −4 To get the energy of the photon, use Eq. (1.5): Ephoton =

hc 1240 eV · nm = = 2.551 eV λ 486.1 nm

4

13. As in Problem 12, the initial state has m = 2, so we use Eq. (1.7) with n = 3: 2 3 λ = (364.56 nm) = 656.2 nm 2 3 −4 14. From Figure 1.6, the ionization energy of a hydrogen atom in the n = 2 state is −3.4 eV . So it takes a photon of 3.4 eV to just ionize this atom. To get the wavelength of light, just invert Eq. (1.5): λ=

hc Ephoton

=

1240 eV · nm = 364.7 nm 3.40 eV

15. Starting with Eq. (1.5) with a wavelength of 200 nm: Ephoton =

1240 eV · nm hc = = 6.20 eV λ 200 nm

From Figure 1.6, a hydrogen atom in the n = 2 state has the electron bound with potential energy (P E) = −3.4 eV . Following Example 1.3, (KE) = 6.20 eV − 3.40 eV = 2.80 eV 16. Starting with Eq. (1.5) with a wavelength of 45 nm: Ephoton =

hc 1240 eV · nm = = 27.6 eV λ 45 nm

For a hydrogen atom in the ground state, the electron is bound with potential energy (P E) = −13.6 eV . Following Example 1.3, (KE) = 27.6 eV − 13.6 eV = 14.0 eV To find the electron’s velocity, convert to SI units (see Example 1.5): (KE) = 14.0 eV ·

1.6 × 10−19 J = 2.24 × 10−18 J 1 eV

From Appendix A, an electron has mass m = 9.11 × 10−31 kg. Using the well known formula KE = (1/2)mv 2 , and solving for v: s r 2(KE) 4.48 × 10−18 J m v= = = 2.22 × 106 −31 m 9.11 × 10 kg s 17. From Figure 1.6, we see that the first transition for the Lyman series is between n = 2 to n = 1, and similarly we can get the transitions for the Balmer and Paschen series. Using Eq. (1.8):

1 1 1 =R − 2 λ m2 n

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with R = 1.0972 × 105 cm−1 , then plugging in for each case: (a) n = 2 to m = 1 gives λ = 1.215 × 10−5 cm, (b) n = 5 to m = 2 gives λ = 4.340 × 10−5 cm, (c) n = 5 to m = 3 gives λ = 1.282 × 10−4 cm.

18. We want to find the maximum wavelength possible for a hydrogen atom transition starting from the E3 state. Using Eq. (1.8) for absorption of a photon, starting at m = 3:

1 1 1 =R − 2 λ 32 n The maximum wavelength occurs when the RHS is at a minimum. Since the photon can only take the atom to a higher state, this requires n > 3. The minimum of the RHS occurs when n = 4.

1 1 1 =R − 2 λ 32 4 Plugging in R = 1.0972 × 105 cm−1 and inverting gives: λ = (9.114 × 10−6 cm)(20.57) = 187.5 × 10−6 cm Converting this to standard units gives λ = 187.5 × 10−8 m = 1875 nm.

Hence, light of wavelength greater than 1875 nm would not be absorbed by a hydrogen atom starting in the E3 state.

19. Starting with Eq. (1.26), p=

6.626 × 10−34 J · s h = = 3.31 × 10−24 J(s/m) λ 0.2 × 10−9 m

The kinetic energy is given by the standard formula: KE =

p2 = 6.024 × 10−18 J 2me

where me = 9.11 × 10−31 kg is the mass of an electron.

20. To get the wavelength of 40 keV photons, solve Eq.(1.5) for λ: λ=

hc Ephoton

=

1240 eV · nm = 0.031 nm 40 × 103 eV

An electron with the same wavelength has, using Eq. (1.26), a momentum of: p=

h 6.626 × 10−34 J · s = = 2.14 × 10−23 J(s/m) λ 0.031 × 10−9 m

6

The electron’s kinetic energy can be calculated in the usual way: KE =

p2 4.57 × 10−46 = = 2.51 × 10−16 J 2me 2(9.11 × 10−31 )

The work done to accelerate the electron is W = (KE). The voltage (or potential difference) is given by W = −qV where q is the charge. Dividing W by minus the electron charge: V =

2.51 × 10−16 J = 1.57 V olts 1.602 × 10−16 C

21. For all three cases, we can use Eqs. (1.26) for the momentum: p=

6.626 × 10−34 J · s h = = 6.626 × 10−23 J(s/m) λ 0.01 × 10−9 m

and calculate the KE as in the problem above, using the appropriate mass from Appendix A. The results are: KE (J) KE (eV) Particle mass (kg) electron 9.11 × 10−31 2.41 × 10−15 15060 proton 1.673 × 10−27 1.312 × 10−18 8.20 neutron 1.675 × 10−27 1.311 × 10−18 8.19 22. From Fig. 1.3, we see that visible light extends from about 400 nm to 700 nm. Using Eq. (1.26) as above and v = p/me for the velocity: Wavelength (nm) momentum (kg·m/s) velocity (m/s) 400 1.657 × 10−27 1818 700 9.466 × 10−28 1039 23. First convert the energy to SI units: E = 40 × 103 eV × Using p =

1.6 × 10−19 J = 6.4 × 10−15 J 1 eV

p 2m(KE) = 1.08 × 10−22 kg(m/s) in Eq. (1.25): λ=

6.626 × 10−34 J · s h = = 6.14 × 10−12 m p 1.08 × 10−22 kg(m/s)

which is the de Broglie wavelength. Next, we observe that Eq. (1.25) does not depend on the mass, so a proton with the same de Broglie wavelength has the same momentum, p = 1.08 × 10−22 kg(m/s). Using the proton mass: (KE)p =

p2 (1.08 × 10−22 kg(m/s))2 = = 3.49 × 10−18 J 2mp 2(1.673 × 10−27 kg)

If desired, the units can be converted giving (KE)p = 21.8 eV .

2 The Schrödinger Wave Equation - Solutions

1. (a) Using Eq. (2.17) for the energies of a particle in an infinite well: E=

n2 h2 2mL2

Solving this for L: L=

r

n2 h2 8mE

Using the given energy E1 = 1.0 eV = 1.6 × 10−19 J and the mass of the electron from Appendix A, the lowest energy is found when the electron is in the ground state (n = 1): s 12 (6.63 × 10−34 J · s)2 L= = 6.14 × 10−10 m 8(9.11 × 10−31 kg)(1.6 × 10−19 J) (b) Now that we have L, the energy for the next excited state (n = 2) is: E2 =

22 (6.626 × 10−34 J · s)2 n2 h2 = = 2.14 × 10−18 J 2mL2 8(9.11 × 10−31 kg)(6.14 × 10−10 m)2

Converting the units, E2 = (2.14 × 10−18 J)/(1.6 × 10−19 J/eV ) = 4.00 eV .

Finally, the energy needed to transition from the ground state (E1 ) to the first excited state (E2 ) is ∆E = E2 − E1 = 4.00 − 1.00 eV = 3.00 eV .

2. Using Eq. (2.20) from Example 2.2, the wave function for an infinite square well with center at x = 0 and odd n is: r nπx 2 ψ(x) = cos L L Plugging in n = 3 and L = 10 nm,

2

ψ(x) =

r

2 cos 10 nm

3πx 10 nm

with boundaries −5 nm < x < 5 nm. This has numerical values: x (nm) ψ(x) 0 0.447 2 −0.139 4 −0.364 8 0 10 0 The last two are zero because they are outside of the infinite square well.

3. Using Eq. (2.17) for the energy levels of a 10 nm wide infinite well with n = 3 and n = 2: 9(6.63 × 10−34 J · s)2 32 h2 = 2 8mL 8(9.11 × 10−31 kg)(10 × 10−9 m)2 1 eV = 5.43 × 10−21 J · = 0.034 eV 1.6 × 10−19 J 4(6.63 × 10−34 J · s)2 22 h2 = E2 = 8mL2 8(9.11 × 10−31 kg)(10 × 10−9 m)2 1 eV = 2.41 × 10−21 J · = 0.015 eV 1.6 × 10−19 J ∆E = E3 − E2 = 0.034 eV − 0.015 eV = 0.019 eV E3 =

where ∆E is the energy of the emitted photon. Using Eq. (1.5) to calculate the wavelength of the light: λ=

1240 eV · nm hc = = 65 × 103 nm = 65 µm Ephoton 0.019 eV

4. To show this, first evaluate the second derivative of ψ(x): dψ mωx −mωx2 /2~ = −A e dx ~ 2 d ψ mω −mωx2 /2~ m2 ω 2 x2 −mωx2 /2~ = −A e + A e dx2 ~ ~2 Substituting this into the LHS of Eq. (2.31) and canceling common factors:

mω m2 ω 2 x2 1 −~2 − + mω 2 x2 = E + 2m ~ ~2 2 Carrying out the algebra: E=

~ω 2

3

which is the same as Eq. (2.38) when n = 0. Hence, the wave function given here satisfies Schrödinger’s equation, Eq. (2.32), for the ground state.

5. The normalized wave function must satisfy Eq. (2.18): Z ∞ |ψ(x)|2 dx = 1 −∞ Z ∞ Z ∞ 2 −mωx2 /2~ 2 A2 e−mωx /~ = 1 |Ae | dx = −∞

−∞

Dividing both sides by A2 and noticing the symmetry of the integral: Z ∞ Z ∞ 2 1 −mωx2 /~ e−mωx /~ dx e dx = 2 = A2 0 −∞ Now substitute a = mω/~ and use the integral given in the problem: r r 1 π π~ = mω = A2 mω ~ mω 14 A= πℏ

6. (a) The wave function is zero for x < 0, then rises linearly for small x, after which it turns over and goes back to zero at large x due to the exponential term in ψ(x). (b) Using the normalization condition, Eq. (2.18): Z ∞ A2 x2 e−2ax = 1 0

Integrating this by parts twice and solving for A gives: A2 = 4a3 (c) We want to find the maximum of the wave function squared, |ψ(x)|2 .

Taking its derivative and setting it equal to zero: d d |ψ(x)|2 = (4a3 x2 e−2ax ) = 8a3 xe−2ax − 8a4 x2 e−2ax = 0 dx dx Solving for x yields 1/a. The maximum probability to find the particle is at x = 1/a. (d) The average value of the position is given by Eq. (2.21): Z ∞ 4a3 x3 e−2ax dx hxi = 0

Integrating by parts three times yields hxi = 3/(2a).

4

7. (a) V = 0 in the region 0 ≤ x ≤ L, so the Schrödinger equation becomes: −

~2 d 2 ψ = Eψ 2m dx2

Now V = V0 in the region x ≥ L so: −

ℏ2 d 2 ψ + V0 = Eψ 2m dx2

(b) Defining: k=

r

2mE and κ = ~2

r

2m(V0 − E) ~2

Then, d2 ψ + k2 ψ = 0 (0 ≤ x ≤ L) dx2 d2 ψ − k2 ψ = 0 (x ≥ L) dx2 For the finite well discussed in section 2.3 of the text, the above equations are satisfied by the general solutions: ψ(x) = A cos(kx) and ψ(x) = A sin(kx) ψ(x) = Be

−κx

(0 ≤ x ≤ L)

(x ≥ L)

(c) The potential is infinite at x = 0, so the wave function must go to zero there. Examining the new boundary conditions at x = 0: Even : ψ(0) = A cos(0) = 0 Odd : ψ(0) = A sin(0) = 0 The even solution requires A = 0, and hence there are no even solutions.

We are left with: ψ(x) = A sin(kx) for 0 ≤ x ≤ L. Next, we examine the boundary conditions at x = L: ψ(L) = A sin(kL) = Be−κL Imposing the continuity of the first derivative: ψ ′ (L) = Ak cos(kL) = −Bκe−κL Dividing the above two equations gives: κ ψ ′ (L) = −cot(kL) = ψ(L) k

5

Inserting κ as defined in part (b): −cot(kL) =

r

2mV0 2mE − 2 2 ~2 k 2 ~ k

Substitute k into the second term of the RHS: r 2mV0 −1 −cot(kL) = ~2 k 2 This can be solved graphically (or numerically) for the values of k for which the LHS and RHS expressions intersect. This happens only for quantized values of k. Then use the definition of k above to solve for the energy E. Each solution for k corresponds to an energy level of a bound state.

8. Starting from Eq. (2.47): ψ(x, t) = Aeikx · e−iωt Taking the second derivative d2 ψ = −Ak 2 eikx e−iωt dx2 Taking the partial derivative ∂ψ = −iωAeikx e−iωt ∂t Plugging into the time-dependent Schrödinger equation, Eq. (2.53), gives: ~2 k 2 ψ(x, t) + V (x, t)ψ(x, t) = ~ωψ(x, t) 2m This equivalence is satisfied for a free particle (V = 0) when ~2 k 2 = ~ω 2m Using de Broglie’s relation p = ~k from Eq. (1.27) and Einstein’s relation E = hf = ~ω from Eq. (1.1), we see that the above is the same as the familiar formula E = p2 /2m. Hence, the traveling wave satisfies the time-dependent Schrödinger equation.

9. The wording of this problem could be confusing, since the infinite potential well represents a situation where the potential energy does not evolve with time. The goal here is to solve the time-dependent Schrödinger equation in the case of a static potential. The time-dependent solutions must then satisfy Eq. (2.49):

6

Ψ(x, t) = ψ(x)e−iωt where ω = E/~. From Eq. (2.20), the normalized even spatial wave functions of a particle in an infinite well are: r nπx 2 cos ψ(x) = L L The corresponding energies are given by Eq. (2.17): E=

n2 h2 8mL2

where n is even for this problem. Calculating ω from the above: ω=

n2 πh E = ~ 4mL2

for even n. The total wave function for even solutions is then: r

nπx 2 n2 πh Ψ(x, t) = t cos exp −i L L 4mL2 for even n.

3 Operators and Waves - Solutions

1. Using Eq. (3.2), p̂ = −i~(d/dx), p̂φ1 (x) = i~k cos kx p̂φ2 (x) = −i~k sin kx These wave functions are not eigenfunctions of the momentum operator. If they were eigenfunctions, then they would have the form p̂φ(x) = Cφ(x) for some constant C. 2. We want to find a linear combination ψ = Aφ1 (x) + Bφ2 (x) such that: p̂ψ = pψ for some constant p. Applying Eq. (3.3), p̂ψ = pψ, p̂ψ = −i~

dψ = i~k A sin(kx) − i~k B cos(kx) dx

Since the constants A and B have switched places compared with ψ(x), we see that an eigenvalue equation can only be formed with something like A=B. Looking more closely at the eigenvalue condition: i~k A sin(kx) − i~k B cos(kx) = p(A cos(kx) + B sin(kx)) then setting the coefficients equal, a solution is found if B = iA and p = ~k.

Hence, ψ is an eigenfunction of the momentum when ψ(x) = A(sin(kx) + i cos(kx)) = Aeikx where Euler’s formula has been used. So, ψ(x) has eigenvalue p = ~k.

3. (a) This one-dimensional lattice looks identical if the coordinate system is shifted by an integer multiple of the lattice spacing a:

2

ψ(x) = ψ(x + a) The momentum eigenvalue equation then becomes p̂ψ(x) = pψ(x) = pψ(x + a) (b) Following the reasoning of Problem 2, the general solution to this equation is: ψ(x) = Aeikx = Aeik(x+a) which has eigenvalue p = ~k (see Problem 2). (c) Using the boundary condition given: ψ(0) = A = ψ(a) = Aeika we see that eika = 1. Using Euler’s identity ei2πn = 1 gives ka = 2πn for integer n. Since the momentum eigenvalue is p = ~k: p=

2π~n nh = a a

Comparing this with de Broglie’s equation (1.26), p = h/λ we see that the electron takes on wavelengths λ equal to integer fractions of the lattice spacing a. For example, if n = 2 then λ = a/2.

4. Following the guidance given in Appendix C, first define the range of x by: x = linspace(0,10,100); which gives 100 equally-spaced points between x = 0 and x = 10. Next, define the function y(x) as: y = x.*exp(-x); just as shown in Appendix C. Note that the ”.” before the multiplication sign is necessary, since x is a vector of length 100. Finally, plot(x,y); shg where the last command (shg, short for show graph) brings the plot window to the foreground.

