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24.1. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:

Substituting into the formula for the Balmer series,

λ=

91.18 nm 91.18 nm ⇒λ= = 410.3 nm 1 1 1 1   − − 2 2  22 n2  2 6

where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, λ = 389.0 nm and λ = 379.9 nm .


24.2. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:

Balmer’s formula is

λ=

91.18 nm  1 − 1  22 n2 

n = 3, 4, 5, 6, …

As n → ∞, 1 n 2 → 0. Thus, λ n→∞ = 4(91.18 nm ) = 364.7 nm .


24.3. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:

Using Balmer’s formula,

λ = 389.0 nm =

91.18 nm 1 1 ⇒ − 2 = 0.2344 ⇒ n = 8 4 n  1 − 1  22 n2 


24.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ , where m = 1, 2, 3, … For first and second order diffraction,

2 d cosθ1 = (1)λ

2 d cosθ 2 = (2)λ

Dividing these two equations,

cosθ 2 = 2 ⇒ θ 2 = cos −1 (2 cosθ1 ) = cos −1 (2 cos 68°) = 41.5° cosθ1


24.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ . For m = 1 and for two different wavelengths,

2 d cosθ1 = (1)λ1

2 d cosθ1′ = (1)λ1′

Dividing these two equations,

cosθ1′ λ1′ cosθ1′ 0.15 nm = ⇒ = ⇒ θ1′ = cos −1 (0.4408) = 63.8° cosθ1 λ1 cos 54° 0.20 nm


24.6. Model: The angles corresponding to the various orders of diffraction satisfy the Bragg condition. Solve:

The Bragg condition for m = 1 and m = 2 gives

2 d cosθ1 = (1)λ

2 d cosθ 2 = (2)λ

Dividing these two equations, cosθ1 =

cosθ 2 cos 45°  cos 45°  = ⇒ θ1 = cos −1 = 69.3°  2  2 2


24.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition.

Solve: The Bragg condition is 2d cosθ m = mλ , where m = 1, 2, 3, … The maximum possible value of m is the number of possible diffraction orders. The maximum value of cosθm is 1. Thus,

2 d = mλ ⇒ m = We can observe up to the fourth diffraction order.

2 d 2(0.180 nm ) = = 4.2 λ (0.085 nm)


24.8. Model: Use the photon model of light. Solve:

The energy of the photon is

Ephoton = hf = h

 3.0 × 10 8 m / s  c −19 = (6.63 × 10 −34 Js)  = 3.98 × 10 J  500 × 10 −9 m  λ

Assess: The energy of a single photon in the visible light region is extremely small.


24.9. Model: Use the photon model of light. Solve:

The energy of the x-ray photon is

 3.0 × 10 8 m / s  c −16 E = hf = h  = (6.63 × 10 −34 Js)  = 1.99 × 10 J  λ  1.0 × 10 −9 m  Assess: This is a very small amount of energy, but it is larger than the energy of a photon in the visible wavelength region.


24.10. Model: Use the photon model of light.

Solve: The energy of a photon with wavelength λ1 is E1 = hf1 = hc λ1 . Similarly, E2 = hc λ 2 . Since E2 be equal to 2E1,

λ hc hc 600 nm ⇒ λ2 = 1 = =2 = 300 nm λ2 λ1 2 2 Assess: A photon with λ = 300 nm has twice the energy of a photon with λ = 600 nm. This is an expected result, because energy is inversely proportional to the wavelength.


24.11. Model: Use the photon model of light. Solve:

The energy of the single photon is

(6.63 × 10 −34 Js)(3.0 × 10 8 m / s) = 1.99 × 10 −19 J c Ephoton = hf = h  =  λ 1.0 × 10 −6 m ⇒ Emol = N A Ephoton = (6.023 × 10 23 )(1.99 × 10 −19 J ) = 1.2 × 10 5 J Assess: Although the energy of a single photon is very small, a mole of photons has a significant amount of energy.


24.12. Solve: Your mass is, say, m ≈ 70 kg and your velocity is 1 m/s. Thus, your momentum is p = mv ≈ (70 kg) (1 m/s) = 70 kg m/s. Your de Broglie wavelength is

λ=

h 6.63 × 10 −34 Js = ≈ 9 × 10 −36 m p 70 kg m / s


24.13. Solve: (a) For an electron, the momentum p = mv = (9.11 × 10 −31 kg)v . The de Broglie wavelength is λ=

h = 0.20 × 10 −9 m ⇒ 0.20 × 10 −9 m = p

(

6.63 × 10 −34 Js ⇒ v = 3.6 × 10 6 m / s (9.11 × 10 −31 kg)v

)

(b) For a proton, p = mv = 1.67 × 10 −9 kg v . The de Broglie wavelength is

λ=

h = 0.20 × 10 −9 m ⇒ 0.20 × 10 −9 m = p

6.63 × 10 −34 Js ⇒ v = 2.0 × 10 3 m / s (1.67 × 10 −27 kg)v


24.14. Solve: The kinetic energy of a particle of mass m is related to its momentum by E = p 2 2 m . Using λ = h p , for an electron we have

(6.63 × 10 −34 Js) h2 −19 E= = J 2 = 2.4 × 10 2 2 mλ 2(9.11 × 10 −31 kg)(1.0 × 10 −9 m ) 2


24.15. Solve: (a) The baseball’s momentum is p = mv = (0.200 kg)(30 m/s) = 6.0 kg m/s. The baseball’s de Broglie wavelength is

λ=

h 6.63 × 10 −34 Js = = 1.1 × 10 −34 m p 6.0 kg m / s

(b) Using λ = h p = h mv , we have

v=

h 6.63 × 10 −34 Js = = 1.7 × 10 −23 m / s mλ (0.200 kg)(0.20 × 10 −9 m )


24.16. Model: The momentum of a wave-like particle has discrete values given by pn = n(h 2 L) where n = 1, 2, 3, …. Solve: Because we want the smallest box and the momentum of the electron can not exceed a given value, n must be minimum. Thus,

p1 = mv =

h h 6.63 × 10 −34 Js = 0.036 mm ⇒L= = 2L 2 mv 2(9.11 × 10 −31 kg)(10 m / s)


24.17. Model: A confined particle can have only discrete values of energy. Solve:

From Equation 24.14, the energy of a confined electron is

En =

h2 2 n 8mL2

n = 1, 2, 3, 4, …

The minimum energy is

E1 =

h2 ⇒L= 8mL2

h = 8mE1

6.63 × 10 −34 Js

8(9.11 × 10

−31

kg)(1.5 × 10

−18

J)

= 2.0 × 10 −10 m = 0.20 nm


24.18. Model: Model the 5.0-fm-diameter nucleus as a box of length L = 5.0 fm. Solve:

The proton’s energy is restricted to the discrete values

(6.63 × 10 −34 Js) n 2 h2 2 En = n = 8mL2 8(1.67 × 10 −27 kg) 5.0 × 10 −15 m 2

(

(

)

2

= (1.316 × 10 −12 J )n 2

)

where n = 1, 2, 3, … For n = 1, E1 = 1.316 × 10 −12 J , for n = 2, E2 = 1.316 × 10 −12 J 4 = 5.26 × 10 −12 J , and for n = 3, E3 = 9 E1 = 1.18 × 10 −11 J .