5. Note that the thickness L and the constant E0 are unchanged from MATLAB Program FiniteWell.m given in Section 3.2. The only change for this problem is that V0 = 0.2 eV whereas it was 0.3 eV in Section 3.2. We need only change one line of the MATLAB program: v0=0.2;

3

After running this program in MATLAB, the ground state eigenvalue for a five-point grid (n = 5 above) returns a value that is not very accurate. Increasing this to a 100-point grid (by changing line 3 of the program to n = 100) returns a more accurate value of 0.0300, in units of eV .

The lowest eigenfunction is plotted automatically by Program FiniteWell.m and the result is shown below:

6. The technique of Gaussian quadrature is described elsewhere (such as Wikipedia), and is often described in terms of symmetric integrals: Z 1

f (x) dx

−1

where f (x) can be approximated by a polnomial of degree n. Then, for N sample points, x1 to xN , we can approximate: Z 1

−1

f (x) dx ≃

N X

ci f (xi )

i=1

where the xi are called the gaussian nodes and the ci are constants called gaussian weights. The nodes and weights can be tabulated for a given N : N 2

qxi

− 31 q 1

3 −

q3

3 5

q0 3 5

ci 1.0 1.0 5 9 8 9 5 9

The results of the above sum is exact if f (x) is a polynomial of order n = 2N − 1. For example, when f (x) = x2 , then N = 2 is sufficient. But when f (x) = x4 , then N = 3 is needed. When P f (x) is not a polynomial, then there is always a small error in the sum ci f (xi ). In that case, it may help to break up the integral into several intervals along x.

4

For definite integrals that are not symmetric about the origin: Z b

f (x) dx

a

b+a b−a then one can transform the variables to ui = b−a 2 xi + 2 and di = 2 ci .

Then

Z b a

f (u) du ≃

N X

di f (ui )

i=1

Now apply these results to the integrals given in this problem. From direct integration: Z 10 x3 10 1000 x2 dx = = 3 3 0 0 Since f (x) is a polynomial, Gaussian quadrature will be exact. Doing the algebra in a MATLAB program x = [-1.0/sqrt(3), 1.0/sqrt(3)] c = [1.0, 1.0] a = 0 b = 10 u = 0.5*(b-a)*x + 0.5*(b+a) d = 0.5*(b-a)*c f = u.^2 ans = sum(d.*f) with the result of ans = 333.33 . . . which is exact. Taking the next integral: Z 10 x5 10 = 20000 x4 dx = 5 0 0 Because the polynomial is of degree n = 4, we need 3 Gauss points. Modifying the above MATLAB program, x = [-sqrt(3/5), 0, sqrt(3/5)] c = [5/9, 8/9, 5/9] a = 0 b = 10 u = 0.5*(b-a)*x + 0.5*(b+a) d = 0.5*(b-a)*c f = u.^4 ans = sum(d.*f) gives the result ans = 2.00e+04 as expected. For the last integral, integrating by parts:

5

Z 10 0

−10 xe−x dx = (xe−x − e−x )|10 0 = 1 + 9e

which equals 1.000409 . . . numerically. Here, the function is not a polynomial, so we expect Gaussian quadrature will not be exact. Modifying the last two lines of the above program: f = u.*exp(-u) ans = sum(d.*f) with result ans = 1.1675, about a 15% error. More accuracy can be obtained by breaking the integral into two parts and adding together the results. First, integrating from a = 0 to b = 2 using Gaussian quadrature with N = 3 gives ans = 0.5941. Next, integrating from a = 2 to b = 10 using the same method gives ans = 0.4094. Adding these gives 1.0035, which is much closer to the exact answer. More precision can be obtained by breaking the integral up into more segments.

7. The only change for this problem is that V0 = 0.2 eV whereas it was 0.3 eV in Section 3.3. This is the same change as for Problem 5, except we now use MATLAB Program FiniteWellcoll.m. We need only change one line of the MATLAB program: v0=0.2; After running this program in MATLAB, and increasing the grid points to n = 100 (line 3 of the program) returns a ground state energy of 0.0309, in units of eV . This is more accurate that what we found in Problem 5, due to the more sophisticated numerical techniques.

8. Following the development for a potential step in Fig. 3.8 (Section 3.4.2): d2 ψ1 + k1 2 ψ1 = 0 dx2 and from Eq. (3.64): k1 =

r

2mE ~2

giving the general solution, Eq. (3.65), for x < 0: ψ1 = Aeik1 x + Be−ik1 x For x > 0, using Eq. (3.75): d2 ψ2 − k2 2 ψ2 = 0 where k2 = dx2

r

2m(V0 − E) ~2

6

gives the physically acceptable solution Eq. (3.76): ψ2 (x) = De−k2 x Now, we impose continuity of the wave equation at x = 0 and its first derivatives at x = 0: ψ1 (0) = ψ2 (0) ⇒ A + B = D ψ1 (0) = ψ2 ′ (0) ⇒ ik1 (A − B) = −k2 D ′

Divide both side of the second equation by ik1 : A+B =D

A−B =−

k2 k2 D=i D ik1 k1

(1)

(2)

Adding these two equations, and solving for A in terms of D: k2 ) k1 1 k2 A = D(1 + i ) 2 k1 2A = D(1 + i

(3)

Subtracting (1) from (2), and solving for B in terms of D: k2 ) k1 k2 1 B = D(1 − i ) 2 k1 2B = D(1 − i

Dividing each of equations (3) and (4) by D: A 1 k2 = (1 + i ) D 2 k1 1 k2 B = (1 − i ) D 2 k1 The reflection coefficient R is defined by Eq. (3.72) giving: R=

|B|2 v1 |B|2 = |A|2 v1 |A|2

Using our ratios above: R=

k2 k2 1 B ∗ B (D ) (D) 4 (1 − i k1 )(1 + i k1 ) = =1 A ∗ A k2 k2 1 ) (D) (D 4 (1 + i k1 )(1 − i k1 )

(4)

7

9. Following the development in 3.4.4, the T -matrix for the case of two barriers is given in Eq. (3.83). Here, we are given that L = 10 nm for each barrier, p and the distance between barriers is also 10 nm. From Eq. (3.64), k1 = 2mE/~2 and using the electron mass √ in GaAs, m = 0.067 · 511 keV along with ~ = 197.3 eV · nm, we get p k1 = 1.759E, where E is the variable here. Similarly, from Eq. (3.75), k2 = 1.759(E − V0 ) where V0 = 0.3 eV for both barriers. All that remains is to program Eq. (3.83) into MATLAB. Following the guide from Function OneBarrier(E) in Section 3.4.4, a program to do the two-barrier case is: function T = twobarrier(E) L=10.0; V0=0.3; k1=sqrt(1.759*E); k2=sqrt(1.759*(E-V0)); A=[(k1+k2) (k1-k2); (k1-k2) (k1+k2)]; B=[(k1+k2) (k2-k1); (k2-k1) (k1+k2)]; C=[exp(i*k2*L) 0; 0 exp(-i*k2*L)]; D=[exp(i*k1*L) 0; 0 exp(-i*k1*L)]; M=A*C*B*D*A*B*C; t=(4*k1*k2)^2/M(1,1); T = abs(t)^2 To plot the transmission coefficient from E = 0.3 eV to 1.0 eV , follow the guide of Program OnebarrierPlot.m of Section 3.4.4, for n=1:140 en=0.3+n/200.; E(n) = en; T(n) = twobarrier(en); end plot(E,T); The result is the following plot:

10. Using the first equation from Section 3.4: ∆x · ∆px ≥

~ 2

8

Taking the given values, ∆x = 1.0 nm and the value of ~ from Appendix A: p ∼ ∆p =

1.055 × 10−34 J · s = 5.28 × 10−26 N · s 2(1.0 × 10−9 m)

11. Using the second equation from Section 3.4: ∆E · ∆t ≥

~ 2

The natural line width is estimated as ∆E =

1.055 × 10 − 34 J · s = 1.32 × 10−25 J 2(4.0 × 10−10 s)

Converting units: ∆E = 1.32 × 10−25 J ·

1 eV = 8.24 × 10−7 eV 1.6 × 10−19 J

12. Using the normalization condition, Eq. (2.18): Z ∞ |ψ(x)|2 dx = 1 −∞

and inserting the given wave function and boundary conditions: Z 1 B 2 e−2x dx = 1 0

Doing the integral:

1 −2 2 e − 1 = −0.5(0.1353 − 1) = 0.432 B − 2 Taking the square root, B = 0.658.

4 The Hydrogen Atom - Solutions

1. The only change needed in MATLAB Program Hydrogen.m is to modify the last few lines: dEnergies(1:2); dVectors = V(:,index(1:2)); The energy of the second d state is -0.0309 (in atomic units) which, after multiplying by 27.2 eV , corresponds to an energy of −0.84 eV .

2. MATLAB Program Hydrogen.m produces an output file called hydrogen.mat. First, load this file, the just plot the dVectors: load(’hydrogen.mat’); plot(dVectors); shg The above lines of code will plot both 3d and 4d functions on the same graph, with the x-axis as the integer index of the vector. If desired, a plot of the 3d function versus r can be shown by loading hydrogen.mat and typing: r = linspace(0,40,800); plot(r,dVectors(:,1); shg Note that there are 800 points in the eigenfunction, since each step has two Gauss quadrature points. A separate plot of the 4d eigenfunction can be shown by changing (:,1) to (:,2) in the above code.

3. To get the average value of 1/r3 for the 2p electron of hydrogen, the following MATLAB code can be run (assuming you have previously run program Hydrogen.m to get the MATLAB output file hydrogen.mat): load(’hydrogen.mat’); vecp2 = pVectors(:,1);

2

r = linspace(0,40,400); inverser3 = r.^(-3); for j = 1:400; average = inverser3(j)*(vecp2(2*j)^2); end average(1) = 0.5*average(1); sum(average) The first line loads the Eigenvectors found from program Hydrogen.m and the second line loads the first p-state (the 2p state) into a 1-dimensional vector with name vecp2, which has length 800. Only the even indices of vecp2 contain the amplitude of the 2p wavefunction, in atomic units. Next, the vector r is generated with length 400, from 0 nm to 40 nm to match the grid points used in Hydrogen.m (first three lines). The for loop gets the integrand of the integral for h1/r3 i, as shown in Section 4.3.5. Finally, using Simpson’s rule for the numerical integration, the result is: sum = h1/r3 i = 0.0414 in atomic units. (To get standard units, multiply by 1/a30 where a0 = 0.0529 nm.) The exact answer (from calculus) gives h1/r3 i = 1/24 = 0.0417 in atomic units.

4. The formula for the spin-orbit constant is given in Section 4.3.5, wee Eq. (4.54):

1 1 1 2 Ze ζ= 2m2 c2 4πǫ0 r3 where mc2 = 511 keV is the electron mass and 1/4πǫ0 is the usual electrostatic constant (see Appendix A). Using Z = 1 for hydrogen, and the result from Problem 3, this can be cast into atomic units as shown in Eq. (4.57), 3 α2 1 a0 ζ= = (27.2 eV ) (27.2 eV )(0.0417) 2 r3 2(1372 ) where α = 1/137 is the fine structure constant and the result of the previous problem has been used. The result is ζ = 3.02 × 10−5 eV .

5. To modify MATLAB Program Hydrogen.m for a hydrogen-like ion with Z = 8, only one line needs to be changed: z = 8; which is found just after the comment % Construct xcol. The resulting eigenvalues are just those of hydrogen multiplied by 82 = 64.

6. After running the modified Hydrogen.m using Z = 8, to plot the 3d eigenfunction, type: load(’hydrogen.mat’); plot(dVectors);

3

shg The wave functions for this hydrogen-like ion are compressed toward r = 0, being pulled inward by the Coulomb force, but have the same shape as for hydrogen.

7. Following Example 4.2 of Section 4.2.2, the radial part of the transition integral for the 3d → 2p transition is: Z ∞ P2p (r) · r · P3d (r) dr 0

Using Simpson’s rule for the numerical integration, similar to Problem 3, the MATLAB code is: load(’hydrogen.mat’); vecp2 = pVectors(:,1); vecd3 = dVectors(:,1); r = linspace(0,40,400); for j = 1:400; radial = r(j)*vecp2(2*j)*vecd3(2*j); end radial(1) = 0.5*radial(1); radial(400) = 0.5*radial(400); sum(radial) The result is that the radial part of the transition integral has a value of 4.75.

More accuracy can be obtained by modifying program Hydrogen load vectors.m in Section 4.1.4, but the above code is sufficient. Note that it is easy to modify the above program to do the radial integral of Example 4.2 (just change dVectors to sVectors) in order to verify the value of 1.29 used there for the 2p → 1s tranision.

8. Using the selection rules from Table 4.3: ∆l = ±1 Hence, a p-state (l = 1) can decay to either a d-state or an s-state with lower n. The possible final states are: 3d, 3s, 2s or 1s. Note that we require l < n so there is no d-state for n = 1, 2 (as shown in Figure 4.5).

9. From Section 4.3.4, and Example 4.3, for the f -state with l = 3, the possible values of ml and ms are: ml = −3, −2, −1, 0, 1, 2, 3 1 1 ms = − , + 2 2

4

10. The total angular momentum of the 4f electron will is the vector sum of the electron’s spin and its orbital angular momentum. Following Section 4.3.4: J = j1 + j2 , j1 + j2 − 1, . . . |j2 − j1 | Here, j1 is the orbital angular momentum and j2 is the spin. So for an electron in the f -state, j1 = 3, with spin j2 = 1/2, the possible values for J are: J=

7 5 , 2 2

Using Eq. (4.50), the spin-orbit coupling energy is: ζ~2 3ζ~2 l= 2 2 −ζ~2 Es−o = (l + 1) = −2ζ~2 2

Es−o =

1 2 1 for j = l − 2 for j = l +

Hence the levels are split by ∆Es−o = (7/2)ζ~2 . The spin-orbit constant ζ for the 2p state of hydrogen was calculated in Problem 4, ζ = 3.02 × 10−5 eV .

11. An electron in the 2p state, with l = 2 and s = 1/2, has possible values: j=

3 1 , 2 2

From Eq. (4.63) and the equations that follow: g3/2 =

2 4 and g1/2 = 3 3

The splitting of the m-levels is: ∆E = gj µB Bmj where µB is the Bohr magneton given in Appendix A and B = 5 T . For j = 3/2, the possible mj values and energy splittings ∆E are: ∆E (eV ) mj +(3/2) 5.796 × 10−4 +(1/2) 1.932 × 10−4 -(1/2) −1.932 × 10−4 -(3/2) −5.796 × 10−4 Similarly, for j = 1/2, the energy splittings are ∆E = ±0.966 × 10−4 eV for mj = ±1/2, respectively.

5

12. Similar to Problem 5, just modify MATLAB Program Hydrogen.m but with Z = 2. Only one line needs to be changed: z = 2; which is found just after the comment % Construct xcol. The resulting eigenvalues are just those of hydrogen multiplied by 22 = 4. For example, in atomic units, the ground state of the He+ ion has eigenvalue −2.00 compared with hydrogen which had eigenvalue −0.500. To change these to regular units, just multiply by the atomic unit of energy, 27.2 eV , see Eq. (4.6).

13. To get the average value h1/r3 i for the 2p state of the He+ ion, first run the program of Problem 12, and then use the same code as in Problem 3. The result is h1/r3 i = 0.326 in atomic units. To get regular units, multiply by 1/a30 where a0 is the Bohr radius.

14. Following Problem 4, but with Z = 2, 3 1 Zα2 a0 = (27.2 eV ) (27.2 eV )(0.326) ζ= 2 r3 1372 where α = 1/137 is the fine structure constant and the result of the previous problem has been used. The result is ζ = 4.72 × 10−4 eV .

15. Using the results from Problems 4 and 14, the ratio of the spin-orbit constant for the 2p state of the He+ ion to that of hydrogen is 4.72 × 10−4 /3.02 × 10−5 = 15.7. With more numerical precision, the exact ratio of 24 = 16 can be obtained.

6 The Emergence of Masers and Lasers Solutions

1. From Figure 6.7, the 4 F2 band ranges from about 17, 000 cm−1 to 19, 000 cm−1 . The units tell us that these quantities are given in terms of 1/λ, the inverse of the light’s wavelength. Using Eq. (1.22): f=

c λ

where c is the speed of light (in units of cm/s). Converting to frequency:

cm 17 × 103 cm−1 3 × 1010 = 5.1 × 1014 Hz s

cm = 5.7 × 1014 Hz 19 × 103 cm−1 3 × 1010 s Similarly, for 4 F1 :

cm 23 × 103 cm−1 3 × 1010 = 6.9 × 1014 Hz s

cm = 8.1 × 1014 Hz 27 × 103 cm−1 3 × 1010 s 2. From Figure 6.9, the 2p to 1s transitions are shown by dashed lines with wavelengths ranging from 5944 Å to 6678 Å. The level splittings are are shown in Figure 6.9 next to the 2p and 1s symbols (note that 2p is repeated twice in the Paschen notation used in this figure–we want the lower-energy one here).

By carefully measuring the energy of the 2p3 and 1s3 states, the wavelength of this transition can be estimated at approximately 6250 Å = 6250×10−10 m.

3. In the JK coupling scheme, we couple the angular momentum of the core, J, to the angular momentum of the electron, L. For neon, with a configuration 2p5 (2 P3/2 )3p1 , where the term in the parenthesis uses spectroscopy notation

2 2s+1

LJ (see Section 5.4.1), the core has J = 3/2 coupled with the excited electron in the 3p orbital with l = 1. Following the rules for addition of angular momentum from Section 4.3.4: ~ = J~ + ~l K then the possible values for K are: ~ = J + l, J + l − 1, . . . |J − l| = |K|

5 3 1 , , 2 2 2

~ +S ~ where |S| ~ = 1/2.

The total angular momentum includes spin, which is K Hence, the possible values for the total angular momentum are: 3,2,1,0.

4. The hyperfine interaction couples the spin of the nucleus to the spin of the electron. Given that I = 1/2 and the electron’s spin S = 1/2, and using the rule for addition of angular momentum from Section 4.3.4: ~ F~ = I~ + S giving the possible values of |F~ | = 0, 1. In a magnetic field, these lines will be split, similar to the drawing shown in Fig. 4.21, with magnetic substate of MF = −1, 0, 1. Of course, the state with |F~ | = 0 has no splitting and is degenerate with the MF = 0 substate.

5. Using Eq. (6.7) with F = 1, I = 1/2 and S = 1/2: gF =

(1)(1 + 1) − 21 ( 12 + 1) + 21 ( 21 + 1) =1 (1)(1 + 1)

The energy splitting in a magnetic field B is given by combining Eqs. (6.3) and (6.5): Emag = gF µB BMF where from the problem above the only non-zero MF values are ±1. Since gF > 0, the states with positive MF have a positive energy in a B-field. From Eq. (6.5), µz = −gF µB MF and so states with positive MF have a negative µz and vice-versa. Hydrogen atoms with a positive µz value will be drawn to regions of high magnetic field outside the magnetic trap (since the trap has a local minimum of B, as described in Section 6.4). These atoms will be lost, decreasing the average kinetic energy and temperature of the remaining atoms caught in the trap.

3

6. We are given that µz = −gS µB Sz + gN µN Iz where Sz and Iz are the projections of the particle spin along the total angular momentum F : Sz = Iz =

~ · F~ )Fz (S F~ · F~ (I~ · F~ )Fz F~ · F~

Using the rules for addition of angular momentum from Section 4.3.4 with ~ + I: ~ F~ = S ~ · F~ ) = 1 (F 2 + S 2 − I 2 ) (S 2 1 2 ~ ~ (I · F ) = (F + I 2 − S 2 ) 2 where the eigenvalue for any angular momentum operator J 2 is J(J + 1) and similarly for Jz the eigenvalue is MJ . Plugging this in gives:

1 F (F + 1) + S(S + 1) − I(I + 1) MF Sz = 2 F (F + 1)

1 F (F + 1) + I(I + 1) − S(S + 1) MF Iz = 2 F (F + 1) The z-component of the magnetic field then becomes:

gs F (F + 1) + S(S + 1) − I(I + 1) µz = − µB MF 2 F (F + 1)

F (F + 1) + I(I + 1) − S(S + 1) gN MF + µN 2 F (F + 1) Using the definition of gF given by Eq. (6.6) and the approximate form (6.7), this simplifies to: µz = −gF µB MF which is the desired result.

7. Given S = 1/2 and I = 5/2, the possible values for F are F = 2, 3. These levels split as follows: F = 3 : MF = −3, −2, −1, 0, 1, 2, 3 F = 2 : MF = −2, −1, 0, 1, 2 The MF = −3 and MF = 3 states are not mixed with any states, and will be “pure” states, with straight lines on a plot like Figure 6.11. The other states of F = 3 and all states where F = 2 will be mixed, and their energy values make curved lines on a plot like Figure 6.11.

7 Diatomic Molecules - Solutions

There are no problems listed in this chapter.

8 Statistical Physics - Solutions

1. There are sixteen possible outcomes of flipping a coin four times. Using Eq. (8.3) to calculate the statistical weight for N = 4, with n equal to the number of heads obtained in each distribution:

4!

4 =1 HHHH : = 4!0!

4

4!

4 =4 HHHT : = 3!1!

3

4!

4 HHT T : = =6 2!2!

2

4!

4 =4 HT T T : = 1!3!

1

4!

4 =1 TTTT : = 0!4!

0 2. Since there is no particular order to the grouping, use Eq. (8.3) with N = 4 and n = 2:

24 4 =6 = 4 2 3. Using Eq. (8.7) for each energy level: n2 n2 N = = eǫ2 /kB T g2 2 Z n1 n1 N = = eǫ1 /kB T g1 1 Z Divide the above two equations:

2

n2 = 2e−(ǫ2 −ǫ1 )/kB T n1 We are given ǫ2 − ǫ1 = 0.025 eV, T = 298K, and kB from Appendix A:

n2 −0.025 eV = 0.755 = 2 exp n1 (8.617 × 10−5 eV /K)(298K) So 75.5% of the electrons are in the first excited state and 24.5% are in the ground state.

4. Integrating over P (p) dp from Eq. (8.20), and letting a = 1/(2mkB T ), Z ∞ Z 4a3/2 ∞ 2 −ap2 P (p) dp = √ p e dp π 0 0 The integral itself is of the form of Eq. (8.18) with u = p and n = 2: Z ∞ 2 u2 e−au du = I2 (a) 0

Using Appendix G: I2 (a) =

√

π 4a3/2

shows that the probability distribution is normalized as given in Eq. (8.21).

5. F1 (u) is defined as: 2 4 F1 (u) = √ u2 e−u π

Maximizing F1 (u) by setting the derivative of F1 (u) equal to zero: 2 2 4 dF1 = √ (2ue−u − 2u3 e−u ) = 0 du π 2

2ue−u (1 − u2 ) = 0 u = 0, ±1

The negative solution has no physical interpretation. Also, F1 (0) = 0, so F1 (u) has a maximum at u = 1. This agrees with the plot shown in Figure 8.2.

6. Following the method of Example 8.2, the average value of v 2 is given by: Z ∞ v 2 P (v) dv hv 2 i = )

=

Z ∞ 0

4π

a 3/2 π

2

v 4 e−av dv

3

where a = m/(2kB T ). Pulling out the constant terms, the integral is of the form given by Eq. (8.18). From Appendix G, r Z ∞ 3 1 π 4 −av 2 v e dv = I4 (a) = · (2a)2 2 a 0 So the average value is: 4a3/2 hv i = √ π 2

√ 3 3kB T 3 π = = 5/2 2a m 8a

7. The root-mean-square (rms) speed is defined as: r p 3kB T vrms = hv 2 i = m

From Problem 5, the most probably speed corresponds to u = 1 or, using Eq. (8.23), r 2kB T vp = m p Comparing the two, vrms > vp by a factor of 3/2 = 1.2247. In Figure 8.2, the rms speed would be plotted at u = 1.225.

8. The problem is to find the percentage of CO2 molecules that have a speed greater than the given escape velocity ve = 5000 m/s, assuming a temperature of T = 240 K. Using the masses from Appendix B:

1 mole kg 10−3 kg = 7.31 × 10−26 m = mC + 2mO ≃ (12.0 + 32.0) 23 mole 6.022 × 10 atom Plugging this into Eq. (8.23), the dimensionless variable is: s r

m 7.31 × 10−26 kg −3 s =v = v 3.32 × 10 u=v 2kB T 2(1.38 × 10−23 J/K)(240K) m and using the value of ve gives ue = 16.6. To get the fraction of atoms with speeds above this value, integrate the probability distribution (given just below Eq. (8.23)): Z ∞ Z ∞ 2 4 P (u) du = √ u2 e−u du π 16.6 ue This integral can be done numerically with MATLAB (see Appendix C) using the following commands: P = @(u) u.*u.*exp(-u.*u);

4

int value = integral(P,16.6,100.0) where infinity has been approximated by 100. The result is about 1.0×10−117 , which means that essentially no CO2 atoms have a speed above the escape velocity.

9. The most probable speed vp is given by Eq. (8.24), which corresponds to up = 1 as shown in Problem 5. Calculating the fraction of atoms with speeds above up follows along the same lines as the previous problem by integrating the probability distribution: Z ∞ Z ∞ 2 4 u2 e−u du P (u) du = √ π 1 up In terms of MATLAB commands,the integral part is found by: P = @(u) u.*u.*exp(-u.*u); int value = integral(P,1.0,100.0) where again infinity has been approximated by 100. The integral has a value √ of about 0.2536. Multiplying by 4/ π gives 0.572. So about 57.2% of atoms have speeds greater than the most probably one. As a check on√this result, the integral from zero to 1.0 gives 0.1895, and multiplying by 4/ π gives 0.428, or 42.8% have speeds less than vp .

10. The average kinetic energy for each degree of freedom is (Example 8.4): ǫav =

1 kB T 2

Plugging in for T = 298 K and using kB from Appendix A: ǫav =

1 (1.38 × 10−23 J/K)(298 K) = 2.06 × 10−21 J 2

In units of eV , this is ǫav = 0.013 eV .

11. As in the problem above: ǫav =

3 kB T 2

The total kinetic energy of a mole of gas is (using NA from Appendix A): ǫ=

3 (1.381 × 10−23 J/K)(298 K)(6.022 × 1023 ) = 3717 J 2

12. The specific heat is defined as the energy divided by the temperature: ǫ 1 = kB T 2

5

for each degree of freedom. Using the information provided: 3 kB NA = 12.5 J/K 2 5 Diatomic : kB NA = 20.8 J/K 2 6 P olyatomic : kB NA = 24.9 J/K 2 Atoms :

13. Using Eq. (8.28): 8πf 2 u(f ) = 3 c

hf ehf /kB T − 1

Setting x = hf /(KB T ) and taking the differential: u(x) = 8π

kB T hc

du(x) = 8π dx

kB T hc

3 3

hx3 (ex − 1)−1

hx3 3x2 (ex − 1)−1 − x3 (ex − 1)−2 · ex

Setting it equal to zero, the constant terms divide out. Multiplying by (ex −1)2 : du = 3x2 (ex − 1) − x3 ex = 0 dx The maximum is not at x = 0, so dividing by x2 : (3ex − 3) − xex = 0 gives the transcendental equation (3 − x)ex = 3. This can be solved by either plotting it or using the technique of successive approximation. The answer is x = 2.821.

14. Integrating the energy density for a frequency range, Eq. (8.28), over all frequencies:

Z ∞ Z ∞ 8πf 2 hf df u(f ) df = c3 ehf /kB T − 1 0 0 Let µ = hf /kb T and pull out the constant terms: 4 4 4 Z ∞ 4 4 µ3 8πkB T 8πkB T π (3!) du = 3 3 µ 3 3 c h e −1 c h 90 0 where the value of the integral is from Appendix G, Eqs. (G.20) and (G.22).

Plugging in the numbers with T = 3000 K:

6

Z ∞ 0

u(f ) df =

48π 5 90

(1.38 × 10−23 · 3000)4 (3.00 × 108 · 6.626 × 10−34 )3

= 0.0611

J m3

15. Using Eq. (8.28): u(f ) =

8πf 2 c3

hf ehf /kB T − 1

Converting to wavelength using f = c/λ.

8π 8π(c/λ)2 h(c/λ) h = u(λ) = c3 λ3 ehc/λkB T − 1 ehc/λkB T − 1 16. Using Eq. (8.39) and given T = 5800 K:

J W I = σT 4 = 5.67 × 10−8 2 (5800 K)4 = 6.42 × 107 2 4 m sK m Note that this is the intensity of light at the Sun’s surface. By the time it reaches the Earth, the intensity has dropped by 1/r2 where r is approximately the Earth-Sun distance.

17. Using Eq. (8.37), the most probably frequency of light is: hf = 2.821 kB T Since we want the energy of the photons, use E = hf , given T = 5800 K: E = hf =(2.821)(kB T ) =(2.821)(8.62 × 10−5 eV /K)(5800 K)

=1.41 eV

The probability of photons in a given energy range ∆E is: P =

I(E)∆E Itotal

where Eq. (8.38) gives the numerator and Eq. (8.39) gives the denominator (found in the previous problem). We are given an energy range, from 1.75 to 1.80 eV . It is most convenient to work with the dimensionless variable u = hf /KB T with E = hf and here T = 5800 K gives kB T = 0.50 eV . So the range 1.75-1.80 eV corresponds to u in the range 3.5-3.6. Plugging the numbers into Eq. (8.38) in terms of u: ~ I(u) du = 4π 2 c2

kB T ~

4 Z 3.6 3.5

u3 du eu − 1

7

For a small enough range in u, we can approximate u by its central value, u = 3.55, and du ≃ ∆u = 0.1. For constant u: 3 c 4 u (k T ) ∆u B 4π 2 (~c)3 eu − 1 3 3.0 × 1017 nm/s 4 (3.55) = (0.5eV ) (0.1) 4π 2 (197.3 eV · nm)3 e3.55 − 1 eV = 8.18 × 106 nm2 · s

I(u) ∆u ≃

Converting this to standard units, I(E) ∆E = 1.31 × 106 J/(m2 · s). Then: P =

1.31 × 106 = 0.0204 6.42 × 107

so about 2% of all photons from the Sun are in the energy range 1.75-1.8 eV .

Repeating the same procedure, but for the energy range 3.05-3.10 eV , the probability is only P = 0.0077 or less than 0.8%.

18. From Eqs. (8.46) and (8.47): dE = −P dV and by conservation of energy for a closed system: dE1 + dE2 = 0 Combining the above equations: P1 dV1 = −P2 dV2 If the pressure of the two parts are equal, then dV1 = −dV2 . In other words the volume of each partition will adjust, since the wall is moveable (keeping the total volume constant), until equilibrium is obtained.

19. Using Eq. (8.50), the work done on the gas is given by: dW = −P dV We are given the ideal gas law. Solving for P : P =

nRT V

Using this P and ntegrating for constant T :

Z V3 dV V3 = −nRT ln W = −nRT V V 2 V2

8

as desired. Next, using Eq. (8.48): dE = dQto + dWon The energy remains constant between points 2 and 3 of Fig. 8.12, so dE = 0 and using the above result for dW :

V3 dQto = nRT ln V2 20. Using the first law of thermodynamics, Eq. (8.48) gives dE = dQto + dWon Between points 1 and 2 of Fig. 8.12, no heat is exchanged, so dQ = 0 and dE = dWon = P dV 1 → 2 where P depends on both T and V . In contrast, between points 2 and 3, the temperature remains constant, so P only depends on V . Since there is an added term for T in the first case, we expect a steeper change in P for an equivalent change in V .

21. Starting from Eq. (8.68) and doing the integral with µ = 0 (Appendix F): 1 2.612

N V

=

2πmkB Tc h2

3/2

Solving for the temperature, which for µ = 0 is just Tc : Tc =

1 N 2.612 V

2/3

h2 2πmkB

Substitute the given density: N atoms atoms = 5.0 × 1020 = 5.0 × 1014 V cm3 m3 Using the value of h and kB from Appendix A: Tc = (3.32 × 1013 )

1.68 × 10−31 kg · K h2 = 2πmkB m

Given the mass of an atom (or a molecule) the gas, m (units of kg), the critical temperature is easily calculated.