24.19. Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve:

(a) The generalized formula of Balmer

λ=

91.18 m  1 − 1  m2 n2 

with m = 1 and n > 1 accounts for a series of spectral lines. This series is called the Lyman series and the first two members are

λ1 =

91.18 m = 121.6 nm 1 − 1   22 

λ2 =

91.18 nm = 102.6 nm 1 − 1   32 

For n = 4 and n = 5, λ 3 = 97.3 nm and λ 4 = 95.0 nm . (b) The Lyman series converges when n → ∞ . This means 1 n 2 → 0 and λ → 91.18 nm. (c) For a diffraction grating, the condition for bright (constructive interference) fringes is d sin θ p = pλ , where p = 1, 2, 3, … For first-order diffraction, this equation simplifies to d sin θ = λ . For the first and second members of the Lyman series, the above condition is d sin θ1 = λ1 = 121.6 nm and d sin θ 2 = λ 2 = 102.6 nm . Dividing these two equations yields

102.6 nm  sin θ 2 =  sin θ1 = (0.84375) sin θ1  121.6 nm  The distance from the center to the first maximum is y = L tan θ . Thus,

tan θ1 =

y1 0.376 m = ⇒ θ1 = 14.072° ⇒ sin θ 2 = (0.84375) sin(14.072°) ⇒ θ 2 = 11.84° L 1.5 m

Applying the position formula once again,

y2 = L tan θ 2 = (1.5 m ) tan(11.84°) = 0.314 m = 31.4 cm


24.20. Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve:

(a) The generalized formula of Balmer

λ=

91.18 m  1 − 1  m2 n2 

with m = 3, and n > 3 accounts for a series of spectral lines. This series is called the Paschen series and the wavelengths are

λ=

91.18 nm 820.62 n 2 = n2 − 9  1 − 1  32 n 2 

The first four members are λ1 = 1876 nm, λ 2 = 1282 nm, λ 3 = 1094 nm, and λ 4 = 1005 nm. (b) The Paschen series converges when n → ∞ . This means

1 91.18 nm →0⇒λ→ = 821 nm 2 n ( 13 )2 (c) For a diffraction grating, the condition for bright (constructive interference) fringes is d sin θ p = pλ , where p = 1, 2, 3, … For first-order diffraction, this equation simplifies to d sin θ = λ . For the first and second members of the Paschen series, the condition is d sin θ1 = λ1 and d sin θ 2 = λ 2 . Dividing these two equations yields

λ  1282 nm  = (0.6834) sin θ1 . sin θ 2 = sin θ1  2  = sin θ1   1876 nm   λ1  The distance from the center to the first maximum is y = L tan θ . Thus,

tan θ1 =

y1 0.607 m = = 0.4047 ⇒ θ1 = 22.03° ⇒ sin θ 2 = (0.6834) sin 22.03° ⇒ θ 2 = 14.85° L 1.5 m

Applying the position formula once again, y2 = L tan θ 2 = (1.5 m ) tan 14.85° = 0.398 m = 39.8 cm


24.21. Model: Use the photon model of light. Solve:

(a) The wavelength is calculated as follows:

(6.63 × 10 −34 Js)(3.0 × 10 8 m / s) = 2.0 × 10 −12 m c Egamma = hf = h  ⇒ λ =  λ 1.0 × 10 −13 J (b) The energy of a visible-light photon of wavelength 500 nm is

(6.63 × 10 −34 Js)(3.0 × 10 8 m / s) = 3.978 × 10 −19 J c Evisible = h  =  λ 500 × 10 −9 m The number of photons n such that Egamma = nEvisible is

n=

Egamma Evisible

=

1.0 × 10 −13 J = 2.51 × 10 5 3.978 × 10 −19 J


24.22. Model: Use the photon model. Solve:

The energy of a 1000 kHz photon is

Ephoton = hf = (6.63 × 10 −34 Js)(1000 × 10 3 Hz) = 6.63 × 10 −28 J The energy transmitted each second is 20 × 10 3 J . The number of photons transmitted each second is 20 × 10 3 J / 6.63 × 10 −28 J = 3.02 × 10 31 .


24.23. Model: Use the photon model for the laser light. Solve:

(a) The energy is

 3 × 10 8 m / s  c −19 Ephoton = hf = h  = (6.63 × 10 −34 Js)  = 3.14 × 10 J  λ  633 × 10 −9 m  (b) The energy emitted each second is 1.0 × 10 −3 J . The number of photons emitted each second is 1.0 × 10 −3 J / 3.14 × 10 −19 J = 3.19 × 1015 .