22. For simplicity, assume there are NA atoms (or 1 mole of atoms), each having energy ǫ = (3/2)kB T . The heat capacity is defined as the energy needed to raise the temperature by one unit. The total energy needed is then,

9

3 Etotal (T ) = NA · kB T 2 The heat capacity, C, is the difference E(T + 1) − E(T ), or 3 C = NA · k B 2 To put this in more familiar terms, the gas constant R is defined by R = NA kB , as shown in many textbooks on Thermal Physics. The molar heat capacity at constant volume is then CV = (3/2)R, in agreement with the above equation.

23. Following MATLAB Program 8.1, except using kB T = 0.2µ: u = linspace(0, 2.0, 300); f = @(u) 1./( exp( (u-1)*5) + 1); plot(u,f(u)) The result looks very similar to Figure 8.11(a), except that the sloped region extends further in both directions around ǫF /µ = 1.0.

24. Using Eqs. (8.78) through (8.80): P rob(u > 1) =

Z ∞

fn p(u) du

1

where fn is the normalization constant and p(u) = problem, kB T = 0.2µ so p(u) =

√

√

uf (u). As for the above

1 u (u−1)/0.2 e +1

Following the method of MATLAB Program 8.1: u = linspace(0, 2.0, 300); f = @(u) 1./( exp( (u-1)*5 ) + 1); p = @(u) sqrt(u).*f(u); fn = 1/integral(p,0,2.0); prob = @(u) fn*p(u); plot( u, prob(u) ) int = integral( prob, 1.0, 2.0 ) The result of the integral is 0.217. So about 21.7% of electrons have an energy greater than the chemical potential when kB T = 0.2µ.

25. Let nc be the number of conduction electrons per atom. Given that M is the molar mass, and ρ is the mass density, then ρNA /M is the number of

10

atoms per volume. Multiplying this by the number of conduction electrons per atom

ρNA electrons atoms (nc ) = M atom volume gives the number of conduction electrons per volume, N/V .

26. Given nc = 1 since gold is monovalent, along with the other given values:

g N electron (19.32 cm3 )(6.022 × 1023 atoms mole ) = 1 g V atom 197 mole electrons = 5.90 × 1022 cm3 27. Using the result from the previous problem, and given V = 1 cm3 :

N (V ) = (5.90 × 1022 cm−3 )(1.0 cm3 ) = 5.90 × 1022 N= V Obviously, there are a lot of conduction electron is a cubic centimeter of gold.

28.(a) Using the result from problem 24 with nc = 1 for sodium:

g N nc ρNA electron (0.971 cm3 )(6.022 × 1023 atoms mole ) = = 1 g V M atom 23.0 mole electrons = 2.54 × 1022 cm3 (b) Using Eq. (8.70) the Fermi energy, along with the above result and the electron mass for m:

2/3 2/3

(hc)2 3 N h2 3 N = ǫF = 2m 8π V 2mc2 8π V

2/3 2 3 (1240 eV · nm) 10−21 cm3 1 22 = (2.54 × 10 )( ) 2(5.11 × 105 eV ) 8π cm3 nm3 = 3.15 eV = 5.05 × 10−19 J

(c) Using the equation at the end of Section 8.7: TF =

ǫF kB

Plugging in the result from part (b): TF =

3.15 eV = 3.66 × 104 K 8.617 × 10−5 eV K

11

29. Following the same procedure as Problem 28, but for Magnesium:

g N nc ρNA electrons (1.7 cm3 )(6.022 × 1023 atoms mole ) = = 2 g V M atom 24.3 mole electrons = 8.43 × 1022 cm3 Then the Fermi energy is: 2/3

(hc)2 3 N ǫF = 2mc2 8π V =

2/3 2 (1240 eV · nm) 3 1 1 (8.43 × 10 ) 2(5.11 × 105 eV ) 8π nm3

= 7.01 eV = 1.12 × 10−19 J giving the Fermi Temperature: TF =

7.01 eV = 8.14 × 104 K 8.617 × 10−5 eV K

9 Electronic Structure of Solids - Solutions

1. The cesium choride lattice is shown in Figure 9.5. Taking one of the lightershaded spheres in that figure to represent a cesium ion, we can count 8 chloride ions that serve as the nearest neighbors. Extrapolating the lines on Figure 9.5 in all directions, there are two next-nearest neighbors (cesium atoms) along the axis for each of 3 dimensions, giving a total of 6 next-nearest neighbors for each cesium atom.

2. In Figure 9.6, a single cube of the face-centered cubic lattice is shown.

Taking the origin at the center of the bottom face of the cube, the four corners are generated for integer values of the primitive vectors a + b, where the distance a is the lattice spacing. For example, the vector a + 0 takes you to one corner and -a + 0 points to the opposite corner of the bottom face.

Similarly, a − b and -a + b point to the other two corners of the bottom face.

Adding integer values of the primitive vector c gives the upper four corners of the cube shown Figure 9.6. Hence, these primitive vectors of the Bravais lattice are an alternate way to describe the face-dentered cubic (fcc) lattice.

3. Referring to Fig. 9.8(b), we will use vectors to denote the locations of the carbon atoms from a fixed origin, and then use vector algebra to determine the bond angle. First, select the origin to be located at point A (a corner of the cell) in Fig. 9.8(b). The other two points of interest are point B and the atom located in the center of the top face shown in Fig. 9.8(b). The primitive vectors î and ĵ will be oriented along the edges of the top face (starting at point A) and the k̂ vector will point directly upward from point A. We may now construct the position vectors to each point. Let a be the position vector from the origin to the top face-centered atom, while b points from point B to the origin (point A): a=

1 1 1 1 1 î + ĵ, b = − î − ĵ + k̂ 2 2 4 4 4

Our strategy will be to find the angle between b and c. By inspection:

2

c=a+b Substituting the expressions for a and b:

1 1 1 1 1 c=a+b= − − î + ĵ − k̂ 2 4 2 4 4 =

1 1 1 î + ĵ − k̂ 4 4 4

The dot (or scalar) product of b and c to find the angle between the two: b · c = |b||c|cos(θ) For the left-hand side: b·c=−

1 1 1 1 − + =− 16 16 16 16

For the right-hand side: |b||c|cos(θ) =

r

3 16

r

3 3 cos(θ) = cos(θ) 16 16

Equating the two: 3 1 = cos(θ) 16 16 1 cos(θ) = − 3 −1 θ = cos (−1/3) = 109◦ 28′ −

4. In a body-centered cubic (or bcc) lattice, as shown in Figure 9.3, the nearest neighbor is from the center to a corner of the cube. If a is the distance along an edge of the cube, then the distance from the center to the face of the cube is a/2. Using the Pythagorean formula, p the distance from the center of the cube to the center of an edge is just (a/2)2 + (a/2)2 . The distance we want, from the center of the cube to a corner is p √ d = (a/2)2 + (a/2)2 + (a/2)2 = 3(a/2) 5. For the simple cubic lattice, as shown in Figure 9.2, the nearest neighbor of any point will be located along the length of a side of the cube. Thus, the nearest neighbor distance is simply a. In the case of the face-centered cubic (or fcc) lattice, shown in Figure 9.6, the nearest neighbor distance is to the midpoint along the diagonal of a face. Using the Pythagorean formula: p √ (a/2)2 + (a/2)2 = 2(a/2)

3

6. For sodium (Na), a body-centered cubic crystal, with density ρ = 0.971 g/cm3 and molar mass 23.0 g/mole, we want to find the lattice constant a. As shown in Figure 9.3, the bcc structure is made of two inter-spaced cubic lattices, each with lattice constant a. Each cube has unit volume a3 , giving a total of n3 = 2/a3 atoms per volume. Then the number of atoms per cm3 is: N 0.971 g/cm3 = × (6.02 × 1023 atoms/mole) = 2.54 × 1022 cm−3 V 23.0 g/mole Taking 1.0 cm3 as the volume, n3 = 2.54 × 1022 cm−3 , so 1 = (1.27 × 1022 )(1/3) cm−1 = 2.33 × 107 cm−1 a The reciprocal is a = 4.3 × 10−8 cm. So, the nearest neighbor distance is: √ 3 a = 3.7 × 10−8 cm 2 7. To find the center-to-center distance of the copper ions, first find the volume of one cell, which gives the length a, then use this to get the face-centered nearest-neighbor distance. The number of moles per unit volume for Cu is: 8.96 g/cm3 mole = 0.141 63.5 g/mole cm3 and the number is: 0.141

mole cm3

6.022 × 1023

atoms mole

= 8.494 × 1022

atoms cm3

In section 9.2, just above Figure 9.7, it states that the face-centered cubic lattice has four atoms per cell: cm3 4 atoms/cell = 4.709 × 10−23 22 3 8.494 × 10 atoms/cm cell One edge of the cube has length: a = (4.71 × 10−22 cm3 )1/3 = 3.611 × 10−8 cm From problem 5, the nearest-neighbor distance is then: √ 2 a = 2.553 × 10−8 cm 2 8. Looking at Figure 9.6(b), the vectors to the center of the other faces can be written as:

4

a v1 = a î + (ĵ + k̂) 2 a v2 = a ĵ + (k̂ + î) 2 a v3 = a k̂ + (î + ĵ) 2 In terms of the primitive vectors from Eq. (9.4): a a (ĵ + k̂) + (k̂ + î) = v3 2 2 a a a2 + a3 = (k̂ + î) + (î + ĵ) = v1 2 2 a a a1 + a3 = (ĵ + k̂) + (î + ĵ) = v2 2 2

a1 + a2 =

9. Looking at Fig. 9.6(b), which corresponds to the primitive vectors of Eq. (9.4), and constructing the position vectors of each of the corners: c1 = ak̂ c2 = aĵ + ak̂ c3 = aî + aĵ + ak̂ c4 = aî + ak̂ The primitive vectors from Eq. (9.4) are:

a a a ĵ + k̂ , a2 = k̂ + î , a3 = î + ĵ a1 = 2 2 2 Next, solve these primitive vectors for î, ĵ, and k̂. First add a1 and a2 : a1 + a2 =

a (î + ĵ) + ak̂ 2

Then subtract a3 : a1 + a2 − a3 = ak̂ and solve for k̂: k̂ =

a1 + a2 − a3 a

î =

a2 + a3 − a1 a

ĵ =

a1 − a2 + a3 a

Similarly:

5

Now substitute into the above: c 1 = a1 + a2 − a3

c2 = a1 − a2 + a3 + a1 + a2 − a3 = 2a1 c 3 = a2 + a3 − a1 + a1 − a2 + a3 + a1 + a2 − a3 = a1 + a2 + a3 c4 = a2 + a3 − a1 + a1 + a2 − a3 = 2a2 10. As in Problem 9, but for the position vectors of the four upper corner in Fig. 9.4(b), corresponding to the vectors of Eq. (9.2), are: c1 = ak̂ c2 = aĵ + ak̂ c3 = aî + aĵ + ak̂ c4 = aî + ak̂ The primitive vectors from Eq. (9.2) are:

a a a a1 = k̂ + î − ĵ , a3 = ĵ + k̂ − î , a2 = î + ĵ − k̂ 2 2 2 Following the same procedure as in Problem 9, the result is: c 1 = a1 + a2 c2 = a1 + 2a2 + a3 c3 = 2a1 + a2 + a3 c4 = a1 + a2 + 2a3 11. Using the results from problems 4 and 5: a simple cubic : d = a ⇒ R = 2 √ √ 3 3 body-centered cubic : d = a⇒R= a 2 √ 4 2 2 face-centered cubic : d = a ⇒ R = a 2 4 substitute these values of R into the packing fraction equation:

a 3 4 π 3π 3 = simple cubic : F = (1) 3 a 6 √ 3 √ 3a 4 3π 4 3π body-centered cubic : F = (2) = a3 8 √ 3 √ 2a 4 π 3 4 2π face-centered cubic : F = (4) = a3 6

6

12. The goal is to prove Eq. (9.21), bi · aj = 2πδij where δij is the Kronecker delta. Starting with i = 1 and j = 1: b1 · a1 = 2π

a1 · (a2 × a3 ) = 2π a1 · (a2 × a3 )

and similarly for i = j = 2 and i = j = 3. Next consider i = 1 and j = 2: b1 · a2 = 2π

a2 · (a2 × a3 ) a1 · (a2 × a3 )

and since (a2 × a3 ) is perpendicular to a2 , the numerator is zero. Similarly, for all i 6= j, the cross product in the numerator is perpendicular to the aj vector, giving zero for the dot product. So Eq. (9.21) is correct.

13. Using the primitive vectors given in this problem and Eqs.(9.17)-(9.19) for the reciprocal lattice’s primitive vectors, first do the cross product: a2 î ĵ k̂ a2 a2 × a3 = √ 0 1 0 = √ î 2 001 2 a2 î ĵ k̂ a2 a3 × a1 = √ 0 0 1 = √ (−î + ĵ) 2 2 110 2 2

a1 × a2 =

a2 î ĵ k̂ a2 k̂ 110 = 2 2 010

Next, do the denominator of Eqs. (9.17)-(9.19): a2 a a3 a1 · (a2 × a3 ) = î + ĵ · √ î = √ 2 2 2 2

Now use Eqs. (9.17)-(9.19) to get the primitive vectors of the reciprocal lattice: √ !

2 2 a2 4π √ î = b1 = 2π î a3 a 2 √ ! 2 2π

a 2 2 √ −î + ĵ = −î + ĵ b2 = 2π 3 a a 2 2 √ ! 2 √ 2 2 a 2π 2 b3 = 2π k̂ = k̂ 3 a 2 a

7

By comparing this result with the reciprocal lattice in Example 9.1, we see that the above primitive vectors are a linear combination of the primitive vectors of the reciprocal lattice from the simple cubic structure.

14. Using the primitive vectors given in this problem and Eqs.(9.17)-(9.19) for the reciprocal lattice’s primitive vectors, first do the cross product: a2 î ĵ k̂ a2 a2 × a3 = √ −1 1 0 = √ (−î − ĵ) 2 0 0 −1 2 a2 a2 î ĵ k̂ a3 × a1 = √ 0 0 −1 = √ (î − ĵ) 2 11 0 2

a1 × a2 =

a2 2

î ĵ k̂ a2 2k̂ 1 10 = 2 −1 1 0

Next, do the denominator of Eqs. (9.17)-(9.19):

a2 √ a a1 · (a2 × a3 ) = √ î + ĵ · √ −î − ĵ = − 2a3 2 2 Now use Eqs. (9.17)-(9.19) to get the primitive vectors of the reciprocal lattice:

2 π

−1 a √ √ −î − ĵ = b1 = 2π î + ĵ a 2a3 2 2

a π −1 √ î − ĵ = −î + ĵ b2 = 2π √ a 2a3 2

2π −1 a2 k̂ = − √ k̂ b3 = 2π √ 2a3 2a As in the previous problem, we see that the above primitive vectors are a linear combination of the primitive vectors of the reciprocal lattice from the simple cubic structure.

15. The body-centered cubic (or bcc) primitive vectors are from Eq. (9.2): a1 =

a a a (ĵ + k̂ − î) , a2 = (k̂ + î − ĵ) , a3 = (î + ĵ − k̂) 2 2 2

Use Eqs. (9.17)-(9.20) to calculate the primitive vectors of the bcc reciprocal lattice. But first do the cross-products:

8

a2 × a3 =

a2 a2 î ĵ k̂ a2 1 −1 1 = ĵ + k̂ 4 2 2 1 1 −1

a3 × a1 =

a2 4

a1 × a2 =

a2 a2 î ĵ k̂ a2 −1 1 1 = î + ĵ 4 2 2 1 −1 1

î ĵ k̂ a2 a2 1 1 −1 = î + k̂ 2 2 −1 1 1

Next, do the denominator of Eqs. (9.17)-(9.19): a1 · (a2 × a3 ) =

a2 a3 a ĵ + k̂ − î · ĵ + k̂ = 2 2 2

Now use Eqs. (9.17)-(9.19) to get the primitive vectors of the reciprocal lattice: 2 2π

2 a b1 = 2π ĵ + k̂ = ĵ + k̂ 3 a 2 a 2

2π 2 a î + k̂ = î + k̂ b2 = 2π 3 a 2 a 2

2π 2 a î + ĵ = î + ĵ b3 = 2π 3 a 2 a So, the bcc reciprocal lattice, when compared with Eq. (9.4), is just the fcc lattice with cell length 4π/a.

16. As in Problem 15, but starting with the fcc lattice given by Eq. (9.4), a2 × a3 =

a2 î ĵ k̂ a2 1 0 1 = (−î + ĵ + k̂) 4 4 110

a3 × a1 =

a2 î ĵ k̂ a2 1 1 0 = (î − ĵ + k̂) 4 4 011

a1 × a2 =

a2 î ĵ k̂ a2 0 1 1 = (î + ĵ − k̂) 4 4 101

9

The denominator of Eqs. (9.17)-(9.19) is the same for each vector: a1 · (a2 × a3 ) =

a3 a2 a −î + ĵ + k̂ = ĵ + k̂ · 2 4 4

Now use Eq. (9.17)-(9.19) to get the primitive vectors of the reciprocal lattice:

2π b1 = −î + ĵ + k̂ a

2π î − ĵ + k̂ b2 = a

2π b3 = î + ĵ − k̂ a

When compared with Eq. (9.2), this a bcc lattice with cell length 4π/a.

17. Use Eqs. (9.17)-(9.19) to calculate the reciprocal lattice primitive vectors of the hexagonal close-packed lattice. First do the cross-products: ĵ k̂ √ √ î ac b × c = −( 3a/2) (a/2) 0 = (î + 3ĵ) 2 0 0 c

c×a=

î ĵ k̂ √ ac 0 0 c = (−î + 3ĵ) √ 2 ( 3a/2) (a/2) 0

√ 2 ĵ k̂ √ î 3a a × b = ( √3a/2) (a/2) 0 = (k̂) 2 −( 3a/2) (a/2) 0 The denominator of Eqs. (9.17)-(9.19) is the same for each vector: √ 2 ac √ 3a c a √ 3î + ĵ · î + 3ĵ = a · (b × c) = 2 2 2

Now use Eq. (9.17)-(9.20) to get the primitive vectors of the reciprocal lattice: √ 2π b1 = √ î + 3ĵ 3a √ 2π −î + 3ĵ b2 = √ 3a 2π k̂ b3 = c

This is the same as the original lattice but with cell lengths 4π/(3a) in the x-y plane and 4π/c along the z axis. The volume of a unit cell is given by:

10

√ 4π 2 √ 2π 16π 3 V = b1 · (b2 × b3 ) = √ 3î + ĵ = √ î + 3ĵ · √ 2 3a 3a 3a3 18. Looking at Eq. (9.4), there is combination of primitive vectors that allow translation by one cell length along the z-axis. Since vectors a and b of this problem translate only in the x-y plane, we need some combination with c that translates along the z-axis: ak̂ = a + b − c = aĵ − c So, c = a(ĵ − k̂) to allow translations along the z-axis.

19. Start with Eqs. (9.17)-(9.19) to show that: a1 = 2π

b2 × b3 b1 · (b2 × b3 )

(1)

a2 = 2π

b3 × b1 b1 · (b2 × b3 )

(2)

a1 = 2π

b1 × b2 b1 · (b2 × b3 )

(3)

where b1 , b2 , and b3 are also given by Eqs. (9.17)-(9.19), directly in terms of a1 , a2 and a3 . First calculate the cross-product in the denominators:

a1 × a2 a3 × a1 × b2 × b3 = 4π 2 a1 · (a2 × a3 ) a1 · (a2 × a3 ) Note that the denominator of each term is a scalar quantity, so it can be brought outside the vector product. For the numerator: (a3 × a1 ) × (a1 × a2 ) The vector triple-product relation, found in many textbooks, is: A × (B × C) = B(A · C) − C(A · B) Comparing the vectors that occur in the relation above: A = a3 × a1 B = a1 C = a2 After substitution:

11

(a3 × a1 ) × (a1 × a2 ) = a1 [(a3 × a1 ) · a2 ] − a2 [(a3 × a1 ) · a1 ] = a1 [a1 · (a2 × a3 )] + a2 [(a1 × a1 ) · a3 ] where, for the second line we used the scaler triple-product: A · (B × C) = (A × B) · C The first part of this term in brackets is now identical to the expression for the denominator. The second term is equal to zero, since we have a vector in a cross-product with itself. The result for b2 × b3 is now: b2 × b3 = 4π 2 a1 Using the same method for the other relevant cross-products: b3 × b1 = 4π 2 a2

b1 × b2 = 4π 2 a3

The denominators of the three equation at the start of this problem may be solved using the cross-products above: b1 · (b2 × b3 ) = 2π

a2 × a3 · 4π 2 a1 a1 · (a2 × a3 )

= 8π 3

Putting together these results, we see that the reciprocal of the reciprocal lattice is, indeed, the same as the direct lattice.

20. Figure 9.15(b) shows part of the lattice plane with Miller indices (110).

The plane perpendicular to the reciprocal lattice vector is: g = b1 + b2 Eq. (9.31) gives the distance between planes: d=

2π |g|

Following Example 9.1, the magnitude of g is: √ 2π 2π 2π î + ĵ ⇒ |g| = 2 a a a √ So, the distance between planes is d = a/ 2. g=

21. Similar to Problem 20, but for the Miller indices (210), the plane perpendicular to the reciprocal lattice vector is:

12

g = 2b1 + b2 Eq. (9.31) gives the distance between planes: 2π |g|

d=

Following Example 9.1,the magnitude of g is: √ 2π 4π 2π î + ĵ ⇒ |g| = 5 a a a √ The distance between planes is d = a/ 5. g=

22 (a) First let g = g ′ . Then n(2π)/a = m(2π)/a or n = m and: Z a Z a Z a (n−m)2πx/a −ig ′ x igx dx = a e dx = e e dx = 0

0

0

On the other hand, if g 6= g ′ , let j = n − m (an integer) and: Z a Z a ′ ej(2πi)x/a dx e−ig x eigx dx = 0

0

The integrand may be written using Euler’s formula: Z a cos(j2πx/a) + i sin(j2πx/a))dx 0

and both functions (sin and cos) are integrated over a full cycle, so Z a ′ e−ig x eigx dx = 0 0

b) Using Eq. 9.13: f (x) =

X

Fg eigx

X

Fg e−ig x eigx

g

′

Multiply by e−ig x : ′

e−ig x f (x) =

′

g

Integrating this: Z a X Z a X ′ −ig ′ x e f (x) dx = Fg e−ig x eigx dx = Fg aδgg′ = aFg′ 0

g

0

g

13

23. First, project the vector l onto g: d=l·

g 2πN = |g| |g|

and similarly for l′ : d′ = l ′ ·

2π(N + 1) g = |g| |g|

Using Eq. (9.31), the distance between the planes containing these points is: d′ − d =

2π 1 (2π(N + 1) − 2πN ) = |g| |g|

24. Starting from Eq. (9.45): ψk (x) = Ak eikx where from Eq. (9.44), k = 2πn/(N a) for integer n. Setting A2k = 1/L: ′ 1 ψk∗′ (x)ψk (x) = ei(k−k )x L x x 1 cos 2π(n − n′ ) + i sin 2π(n − n′ ) = L Na Na

where the crystal length is L = N a. Integrating: 1 L

Z L 0

cos 2π(n − n′ )

x x dx = 0 + i sin 2π(n − n′ ) Na Na

since each function (sin and cos) is integrated over a full cycle. So, when n 6= n′ , the orthogonality condition of Eq. (9.48) is satisfied.

25 (a) From Eq. (9.39), Block’s Theorem is: ψk (r + l) = eik·l ψk (r) Let l = N1 a1 , giving: ψk (r + N1 a1 ) = eik·N1 a1 ψk (r) Using the Born-von Karman boundary condition given here: ψk (r) = eik·N1 a1 ψk (r) resulting in eik·N1 a1 = 1, and similarly for N2 a2 and N3 a3 . b) For k given by Eq. (9.50):

14

k=

n2 n3 n1 b1 + b2 + b3 N1 N2 N3

Then the dot product gives: k · (N1 a1 ) =

n1 N1

N1 b 1 · a 1

and comparing this to Eqs. (9.32) - (9.35) gives: k · (N1 a1 ) = n1 (2π) So, we get e2πi = 1, which is just Euler’s identity, so this choice of k satisfies the conditions given in part (a).

26. Starting with Eqs. (9.51): ψk (r) =

1 ik·r e V 1/2

gives: ψk′ (r) =

1 ik′ ·r 1 e and ψk∗ (r) = 1/2 e−ik·r V 1/2 V

Then from Eq. (9.54): V (r) =

X

Vg eig·r

g

and substituting these into the integral: X Vg Z ′ ei(g+k −k)·r dV Ik∗′ , k = V g Separating the exponential terms: Ik∗′ , k =

X Vg Z g

This can be rewritten as: X Vg Z g

V

V

′

ei(g+k )·r e−ik·r dV

ψk∗ (r)ψ(k′ +g) (r) dV

and using the orthogonality condition of Eq. (9.58): k = k′ + g and the result of Problem 24 gives the desired orthogonality properties.

15

27. The band gap ∆E between the occupied and unoccupied states is at energies that correspond to infrared light. Since electrons must jump over the band gap to get into the conduction band, infrared light does not have enough energy to do this. Hence, infrared light will not create a quantum excitation and has a low probability of being absorbed by a semiconductor. On the other hand, visible light has enough energy to excite an electron into the conduction band. So, photons of visible light frequencies will cause quantum excitations in a semiconductor and are readily absorbed. Hence, the semiconductor is opaque to visible light.

10 Charge carriers in semiconductors - Solutions

There are no problems listed in this chapter.

11 Semiconductor Lasers - Solutions

1. Starting with Eq. (1.5) from Chapter 1: E=

hc λ

and putting in the given energy needed to hop over the band gap: λ=

1240 eV · nm hc = = 816.3 nm E 1.519 eV

2. Given for GaAs the lattice constant a = 5.65Å and effective mass m0 = 0.067, then Eq. (11.2) becomes: ε(k) = ǫc +

~2 k 2 2(0.067)me

Multiplying the top and bottom of the fraction by c2 and given k = 0.1(π/a): ε(k) − ǫc =

(hc)2 (0.1)2 0.134(0.565 nm)2 (me c2 )

From Appendix A, hc = 1240 eV · nm and me c2 = 511 × 103 eV , giving ε(k) − ǫc = 0.703 eV 3. From Section 11.3.1, heterostructures are formed using semiconductors with similar lattice constants. The heterostructures grown on a substrate of InP must have the same lattice constant as InP. Using linear interpolation: a(Inx Ga1−x As) = a(InP) = xa(InAs) + (1 − x)a(GaAs) 5.869 = 6.058x + 5.653(1 − x) Solving for x gives the value 0.53. The alloy has composition: In.53 Al.48 As

2

4. We use the following procedure for these matrix multiplication operations:      A1 D1 + A2 D2 + A3 D3 D1 A1 A2 A3  B 1 B 2 B 3   D2  =  B 1 D1 + B 2 D2 + B 3 D3  C1 D1 + C2 D2 + C3 D3 D3 C1 C2 C3     120 3 1 1 1 21 = 2 131 4 0 



    101 2 1 1 2 10 = 2 113 4 1 

    120 2 4 1 1 21 = 5 131 1 6 

    211 5 1 1 0 12 = 2 110 3 1 5. Applying the standard methods of matrix multiplication, the first case is:

01 0 −i (0 + i) (0 + 0) i 0 = = 10 i 0 (0 + 0) (−i + 0) 0 −i The second case gives:

01 10

2 0 0 −2 = 0 −2 2 0

The final case is:     413 120 211 1 1 21 0 1 = 5 3 2 624 131 110 

6. First, consolidate the last two terms:

ik̄L

ik̄L e r21 eik̄L 1 r21 e 0 = r21 1 r21 e−ik̄L e−ik̄L 0 e−ik̄L Using this result in Eq. (11.28):

3

1 t12 t21

T =

1 r12 r12 1

ik̄L e r21 eik̄L r21 e−ik̄L e−ik̄L

Carrying this through: T =

T11 T12 T21 T22

where the first element element is: T11 =

1 [eik̄L + r12 r21 e−ik̄L ] t12 t21

The same may be done for the remaining three elements.

7. It is only necessary to make a small change to MATLAB Function 11.1.

For a two-strip Fabry-Perot laser, we just need to extend Eq. (11.42) in a similar way as was done for the two-barrier case of Chapter 3 (see Problem 9 of Chapter 3 solutions). Hence, if both strips have the same reflection coefficient, we change only one line of Function 11.1: M=A*C*B*A*C*B This assumes that there is essentially no space between the two strips. In other words, the matrix D in Chapter 3’s Problem 9 becomes the unity matrix for sufficiently small L. In addition, if the second cavity has a different 2 reflection coefficient, r32 = −r13 , with t31 = t13 = 1 − r32 . Then one can easily define new matrices: E = (1/t31)*[1 r13; r13 1]; F = (1/t13)*[1 r31; r31 1]; and then multiply through, left to right, M=A*C*B*E*C*F

8. MATLAB Program 11.1 can be used, along with the changes given to MATLAB Function 11.1 in Problem 7 above. To plot the transmission through the Fabry-Perot laser of the previous problem as a function of 2kL/2π, just rerun MATLAB Program 11.1 for the given reflection coefficients.

9(a) The total rate of change in the amount of water in the reservoir will equal the rate of water entering minus the rate in which it is leaving. Letting V equal the volume of water in the reservoir: dV = Rf − RD dt

4

where RD is the total rate at which water drains. Since the reservoir is draining from two ports: RD = Rd1 + Rd2 Plugging this in, the total rate equation is: dV = Rf − h(C1 + C2 ) dt b) To find the steady state height, set: dV =0 dt And solve for h: h=

Rf C1 + C2

12 The special theory of relativity - Solutions

1. Einstein’s second postulate on the special theory of relativity states that the speed of light is independent of the motion of the source, therefore the velocity of the light in the reference frame of the spaceship is c.

2.a) Starting with Eq. (12.16), calculate the gamma factor: γ=p

1 1 − u2 /c2

=p

1 1 − (0.2)2

= 1.0206

Using the Lorentz transformation from Eq. (12.15): x′ =γ(x − ut) = (1.0206)(2 m − (0.2 c)(4 m/c)) = 1.2247 m y ′ =0 m z ′ =0 m t′ =γ(t −

ux m m 2m = 3.6742 ) = (1.0206) 4 − (0.2 c) c2 c c2 c

b) Using Eq. (12.17) for the inverse transformation: x =γ(x′ + ut′ ) = (1.0206)(1.2247 m − (0.2 c)(3.6742 m/c)) = 2.000 m y =0 m z =0 m t =γ(t′ +

ux′ m m 1.2247 m = 4.000 ) = (1.0206) 3.6742 − (0.2 c) c2 c c2 c

3.a) Starting with Eq. (12.16), calculate the gamma factor: γ=p

1 1 − u2 /c2

=p

1 1 − (0.25)2

Transforming from S ′ to S using Eq. (12.17):

= 1.033

2

x =γ(x′ + ut′ ) = (1.033)(2 m + (0.25 c)(4 m/c)) = 3.10 m y =0 m z =0 m

m m 2m ux′ = 4.65 t =γ(t + 2 ) = (1.033) 4 + (0.25) 2 c c c c ′

b) To transform back to S ′ , use Eq. (12.15): x′ =γ(x − ut) = (1.033)(3.10 m − (0.25 c)(4.65 m/c)) = 2.00 m y ′ =0 m z ′ =0 m

ux m m 3.10 m t =γ(t − 2 ) = (1.033) 4.65 − (0.25 c) = 4.00 c c c2 c ′

4. Using the length contraction formula, Eq. (12.23): r u2 LM = 1 − 2 (LR ) c r u2 LM = 1− 2 LR c Setting the length ratio equal to 1/2 and squaring both sides: 1 u2 =1− 2 4 c Solving for the speed u: u=

√

3 c 2

5.a) To calculate the length as perceived in your reference frame, use Eq. (12.23) with LR = 1 m: r (0.5 c)2 (1 m) = 0.866 m LM = 1 − c2 b) Using the above result, calculate the time it would take in your frame: ∆tM =

d 0.866 m = = 5.773 × 10−9 s v (0.5)(3.0 × 108 m/s)

6. Length contraction occurs only along the axis of motion. So the xcomponent will contract but the y-component is not changed. The x-component in the rest frame is: (LR )x = (1.0 m) cos 30◦ = 0.866 m

3

Converting to the moving frame: (LM )x = (0.866 m)

p

1 − (0.8 c)2 /c2 = 0.520 m

The y-component (in both frames) is just Ly = (1.0 m) sin 30◦ = 0.5 m. So, q p LM = (Lx )2 + (Ly )2 = (0.52)2 + (0.5)2 = 0.721 m

7. The percentage of length contraction is 1 − (LM /LR ) × 100%. Using Eq. (12.23), and recalling that (1 − ǫ)x ≃ (1 − xǫ) for ǫ small: r p u2 LM = 1 − 2 = 1 − (3 × 10−6 )2 ≃ 1 − 4.5 × 10−12 LR c

The percentage of length contraction is about (4.5 × 10−10 )%. Now, using Eq. (12.26), the time difference after 1 year is: ∆tR − ∆tM = 1 yr − p

1 yr

1 − (3 × 10−6 )2

= 3.1536 × 107 s 1 − (1 − 9 × 10−12 )−1/2

Using the above approximation formula, the clocks are different by 0.14 ms.