24.24. Model: Use the photon model for the incandescent light. Solve: The photons travel in all directions. At a distance of r from the light bulb, the photons spread over a sphere of surface area 4πr2. The number of photons per second per unit area at the location of your retina is

3 × 1018 s −1

4π (10 × 103 m )

2

= 2.387 × 10 9 s −1 m −2

The number of photons that enter your pupil per second is 2.387 × 10 9 s −1 m −2 × π (3.0 × 10 −3 m ) = 6.75 × 10 4 s −1 2


24.25. Model: Use the photon model of light and the Bragg condition for diffraction.

Solve: The Bragg condition for the reflection of x-rays from a crystal is 2d cosθ m = mλ . To determine the angles of incidence θm, we need to first calculate λ. The wavelength is related to the photon’s energy as E = hc λ . Thus,

λ=

−34 8 hc (6.63 × 10 Js)(3.0 × 10 m / s) = = 1.326 × 10 −10 m E 1.50 × 10 −15 J

From the Bragg condition,

 1.326 × 10 −10 m )m   mλ  −1 ( θ m = cos −1   = cos −1 (0.3157m) ⇒ θ1 = cos −1 (0.3157) = 71.6°  = cos  −9  2d  . × 2 0 21 10 m ( )   −1 Likewise, θ 2 = cos (0.3157 × 2) = 50.8° and θ 3 = 18.7° . Note that θ 4 = cos −1 (0.3157 × 4) is not allowed because the cosθ cannot be larger than 1. Thus, the x-rays will be diffracted at angles of incidence equal to 18.7°, 50.8°, and 71.6°.


24.26. Model: The angles for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: We cannot assume that these are the first and second order diffractions. The Bragg condition is 2d cosθ m = mλ . We have

2 d cos 45.6° = mλ

2 d cos 21.0° = ( m + 1)λ

Notice that θm decreases as m increases, so 21.6° corresponds to the larger value of m. Dividing these two equations,

cos 45.6° m = = 0.7494 ⇒ m = 3 cos 21.0° m + 1 Thus these are the third and fourth order diffractions. Substituting into the Bragg condition, 3 × 0.070 × 10 −9 m d= = 1.50 × 10 −10 m = 0.150 nm 2 cos 45.6°


24.27. Model: The x-ray diffraction angles satisfy the Bragg condition.

Solve: (a) The Bragg condition ( 2d cosθ m = mλ ) for normal incidence, θm = 0°, simplifies to 2d = mλ. For a thin film of a material on a substrate where nair < n material < nsubstrate, constructive interference between the two reflected waves occurs when 2d = mλ, where λ is the wavelength inside the material. (b) From a thin film with a period of 1.2 nm, that is, with d = 1.2 nm, the two longest x-ray wavelengths that will reflect at normal incidence are 2d 2d λ1 = λ2 = 1 2 This means that λ1 = 2(1.2 nm ) = 2.4 nm and λ 2 = 1.2 nm .


24.28. Solve: A small fraction of the light wave of an appropriate wavelength is reflected from each little “bump” in the refractive index. These little bumps act like the atomic planes in a crystal. The light will be strongly reflected (and hence blocked in transmission) if it satisfies the Bragg condition at normal incidence (θ = 0).

2d = mλ glass =

2 dnglass 2(0.45 × 10 −6 m )(1.50) mλ ⇒λ= = = 1.35 µm nglass 1 m


24.29. Model: Particles have a de Broglie wavelength given by λ = h p . The wave nature of the particles causes an interference pattern in a double-slit apparatus. Solve: (a) Since the speed of the neutron and electron are the same, the neutron’s momentum is m m m p n = m n v n = n m e v n = n m e v e = n pe me me me where mn and me are the neutron’s and electron’s masses. The de Broglie wavelength for the neutron is m h h pe λn = = = λe e pn pe p n mn From Section 22.2 on double-slit interference, the fringe spacing is ∆y = λL / d . Thus, the fringe spacing for the electron and neutron are related by