8. Using Eq. (12.39), with v and u as the velocities of rockets A and B: v′ =

(0.8 c) − (0.6 c) = 0.385 c 1 − (0.8)(0.6)

An observer on rocket B sees rocket A to be moving away at speed 0.385 c.

9.a) We are given the position of the center of mass (CM), is: X=

mxe + mxp 2m

and both particles have the same mass m. Taking the differential: uCM =

dX 1 = (ue + up ) dt 2

where ue and up are the velocities of the electron and positron, respectively.

Substituting in the given velocities (with the correct sign): uCM = (0.5)(0.95 c − 0.2 c) = 0.375 c b) Given a lifetime of 2.0 × 10−8 s in the rest frame, Eq. (12.26) gives: 2.0 × 10−8 s = 2.157 × 10−8 s ∆tM = p 1 − (0.375)2

4

10.a) An observer on Earth measures the time to be: ∆tM =

d 8.6 c · yr = = 8.687 yr v 0.99 c

b) An observer on the spaceship measures using proper time: p ∆tR = ∆tM 1 − (0.99)2 = 1.225 yr

11.a) First, change the wavelength to frequency using the equation from Chapter 1: f0 =

c 3.0 × 108 m/s = = 5.09 × 1014 Hz λ0 589.0 × 10−9 m

Using Eq. (12.41) (and above) for an approaching light source: s 1+β f0 f= 1−β where β = v/c = 0.3. Then: r 1.3 f= (5.09 × 1014 Hz) = 6.94 × 1014 Hz 0.7 The correspond wavelength is: λ=

3.0 × 108 m/s c = = 432.2 nm f 6.94 × 1014 Hz

and the shift in wavelength is: λ − λ0 = 432.2 nm − 589.0 nm = −156.8 nm b) Using Eq. (12.41) for a receding light source: s r 1−β 0.7 f0 = (5.09 × 1014 Hz) = 3.74 × 1014 Hz f= 1+β 1.3 c λ = = 802.1 nm f λ − λ0 = 802.1 nm − 589.0 nm = 213.1 nm c) Figure 12.12 shows how to calculate the Doppler shift for transverse motion: f= where the gamma factor is:

f0 γ

5

Substituting:

γ=p f=

1 1 − u2 /c2

=p

1 1 − (0.3)2

= 1.05

5.09 × 1014 Hz = 4.86 × 1014 Hz 1.05

Now the wavelength and the wavelength shift are: 3.0 × 108 m/s c = = 617.2 nm f 4.86 × 1014 Hz λ − λ0 = 617.2 nm − 589.0 nm = 28.2 nm λ=

12. Change the wavelength to frequency using the equation from Chapter 1: f0 =

3.0 × 108 m/s c = = 4.57 × 1014 Hz λ0 656.5 × 10−9 m

Using Eq. (12.41) for an receding light source: s 1−β f= f0 1+β where β = v/c = (1.2 × 108 )/(3.0 × 108 ) = 0.4. Then: r 0.6 f= (4.57 × 1014 Hz) = 2.99 × 1014 Hz 1.4 The correspond wavelength is: λ=

3.0 × 108 m/s c = = 1003 nm f 2.99 × 1014 Hz

and the relative shift in wavelength is: 1003 nm − 656.5 nm λ − λ0 = = 0.528 λ0 656.5 nm 13. Using Eq. (12.41) for the receding source with β = 0.8: r 1 − .80 f= × 100 M Hz = 33 M Hz 1 + .80 14. Plugging the relation f = c/λ into Eq. (12.41) for a receding source: s c 1−β c = λ λ0 1 + β

6

Canceling the common factor c and multiplying through by λ0 : s 1−β λ0 R= = λ 1+β Solving for β in terms of the ratio R: β=

1 − R2 1 + R2

We want to find β when λ = 650 nm. Plugging this into the above equations, the results are: λ0 = 590 nm : β = 0.0965 λ0 = 525 nm : β = 0.210 λ0 = 460 nm : β = 0.333

15. Starting with Eq. (12.38): v ′ = γ(v − u) · γ(1 + uv ′ /c2 ) we want to solve for v ′ . Bring the terms containing v ′ to the left:

u2 v ′ vuv ′ ′ 2 v =γ v−u+ 2 − 2 c c v ′ vu u2 v ′ v′ − 2 + 2 =v−u γ2 c c

uv u2 1 ′ − 2 + 2 =v−u v γ2 c c Now plug in for γ from Eq. (12.16):

u2 uv u2 ′ v 1− 2 − 2 + 2 =v−u c c c v − u v′ = uv 1 − c2 16. The proper time is given by the equations below Fig. 12.16: (proper time)2 = (ct)2 − x2 where the distance is given as x = 3 m and the lab time is (ct) = 8 m. Hence, p p proper time = (ct)2 − x2 = (8 m)2 − (3 m)2 = 7.4 m

7

17. The world line for this traveler can be plotted in a similar way as done in Fig. 12.16. For the first leg, the traveler remains stationary for 1 m of time: (proper time)2 = 12 − 02 proper time = 1 m Next, the traveler goes a distance of 4 m in a time of 5 m: (proper time)2 = 52 − 42 proper time = 3 m Finally, the traveler goes 1 m in a time of 2 m: (proper time)2 = 22 − 12 √ proper time = 3 m √ The total proper time is (4 + 3) m. To get units of time, just divide by c.

18. The world line is drawn similar way to Fig. 12.16. Since u = 0.2 c, the slope of the x′ -axis is 0.2 and the slope of the ct′ -axis is 1/0.2 = 5. The coordinates of x′ and t′ are the same as for Problem 2.

19. The world line is drawn similar way to Fig. 12.16. Since u = 0.25 c, the slope of the x′ -axis is 0.25 and the slope of the ct′ -axis is 1/0.25 = 4. The coordinates of x′ and t′ are the same as for Problem 3.

20. Using Eq. (12.44) with the usual convention x0 = ct, x1 = x, x2 = y, x3 = z and a velocity of β = 0.2, giving γ = 1.02: x′0 =γ(x0 − βx1 ) = (1.02)(4 m − (0.2)(2 m)) = 3.672 m

x′1 =γ(x1 − βx0 ) = (1.02)(2 m − (0.2)(4 m)) = 1.224 m x′2 =x2 = 0

x′3 =x3 = 0

21. Using Eqs. (12.15), solve for x: x′ + ut γ

x= Again using Eqs. (12.15), solve for t: t=

t′ ux + 2 γ c

Using this, substitute t into the first equation:

8

x=

ut′ u2 x x′ + + 2 γ γ c

Collecting terms of x u2 x 1− 2 c

=

1 ′ (x + ut′ ) γ

and using the definition of γ 2 = (1 − u2 c2 ): x = γ(x′ + ct′ ) as desired. Next, substitute x from the first equation into the second one: t=

u2 t t′ ux′ + γ γc2 c2

Collecting terms for t and using the definition of γ: t = γ(t′ +

ux′ ) c2

22. Given Eq. (12.57), tµ = gµν tν , and Eq. (12.59), tσ = g σµ tµ , we see that the g matrices serve to raise or lower the indices of a vector. Given here that Λνµ = gµρ Λρσ g σν is to operate on vν , we start with the latter two terms: Λρσ g σν vν = Λρσ v σ = v ′ρ where we use the given transformation equation: v ′µ = Λµν v ν with the dummy indices rewritten as ν = σ and µ = ρ. Now use gµρ to lower the index: Λνµ vν = gµρ Λρσ g σν vν = gµρ v ′ρ = vµ′ as desired. This shows how to transform the covariant components of a vector.

23. First recall that gµρ and g σν will have the form:   1 0 0 0  0 −1 0 0     0 0 −1 0  0 0 0 −1

The first operation, gµρ Λρσ :       γ −βγ 0 0 γ βγ 0 0 1 0 0 0  0 −1 0 0   −βγ γ 0 0   βγ −γ 0 0         0 0 −1 0   0 0 1 0  =  0 0 −1 0  0 0 0 −1 0 0 01 0 0 0 −1

9

Now, gµρ Λρσ g σν :       γ −βγ 0 0 1 0 0 0 γ βγ 0 0  βγ −γ 0 0   0 −1 0 0   βγ γ 0 0         0 0 −1 0   0 0 −1 0  =  0 0 1 0  0 0 0 −1 0 0 0 −1 0 0 01 24. Using Eq. (12.60), xν = Λµν x′µ , we see that

Λρµ x′ρ Λσν x′σ = xµ xν where the dummy indices are substituted appropriately into Eq. (12.60). Because the summation over indices is preserved, the terms communte. So, g µµ Λρµ Λσν x′ρ x′σ = g µν xµ xν = |x|2 We are given that |x|2 = g ρσ x′ρ x′σ and so the desired relation is obtained.

13 The relativistic wave equations and general relativity - Solutions

1. The components of the four-velocity for velocity are found between Eqs. (13.3) and (13.4). Since the electron is traveling only in the x-direction: v2 = v3 = 0 The relevant velocity components are: v 0 = γc, v 1 = γ

dx1 = γ(0.2 c) dt

Calculating γ: 1

The results are:

p

1 − u2 /c2

=p

1 1 − (0.2)2

= 1.02

v 0 = 1.02 c v 1 = 0.204 c b. Using Eq. (13.6) with the electron’s mass (me = 0.511 M eV /c2 ): p0 = 1.02 me c = 0.521 M eV /c p1 = 0.204 me c = 0.104 M eV /c p2 = p3 = 0

2. Using Eq. (13.13): KE = (γ − 1)mc2 p Plug in γ = 1/ 1 − (0.2)2 = 1.02 and the electron mass:

2

KE = (1.02 − 1)(0.511 M eV ) = 0.011 M eV

3. Using Eqs. (13.12) and (13.13) for the rest energy and kinetic energy, respectively: (γ − 1)mc2 = mc2 gives γ = 2.0. Now using Eq. (13.3): 1 p

1 − v 2 /c2

=2

1 (1 − v 2 /c2 ) = 2

v2 1 1− 2 = c 4 2 3 v = c2 4

p

√ The particle must have a speed of v = ( 3/2)c.

4. Using Eqs. (13.9) and (13.12) for the total energy and rest energy, respectively: γmc2 = 2mc2 √ gives γ = 2.0. This gives the same speed as the previous problem, v = ( 3/2)c.

5. We first evaluate the derivatives of f (x) = γ called for in the Taylor series: f ′ (x) =

1 1 2 (1 − x)3/2

f ′′ (x) =

1 3 4 (1 − x)5/2

We now evaluate at x=0 and substitute into the given form of the Taylor series: 1 3 f (x) = 1 + x = x2 + . . .

2 8 Now, let x = v 2 /c2 :

3

3 v4 1 v2 1 p = =1+ 2 2c 8 c4 1 − u2 /c2

This result is identical to eq. 13.10.

6. From Eq. (13.13) with the mass of the electron: KE = (γ − 1)(0.511 M eV ) p where γ = 1/ 1 − (v/c)2 . This gives the following results: v = 0 : γ = 1.00 : KE = 0

v = 0.99 c : γ = 7.09 : KE = 3.11 M eV v = 0.999 c : γ = 22.37 : KE = 10.9 M eV 7. Using Eq. (13.13) with m as the electron’s mass: KE = 100 M eV = (γ − 1)(0.511 M eV ) Solving for γ gives γ = 196.7. Now use the definition of γ in Eq. (13.3): 196.7 = p

1

1 − u2 /c2 1 38688 = 2 1 − vc2

1−

v2 = 2.58 × 10−5 c2 v2 0.999974 = 2 c v = 0.999987c

8.a) Using Eq. (13.9) with the mass of the proton (mp = 938.3 M eV ) and a total energy of E = 1500 M eV : 1500 M eV = γ(938.3 M eV ) giving γ = 1.60, and plugging into Eq. (13.3) gives v = 0.78 c. b) Using Eq. (13.14), p p pc = E 2 − (mc2 )2 = (1500)2 − (938.3)2 M eV p = 1170 M eV /c

9.a) Carrying out the four-vector addition (see Chapter 12) in the given expression for s: |pA + pB |2 [(EA + EB )2 − (pA + pB )2 ] = 2 c c2

4

In the laboratory frame, EB equals the rest energy of the particle and pB = 0: s=

[(EA + mB c2 )2 − |pA |2 ] c2

In the center of mass frame, we take the colliding particles to have equal and opposite momenta: s=

[(EA + EB )2 − (pA + pB )2 ] (EA + EB )2 = c2 c2

b) Substituting the sums of the four-vectors into the sum of the Mandelstam variables: 1 h (EA + EB )2 (EA − EC )2 s+t+u= 2 − (pA + pB )2 + − (pA − pC )2 2 c c c2 i (EA − ED )2 2 − (p − p ) + A D c2 Using conservation of energy and momentum: E A − E D = EC − E B E A − E C = ED − E B

pA − pD = pC − pB pA − pC = pD − pB

Substitute these quantities into the equation above: s+t+u=

(ED − EB )2 1 h (EA + EB )2 2 − (p + p ) + − (pD − pB )2 A B c2 c2 c2 i (EC − EB )2 2 − (p − p ) + C B c2

In the laboratory frame, pB = 0 and EB = mB c2 and so: 2ED mB 2EC mB 2EA mB − − + m2A + 3m2B + m2c + m2D 2 2 c c c2 2mB = 2 (EA − ED − EC ) + m2A + 3m2B + m2c + m2D c

s+t+u=

From the conservation of energy, EB = EA − ED − EC . Making this substitution and using EB = mB c2 : s + t + u = −2m2B + m2A + 3m2B + m2c + m2D = m2A + m2B + m2c + m2D

10. Starting from conservation of energy and momentum in the center of mass frame for the decay ρ → 2π 0 (where the ρ meson is at rest):

5

Eρ = Eπ1 + Eπ2 pπ1 = −pπ2 and Eρ is just the given rest mass, mρ c2 = 775.5 M eV . Since the pions are symmetric, Eπ1 = Eπ2 . Using Eq. (13.14) for the RHS: p mρ c2 = 2 p2π c2 + m2π c4

where the π subscript refers to either pion. Squaring both sides: (775.5 M eV )2 = (4)[(pπ c)2 + (135.0 M eV )2

gives pπ = 365.5 M eV /c. To get the velocity of the pions, we need γ: p p2π c2 + m2π c4 Eπ = γπ = 2 mπ c mπ c 2 and plugging in the above values gives γπ = 2.87. So: 2.87 = p

1 1 − v 2 /c2

and solving for v gives v = 0.937 c.

11. The decay here is Λ0 → p + π − . Using four-vectors to conserve both momentum and energy: pΛ = pp + pπ First, isolate the proton momentum on the LHS: pp = pΛ − pπ Squaring both sides: p2p = p2Λ + p2π − 2pΛ · pπ As shown above Eq. (13.14), the square of a momentum four-vector gives just its mass: m2p c2 = m2Λ c2 + m2π c2 − 2pΛ · pπ Carrying out the dot product: m2p c2 = m2Λ c2 + m2π c2 − 2(

E Λ Eπ − pΛ · pπ ) c2

where the second term of the dot product is zero since p~Λ = 0 in its rest frame. Collecting terms and noting EΛ = mΛ in this frame:

6

m2Λ c4 + m2π c4 − m2p c4 2mΛ c2 (1115.7)2 + (139.6)2 − (938.3)2 M eV = 172.0 M eV = 2(1115.7)

Eπ =

Using just energy conservation: Ep = EΛ − Eπ = 1115.7 M eV − 172.0 M eV = 943.7 M eV To get the velocity of the each particle, find γ using Eq. (13.9): γ=

E mc2

giving γπ = 172.0/139.6 = 1.232 and γp = 943.7/938.3 = 1.0058. The velocity is now found using Eq. (13.3): γ=p

1 1 − v 2 /c2

which results in the pion’s speed vπ = 0.584 c and the proton’s speed vp = 0.107 c.

12. The particle collision here has two particles with equal mass and speed traveling in opposite directions. Using the conservation of momentum in the center of mass frame: pA + pB = pC = 0 This says that the resulting particle is at rest. Using conservation of energy: EA + EB = EC 2γmc2 = mC c2 mC = 2γm Using Eq. (13.3) with the given speed of v = (2/3)c, mC = p

2m 1 − (2/3)2

= 2.68m

Since the mass of the resulting particle is greater than the sum of the colliding particles, some of their kinetic energy was converted into the resulting particle’s rest-energy.

13. Using the result of the Problem 12, but for electron-positron collisions:

7

mC = 2γme where in this case, we find γ using Eq. (13.9) γ=

E me c 2

Plugging this in above gives mC = 20 GeV /c2 . Now we want to get the reaction mC → µ + µ where µ is the symbol for a muon. Following the same steps as for Problem 10: q m2C = 2 p2µ c2 + m2µ c4 where the µ subscript refers to either muon. Squaring both sides: (20.0 GeV )2 = (4)[(pµ c)2 + (0.1057 GeV )2 gives pµ = 20 GeV /c to high precision. To get the velocity of the muons, we need γ. Using Eq. (13.9): γµ =

Eµ 20.0 GeV = = 189.2 mµ c 2 0.1057 GeV

Plugging in for γ from Eq. (13.3): 189.2 = p

1 1 − v 2 /c2

and solving for v gives v = 0.999986 c. The kinetic energy is, from Eq. (13.13): KE = (γ − 1)mµ c2 = (189.2 − 1)(0.1057 GeV ) = 19.89 GeV

14. In the center of mass frame where the proton and antiproton have equal and opposite momenta: E p + E p = EX where EX is the energy of the resulting particle. As in Problem 12: 2γmp c2 = mX c2 2γ(938 M eV ) = 9700 M eV

giving γ = 5.17. Using Eq. (13.13) for the kinetic energies: KE = (γ − 1)mp c2 = 3912 M eV

8

Using Eq. (13.3) to get velocities: γ = 5.17 = p

1 1 − v 2 /c2

Solving for v, the proton and antiproton will each have a speed of v = 0.981 c in the center of mass frame.

15.a) The reaction here is π + → µ + νmu where the neutrino is assumed to have zero mass. From conservation of energy in the center of mass frame: E π = m π c 2 = Eµ + E ν where Eν = |pν |c since mν = 0. From conservation of momentum: |pν | = |pµ | Using the above and Eq. (13.14) for Eµ : q mπ c2 = p2µ c2 + m2µ c4 + pµ c Isolating the square root and then squaring both sides: (mπ c2 − pµ c)2 = p2µ c2 + m2µ c4

m2π c4 − 2mπ c2 pµ c + p2µ c2 = p2µ c2 + m2µ c4 Canceling terms and solving for pµ : m2π c4 − m2µ c4 2mπ c2 (139.6 M eV )2 − (105.7 M eV )2 = = 29.8 M eV 2(139.6 M eV )

pµ c =

To get the velocity, we need to get γ for the muon: q p2µ c2 + m2µ c4 Eµ = γµ = mµ c 2 mµ c 2 giving γµ = 1.039 Using Eq. (13.3) and solving for v results in v = 0.271 c. b) The mean distance traveled by the muon is given by d = v∆tM where ∆tR = 2.2 × 10−6 s is the muon lifetime in the rest frame. Using Eq. (12.26) to relate ∆tM and ∆tR : d = v∆tM = vγµ ∆tR = (0.271)(3.0 × 108 m/s)(1.039)(2.2 × 10−6 s) = 186 m

9

16. First, test the relation: αi αj + αj αi = 2δij I The Dirac-Pauli representation is given in Eq. (13.42):

i j

σσ 0 0 σj 0 σi i j = αα = σj 0 0 σi σj σi 0

αj αi =

αi αj + αj αi =

0 σj σj 0

0 σi σi 0

=

σj σi 0 0 σj σi

σi σj + σj σi 0 0 σi σj + σj σi

= (σ i σ j + σ j σ i )I

The Pauli matrices satisfy the following condition:

I if i = j i j σσ = −σ j σ i if i 6= j Therefore: σi σj + σj σi =

2 if i = j 0 if i = 6 j

Or, more concisely, σ i σ j + σ j σ i = 2δij I. Thus, αi αj + αj αi = 2δij I. Next, check β 2 = I. From Eq. (13.42):

I0 β= 0 −I We carry out the multiplication:

2 I0 I0 I 0 = 0 −I 0 −I 0 I2 Since I 2 = I, β 2 = I. 17. The definition of the co-variant γ µ matrices are given in Eq. (13.47): γ i = βαi , γ 0 = β where αi and β matrices are defined in Eq. (13.42). From Example 13.6, it is shown by Eqs. (13.58)-(13.59) that:

10

α·p=

0 σ·p σ·p 0

, βm =

mI 0 0 −mI

where the Pauli spin matrices, σ i are given by Eq. (13.43) and I is the 2 × 2 unit matrix. Since the zeroth component of the momenta is the energy, p0 = E, and the zeroth component of the Dirac matrices is just γ 0 = β, then

EI 0 γ 0 p0 = 0 −EI Similarly, we can use the above result for α · p and multiply by β giving:

I 0 0 σ·p 0 σ·p i γ pi = −βα · p = − =− 0 −I σ·p 0 −σ · p 0 where the minus sign comes from lowering the index, recall pµ = (E, p) but pµ = (E, −p). Adding the results together:

(E + m) −σ · p µ γ pµ + m = σ · p (−E + m) which is written in two-component form. Similarly,

(E − m) −σ · p γ µ pµ − m = σ · p (−E − m) 18. Muons and electrons are both Dirac particles. The Feynman diagram for muon-electron scattering is the same as the Feynman diagram shown in Fig. 13.4 with one of the incoming and outgoing lines corresponding to an electron and the other corresponding to a muon.

14 Particle Physics - Solutions

1.a) This process can not occur. It violates the conservation of baryon number and lepton number. The baryon number is one of the LHS but zero on the RHS. The electron lepton number is zero on the LHS but −1 on the RHS. b) This process can not occur due to violation of energy conservation. The Σ◦ has a rest energy of 1192 MeV while the Λ and the π ◦ have a total rest energy of 1250 MeV. c) This process is allowed. Conservations of charge, energy, baryon number, and lepton number are all obeyed. This process conserves strangeness and proceeds by the strong interaction. d) This process can not occur. The conservation of muon lepton number is violated. The total number is zero on the LHS, but is −1 on the RHS. e) This process is allowed. K − has a strange quantum number S= −1 where the π − π ◦ system has S=0. Since charge, energy, lepton number, and baryon number are all conserved, so this process proceeds via the weak interaction. f ) This process can not occur. The ρ0 , being a boson, with total spin S=1, must have a symmetric wave function. The ρ0 has a total angular momentum of J=1. By the conservation of angular momentum, the spinless pions would have an antisymmetric total wave function with L=1. Pions are bosons, so this is forbidden. g) This process is not seen experimentally. Strangeness is not conserved while the other conservation laws are, so the process can only occur via the weak interaction. But since no neutrinos or lepton pairs are present, indications of a weak decay are lacking in the final state. The photon does not couple directly to the neutral weak current (the photon couples only to charged particles).

2

2.a) This reaction is allowed, assuming that the K − and proton collision has enough energy. b) This reaction is allowed only if the neutrino and antineutrino are of the electron type. c) This reaction can not occur. On the LHS, the lepton number is Lµ = +1, but on the RHS it is Lµ = −1. d) This reaction is allowed. e) This reaction is allowed. Both Le and Lµ are conserved separately. f ) This reaction can not occur. On the LHS, the lepton number is Le = +1, but on the RHS it is Le = −1. (Strangeness is conserved.) g) This reaction is not seen experimentally. Strangeness is not conserved.

(This can not be a weak decay, since the RHS has more mass than the LHS. The weak process is too weak to be seen in typical beam collision experiments.)

3. The rest energies may be found in Tables 14.8 through 14.10. a) π − + p → n + π 0 Initial state: 140 + 938 = 1078 MeV Final state: 940 + 135 = 1074 Mev b) Λ → p + π − Initial state: 1116 MeV Final state: 938 + 140 = 1078 Mev c) π + + p → K + + Σ + Initial state: 140 + 938 = 1078 MeV Final state: 494 + 1189 = 1683 Mev The kinetic energy of this initial state must be greater than that of the final state to ensure conservation of energy.

4.a) The decay is allowed. b) This reaction violates electron lepton number. It has Le = +1 on the LHS, but Le = −1 on the RHS. c) This decay violates electron lepton number. It has Le = 0 on the LHS, but Le = 2 on the RHS. d) This decay is allowed.

3

5. See Tables 14.8-14.10 to find the quark content, then use Eq. (14.26). a) The Ω− has a quark content of sss, therefore S(Ω) = −3. For the particles on the RHS, S(Λ) = −1 and S(K − ) = −1. Since ∆S = 1, the process must occur via the weak interaction. b) On the LHS, S(Σ◦ ) = −1 and on the RHS, S(Λ) = −1, and S(γ) = 0.

There is no change in strangeness. The presence of the photon signals that this is decay via the electromagnetic interaction. c) On the LHS, S(Λ) = −1 and on the RHS, S(p) = 0, and S(π − ) = 0.

Since ∆S = 1, the decay must occur via the weak interaction.

6.a) Here, ∆S = 1, so this process occurs via the weak interaction. b) ∆S = 0. The presence of the antineutrino tells us this must be a weak interaction. c) Again, ∆S = 1, so this process occurs via the weak interaction.

7.a) This process occurs via the strong interaction. All conservation laws are met. b) This process occurs via the weak interaction. The strangeness of the initial state is zero, but has strangeness of −1 in the final state. The other conservation laws are met. c) This process is strictly forbidden. The baryon number of the initial state is +1 while that of the final is zero. Note that the antiparticle has a baryon number of −1. d) This process is strictly forbidden. The muon lepton number and the electron lepton number must each be conserved separately. In this case, both lepton numbers are violated. The muon lepton number is −1 on the LHS but zero on the RHS. Similarly, the electron lepton number is zero on the LHS but −1 on the RHS. e) This process is strictly forbidden. Here again, lepton number conservation is violated. The electron neutrino on the left-hand side has an electron lepton number of +1 whereas the positron on the right-hand side has an electron lepton number of −1.

8. The quark compositions for most particles are given in Tables 14.3-14.4 and 14.8-14.10.

4

a) The Lambda is uds, the proton is uud and the π 0 is a mixture of uū ¯ and dd. ¯ Λ(uds) → p(uud) + π 0 (uū or dd) The weak interaction has changed one s quark to a d quark and produced a ¯ pair in the final state. uū (or dd) b) The π − is dū, the proton is uud, the neutron is udd, and the π 0 is a ¯ mixture of uū and dd. ¯ π − (dū) + p(uud) → n(udd) + π 0 (uū or dd) All quark numbers are conserved. This is a strong interaction. ¯ the proton is uud, the K + is us̄ and the Σ + is uus. c) The π + is ud, ¯ + p(uud) → K + (us̄) + Σ + (uus) π + (ud) All quark numbers are conserved. This is a strong interaction, where a uū pair annihilated in the initial state and a ss̄ pair created in the final state from the kinetic energy of the collision.

9. The quark composition may be found in Tables 14.3-14.4 and 14.8-14.10: a) Ω − (sss) → Λ(uds) + K − (sū) There are only strange quarks in the initial state. The weak interaction has changed one s quark to a d quark, and produced a uū pair in the final state. b) ¯ π − (dū) + p(uud) → Λ (uds) + K 0 (sd) There are no strange quarks in the initial state, and a ss̄ pair in the final state. All quark numbers are conserved. This is a strong interaction. c) p(uud) + K − (sū) → Ξ − (dss) + K + (us̄) There is no change in strangeness from initial to final state. All quark numbers are conserved. This is a strong interaction.

10. The isospin quantum numbers for various particle are given in Tables 14.8-14.10.

5

a) The isospin of the π − is I = 1, I3 = −1. For the proton, I = 1/2, I3 = +1/2.

The possible values of the combined isospin range from |Iπ − Ip | to |Iπ + Ip |, or I = 1/2, 3/2. The isospin projections add linearly: I3 = −1/2. Hence, the possible values are: I = 1/2, I3 = −1/2

I = 3/2, I3 = −1/2

b) The isospin of the π − is I = 1, I3 = −1. For the neutron, I = 1/2, I3 = −1/2. The possible values of the combined isospin range from |Iπ − Ip | to |Iπ + Ip |, or I = 1/2, 3/2. The isospin projections add linearly: I3 = −3/2.

Since the isospin I cannot be less than |I3 |, the only possible values are: I = 3/2, I3 = −3/2 c) Similarly to the above two cases, there is only one possible value of isospin numbers: I = 3/2, I3 = +3/2 d) Here, we have only one type of particle, with I = 1/2 and I3 = −1/2. The total isospin projection is just the sum of the two, I3 = −1. Now I can range from I = 0 to I = 1, but only the latter is possible for the given total I3 . The possible values are: I = 1, I3 = −1 e) Similarly to the above problems, the total I3 = 0 and I = 0, 1. The allowed values are: I = 0, I3 = 0 I = 1, I3 = 0 11. The quark contents of various particles is given in Tables 14.8-14.10. The relation for the Cabibbo angles are given by Eq. (14.48) and Fig. 14.22. a) Λ(uds) → π − (dū) + p(uud) The weak interaction has changed a s quark to a d quark, via Z 0 exchange, creating a uū in the final state. This has no Cabibbo angle. b)

6

n(udd) → p(uud) + e− + ν̄e The weak interaction has changed a d quark to a u quark, via W − exchange, creating a e− ν̄e lepton pair in the final state. The Cabibbo angle is cos θc at the quark vertex. c) Ξ 0 (uss) → Σ − (dds) + e+ + νe This process is difficult to draw as a Feynman diagram, as it involves the change of two quarks, a u → d and a s → d, at the same time. The u → d goes via W + exchange, since the quark charge changes by one unit, whereas the s → d goes via Z 0 exchange, since the quark charge is the same for both quarks. In other words, two weak interactions must occur, and this has a very low probability. This process has not been seen experimentally. d) Ω − (sss) → Σ 0 (uds) + e− + ν̄e As in the previous problem, this is difficult to draw as a Feynman diagram, as it involves the change of two quarks, s → u and s → d. The former goes via W − exchange and the latter goes via Z 0 exchange. This process changes strangeness by two units, and has a very small probability to occur.

12. Figure 14.30 serves as a guide to obtain the remaining weights from the highest weight. In Figure 14.30(B), the highest weight is 3µ1 , where µ1 is given

by Eq. (14.83). Along a diagonal, such as the one with the weights labeled in Fig. 14.30(B), just subtract an additional α1 . Going right-to-left along a row, just subtract (α1 + α2 ) for each step. Below, the results go right-to-left across a row, and after completing rows, from the top down to the last element, 3µ1 − 3α1 . Matrices for α1 and α2 are given by Eq. (14.77):

3/2 1/2 √ √ = 3µ1 = 3 3/6 3/2

1/2 3µ1 − α1 − α2 = √ 3/2

−1/2 3µ1 − 2α1 − 2α2 = √ 3/2

−3/2 3µ1 − 3α1 − 3α2 = √ 3/2

7

3µ1 − α1 =

1 0

3µ1 − 2α1 − α2 =

0 0

−1 3µ1 − 3α1 − 2α2 = 0

1/2 √ 3µ1 − 2α1 = − 3/2 3µ1 − 3α1 − α2 =

3µ1 − 3α1 =

−1/2 √ − 3/2 0 √

− 3

15 Nuclear Physics - Solutions

1. From Section 15.2, the notation for isotopes is: A ZX

where A is the atomic mass number, Z is the atomic number, and X is the element symbol. The number of neutrons is equal to A minus Z: 3 7 Li : 3 protons, 4 neutrons 63 29 Cu : 29 protons, 34 neutrons 238 92 U : 92 protons, 146 neutrons

2. Using the notation explained above, the atomic numbers add up, 92 = 36 + 56. So any loss of atomic mass number must be due to neutrons: Nneutrons = 236 − (90 + 144) = 2 With 2 neutrons per fission, a chain reaction can develop if the neutrons are mostly contained within the volume of enriched 236 U. 3. Using Eq. (15.3), an estimate of the nuclear radius is: R = 1.12A1/3 f m where A is the atomic mass number. Using this formula: 4 1/3 f m = 1.78 f m 2 He : R = 1.12(4) 1/3 16 f m = 2.82 f m 8 He : R = 1.12(16) 56 1/3 f m = 4.28 f m 26 F e : R = 1.12(56) 208 1/3 f m = 6.64 f m 82 P b : R = 1.12(208) 237 1/3 f m = 6.93 f m 93 N p : R = 1.12(237)

2

4. Using Eq. (15.5) for the binding energy: 4 2 He

: B = [2(1.007825) + 2(1.008665) − 4.002602]u = 0.030378(931.5 M eV ) = 28.3 M eV

16 8 O

: B = [8(1.007825) + 8(1.008665) − 15.994915]u = 0.137005(931.5 M eV ) = 127.62 M eV

56 26 F e

: B = [26(1.007825) + 30(1.008665) − 55.934938]u = 0.528462(931.5 M eV ) = 492.26 M eV

208 82 P b

: B = [82(1.007825) + 126(1.008665) − 207.97665]u = 1.75679(931.5 M eV ) = 1636.45 M eV

238 92 U

: B = [92(1.007825) + 146(1.008665) − 238.05079]u = 1.9342(931.5 M eV ) = 1801.71 M eV

To get the binding energy per nucleon, just divide by A: 4 2 He 16 8 O 56 26 F e 208 82 P b 238 92 U

: B/A = (28.3/4) M eV = 7.08M eV /nucleon : B/A = (127.62/16) M eV = 7.98M eV /nucleon : B/A = (492.26/56) M eV = 8.79M eV /nucleon : B/A = (1636.45/208) M eV = 7.87M eV /nucleon : B/A = (1801.71/238) M eV = 7.57M eV /nucleon

5. To find the binding energy of a proton, take the difference between the nuclear masses and subtract the proton mass: A−1 Bp = m(A Z X) − m(Z−1 Y ) − m(H)

where X and Y are the chemical symbols for the original and final nucleus, respectively, and m(H) is the mass of hydrogen. The result is: 4 4 3 2 He : Bp = m(2 He) − m(1 H) − m(H) = [4.002603 − 3.016049 − 1.007825]u

= −0.021271(931.5 M eV ) = −19.81 M eV 56 56 55 26 F e : Bp = m(26 F e) − m(25 M n) − m(H) = [55.93494 − 54.93805 − 1.00783]u

= −0.01094(931.5 M eV ) = −10.2 M eV 208 208 207 82 P b : Bp = m(82 P b) − m(81 T l) − m(H) = [207.9767 − 206.9774 − 1.0078]u

= −0.0085(931.5 M eV ) = −7.9 M eV The energy to remove the proton is just E = −Bp . Note that another way to do this is simply to subtract the binding energies of the parent and daughter nucleus.

3

6. To find the binding energy of a neutron, take the difference between the nuclear masses and subtract the neutron mass: A−1 Bn = m(A X) − mn Z X) − m(Z

where X is the chemical symbol for the original nucleus and mn is the neutron mass. The result is: 3 4 4 2 He : Bn = m(2 He) − m(2 He) − mn = [4.002603 − 3.016029 − 1.008665]u

= −0.022091(931.5 M eV ) = −20.58 M eV 56 56 55 26 F e : Bn = m(26 F e) − m(26 F e) − mn = [55.93494 − 54.93829 − 1.00867]u

= −0.01202(931.5 M eV ) = −11.2 M eV 208 208 207 82 P b : Bn = m(82 P b) − m(82 P b) − mn = [207.9767 − 206.9759 − 1.0087]u

= −0.0079(931.5 M eV ) = −7.4 M eV The energy to remove the neutron is just E = −Bn .

7. Using Eq. (15.6) to calculate the binding energies, with the parameters of Eq. (15.7)-(15.8): 4 2 He 16 8 O 56 26 F e 208 82 P b 238 92 U

: B = 9.75 M eV, B/A = 2.44 M eV : B = 116.04 M eV, B/A = 7.25 M eV : B = 484.18 M eV, B/A = 8.65 M eV : B = 1623.2 M eV, B/A = 7.80 M eV : B = 1804.6 M eV, B/A = 7.58 M eV

Except for the first case of 4 He, the agreement with B/A of Problem 4 is good to an accuracy of about 1%.

8. The semi-empirical formula, Eq. (15.6), but without the pairing term is as follows: B(N, Z) = aA − bA2/3 −

dZ 2 (N − Z)2 −s 1/3 A A

When N=Z, the fourth term is zero. To get the binding energy per nucleon, divide by A: dZ 2 B(N, Z) = a − bA−1/3 − 4/3 A A Now let Z=A/2: d Binding energy per nucleon = a − bA−1/3 − A2/3 4

4

To find the maximum, differentiate with respect to A and set equal to zero: 1 −4/3 1 −1/3 A − dA =0 3 2 Solving for A (assuming A 6= 0): A=

2b d

Plugging in the values for b and d from Eq. (15.7), the maximum is at A = 51.3.

Rounding up, this gives A/2 ≈ 26 as given in the problem.

9. Eq. (15.13) relates the half-life to the proportionality constant λ: t1/2 = 2 min = 120 s = λ=

ln(2) λ

0.693 = 0.006 s−1 120 s

The decay rate, given just below Eq. (15.12), and the given initial rate R = 1200 s−1 at t = 0: R = 1200 s−1 = (0.006 s−1 )N0 (1) giving N0 = 2.0 × 105 . The same equation gives the decay rates at later times: t = 4 min = 240 s : R = (0.006 s−1 )(2 × 105 )e−0.006(240) = 284 s−1 t = 6 min = 360 s : R = (0.006 s−1 )(2 × 105 )e−0.006(360) = 138 s−1 t = 8 min = 480 s : R = (0.006 s−1 )(2 × 105 )e−0.006(480) = 67 s−1 10. Using Eq. (15.13) to calculate the decay constant: λ=

0.693 ln(2) = = 0.0563 yr−1 t1/2 12.3 yr

Using Eq. (15.12) to calculate the fraction left after 40 years: N = e−(0.0563)(40) = 0.105 N0 About 10% of the original amount is left after waiting 40 years.

11.a) Using Eq. (15.13) to get the decay constant: λ=

ln(2) 0.693 = = 2.30 × 10−10 min−1 5730 yr 3.012 × 109 min

Using this in the rate equation to get the initial amount:

5

15.3 min−1 = (2.30 × 10−10 min−1 )N0 e−(2.30×10

−10

)(1)

N0 = 6.65 × 1010 atoms of 14 6 C The total number of carbon atoms in the sample is: 1g·

6.022 × 1023 atoms = 5.018 × 1022 atoms of C 12 g/mol

The proportion is: 6.65 × 1010 1.32 × 10−12 atoms 14 6 C = 22 5.018 × 10 total atoms C This indicates the rarity of 14 C for isolated objects and thus its reliability for dating ancient relics since these nuclear decays can be easily measured. b) Using the above values for λ and N0 when the organism was alive and the given time period: 20, 000 years = 1.051 × 1010 minutes then the rate of decays is: R = (2.30 × 10−10 min−1 )(6.65 × 1010 decays) e−(2.30×10 = 1.36 decays per minute

−10

)(1051×1010 )

12. After calculating A and Z for the missing nucleus, use the semi-empirical mass formula, Eq. (15.6), to calculate the mass of each nucleus, with u = 931.5 M eV from Appendix A. a) 209 205 83 Bi → 81 Tl

+ 42 He

∆E = (208.98037 − 204.97440 − 4.00260) u = 0.00337u = 3.14 M eV b) 238 234 92 U → 90 Th

+ 42 He ∆E = (238.05078 − 234.04360 − 4.00260) u = 0.00458u = 4.27 M eV

c) For the next two reactions, the Q-value of beta-decay is given, see Examples 15.5 and 15.6. The Q-value is equal to the energy release (including the positron mass of 0.511 MeV).

77 77 36 Kr → 35 Br

+ e+ + ν e

Q = (71655.330 − 71652.265) M eV = 3.065 M eV

6

d) 77 77 35 Br → 34 Se

+ e+ + ν e

Q = (71652.265 − 71650.900) M eV = 1.365 M eV

13. The energy release can be found by taking the difference in masses between and 73 Li. The mass of the electron is irrelevant, since the masses are tabulated for neutral atoms, and Li has one fewer electron than Be. (The mass of the electron has been converted to energy via the weak decay mechanism.)

Using Table 15.2 for the 73 Li mass and looking up the mass of 74 Be:

7 4 Be

∆E = (7.016930 − 7.016004) u = 0.000926 u = 0.86M eV 14. The most likely decay of 62 He is beta-decay, since it is a light nucleus and neutron-rich: 6 2 He

→63 Li + e− + ν̄e Q = (6.01889 − 6.01512)u = 3.512 M eV

The case of 84 Be is unusual, since it can split exactly into two 42 He nuclei: 8 4 Be

→42 He +42 He Q = (8.005305 − 2 · 4.002603)u = 0.092 M eV

With 12 nucleons, 12 4 Be quickly decays via beta-decay: 12 4 Be

− →12 + ν̄e 5 B +e

Q = (12.02692 − 12.01435)u = 11.71 M eV The nucleus 15 8 O has more protons than neutrons, so it will emit a positron: 15 8 O

+ →15 + νe 7 N + e Q = (15.00307 − 15.00011)u = 2.76 M eV

Although a heavy nucleus like 240 95 Am could α-decay, this nucleus prefers to emit a positron: 240 95 Am

+ →240 + νe 94 P u + e Q = (240.05529 − 240.05381)u = 1.38 M eV

The energy releases for beta decay here are the Q-value (see Examples 15.5 and 15.6) which includes the electron or positron mass.

7

15.a) Start by looking up the masses of each nucleus: m(234 92 U ) = 234.04095 u m(230 90 T h) = 230.03313 u m(42 He) = 4.00260 u Using u = 931.5 M eV , the energy release of the decay is: ∆E = (234.04095 − 230.03313 − 4.00260)u = 4.86 M eV b) The kinetic energy of the α-particle comes from conservation of momentum: mT h vT h = mHe vHe and the sum of the kinetic energies must add to the total energy release: 1 1 2 mT h vT2 h + mHe vHe = 4.86 M eV 2 2 Solving these equations for vT h in the first equation gives:

1 mHe 2 + 1 mHe vHe = 4.86 M eV 2 mT h so the kinetic energy of the α-particle is: KEα =

4.86 M eV = 4.78 M eV 1 + mHe /mT h

16.a Using the information in Table 15.5, along with the example of mirror nuclei in Fig. 15.16, and the text below it. The angular momentum, j and parity, π, of the ground states are: 5+ reason : one extra nucleon in 1d5/2 orbit 2 5+ 17 π reason : mirror nucleus 9 F :j = 2 1+ 31 π reason : mirror nucleus 15 P : j = 2 1+ 31 π reason : one nucleon missing in 2s1/2 orbit 16 S : j = 2 32 π + 16 S : j = 0 reason : even − even nucleus

17 8 O

: jπ =

40 20 Ca

: j π = 0+ reason : even − even nucleus

45 21 Sc

: jπ =

7− reason : one nucleon in 1f7/2 orbit 2

8

b) The single-particle excited states have one nucleon from the closed shell moving up into the next shell. The angular momentum and parity of the excited states are not easy to surmise, but here they are: 1+ reason : one extra nucleon in 2s1/2 shell 2 1+ 17 π reason : mirror nucleus 9 F : excited j = 2 3+ 31 π 15 P : excited j = 2 3+ 31 π 16 S : excited j = 2 32 π + S : excited j = 2 reason : even − even nucleus 16

17 8 O

: excited j π =

40 20 Ca

: excited j π = 2+ reason : even − even nucleus

45 21 Sc

: excited j π =

3+ 2

For the case of 17 8 O, one nucleon gets promoted from the 1d5/2 shell to the 2s1/2 shell. This single nucleon determines the spin-parity of the nucleus. The other 16 nucleons in the core are tightly bound, and not easy to excite. One could imagine that for 31 16 S, the unpaired nucleon in the 2s1/2 shell of the ground stategets promoted to the 1d5/2 shell, which would result in j π = (5/2)+ , but this is not what actually happens in nature. The reason for the (3/2)+ excited state can be explained by the spin-coupling between two excited nucleons, which takes us beyond the scope of this textbook.

17. Starting with Eq. (15.23), but using notation where subscript D represents the deuteron, subscript X represents the excited nucleus, and subscript p represents the proton, then using conservation of momentum: pD = pp cosθ + Px 0 = pp sinθ + Py Solving for Px and Py : Px = pD − pp cosθ Px = −pp sinθ The total kinetic energy before the collision is just from the deuteron: Ei =

p2D 2mD

The total kinetic energy after the collision is:

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Ef =

p2p 1 (pD − pp cosθ)2 + (pp sinθ)2 + 2mp 2mX

The energy absorbed by the nucleus equals the loss of kinetic energy: E = Ei − E f =

p2p p2D 1 − − (p2 + p2p − 2pD pp cosθ) 2mD 2mp 2mX D

Substitute the corresponding energies for the first two terms: p2D = ED 2mD p2p = Ep 2mp p pD pp = 4ED Ep mD mp

The energy of the excited nucleus due to deuteron stripping is: p

2 ED Ep mD mp mD mp E = ED 1 − − Ep 1 − − cosθ mX mX mX 18. Using conservation of momentum: pα = −pd Substituting this into the given energy: E=

1 2 1 2 p + p 2mα α 2md α

Collecting terms: 1 E= 2

1 1 + mα md

p2α

Adding the fractions to get a common denominator:

1 md + mα p2α E= 2 mα md which is the desired result.

19. The Q-value for this reaction has already been done in problem 14 above.

The result is: Q = 0.092M eV The mean lifetime can be estimated using the uncertainty principle:

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∆E · ∆t ∼ ~ where ∆E ≃ Q. Using ~c = 197.3 M eV · f m and c = 3.0 × 108 m/s = 3.0 × 1023 f m/s: ∆t ∼

197.3 s = 7 × 10−21 s 0.092(3.0 × 1023 )

This is shorter than the observed lifetime, but this is a very crude estimate.

A better calculation of the lifetime can be obtained using the WKB approximation, which is presented in several advanced textbooks on quantum theory.

20. Start by looking up the masses of each nucleus: m(238 94 P u) = 238.04955 u m(234 92 U ) = 234.04095 u m(42 He) = 4.00260 u Using u = 931.5 M eV , the Q-value of the decay is: Q = (238.04955 − 234.04095 − 4.00260)u = 5.6 M eV The half-life can be estimated using the uncertainty principle: ∆E · ∆t ∼ ~ where ∆E ≃ Q. Using ~c = 197.3 M eV · f m and c = 3.0 × 1023 f m/s: ∆t ∼

197.3 s = 1 × 10−22 s 5.6(3.0 × 1023 )

In this case, the lifetime is off (compared with the observed lifetime) by many orders of magnitude. The reason is that the α must penetrate the Coulomb barrier, which is substantial in a high-Z nucleus like 238 94 Pu but much smaller in a light nucleus (as in the previous problem).

21. Start by looking up the masses of each nucleus: m(194 79 Au) = 193.96534 u m(194 78 P t) = 193.96266 u m(190 77 Ir) = 189.96059 u m(42 He) = 4.00260 u Using u = 931.5 M eV , the Q-value of the β-decay is: Q = (193.96534 − 193.96266)u = 2.5 M eV

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The Q-value for the α-decay is: Q = (193.96534 − 189.96059 − 4.0026)u = 2.0 M eV The lifetime for the β-decay can be estimated as in Problem 19, and is on the order of 10−22 s, but again this is a very crude estimate. As shown in Problem 20, it is not possible to get a good estimate on the lifetime of α-decay since the lifetime is very sensitive to the height of the Coulomb barrier.

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