∆yn =

 9.11 × 10 −31 kg  m λn −3 −7 ∆ye = e ∆ye =   (1.5 × 10 m ) = 8.18 × 10 m = 0.818 µm λe mn  1.67 × 10 −27 kg 

(b) If the fringe spacing has to be the same for the neutrons and the electrons,

∆ye = ∆yn ⇒ λ e = λ n ⇒

 9.11 × 10 −31 kg  m h h 3 = ⇒ vn = ve e = (2.0 × 10 6 m / s)  = 1.09 × 10 m / s me ve m n v n mn  1.67 × 10 −27 kg 


24.30. Model: Electrons have a de Broglie wavelength given by λ = h p . The wave nature of the electrons causes a diffraction pattern. Solve: The width of the central maximum of a single-slit diffraction pattern is given by Equation 22.22:

w=

2(1.0 m )(6.63 × 10 −34 Js) 2λL 2 Lh 2 Lh = = = = 9.70 × 10 −4 m = 0.970 mm −6 −31 6 a ap amv × × × . m . kg . m / s 1 0 10 9 11 10 1 5 10 )( ( )( )


24.31. Model: Neutrons have a de Broglie wavelength given by λ = h p . The wave nature of the neutrons causes a double-slit interference pattern. Solve: Measurements show that the spacing between the m = 1 and m = –1 peaks is 1.4 times as long as the length of the reference bar, which gives the real fringe separation ∆y = 70 µm. Similarly, the spacing between the m = 2 and m = –2 is 2.8 times as long as the length of the reference bar and yields ∆y = 70 µm. The fringe separation in a double-slit experiment is ∆y = λL d . Hence,

λ=

(6.63 × 10 −34 Js)(3.0 m) ∆y d h h ∆y d hL ⇒ = = ⇒v= = = 170 m / s L p mv L ∆y md (70 × 10 −6 m )(1.67 × 10 −27 kg)(0.10 × 10 −3 m )


24.32. Model: Electrons have a de Broglie wavelength given by λ = h p .

Visualize: Please refer to Figure 24.11. Notice that a scattering angle φ = 60° corresponds to an angle of incidence θ = 30° . Solve: Equation 24.6 describes the Davisson-Germer experiment: D sin(2θ m ) = mλ . Assuming m = 1, this equation simplifies to Dsin 2θ = λ . Using λ = h mv , we have

D=

h 6.63 × 10 −34 Js = = 1.95 × 10 −10 m = 0.195 nm mv sin 2θ (9.11 × 10 −31 kg)( 4.30 × 10 6 m / s) sin(60°)


24.33. Model: A confined particle can have only discrete values of energy. Solve:

(a) Equation 24.14 simplifies to

(6.63 × 10 −34 Js) h2 2 −19 En = n = J )n 2 2 = (1.231 × 10 2 8mL 8(9.11 × 10 −31 kg)(0.70 × 10 −9 m ) 2

(

)( )

(

)( )

Thus, E1 = 1.231 × 10 −19 J 12 = 1.23 × 10 −19 J , E2 = 1.231 × 10 −19 J 2 2 = 4.92 × 10 −19 J , and E3 = 1.11 × 10 −18 J. (b) The energy is E2 − E1 = 4.92 × 10 −19 J − 1.23 × 10 −19 J = 3.69 × 10 −19 J . (c) Because energy is conserved, the photon will carry an energy of E2 − E1 = 3.69 × 10 −19 J . That is,

E2 − E1 = Ephoton = hf =

(6.63 × 10 −34 Js)(3.0 × 10 8 m / s) = 539 nm hc hc ⇒λ= = 3.69 × 10 −19 J E2 − E1 λ


24.34. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:

(a) Equation 24.14 for the n = 1 state is

(6.63 × 10 −34 Js) = 5.50 × 10 −64 J h2 En = = 8mL2 8(10 × 10 −3 kg)(0.10 m ) 2 The minimum energy of the Ping-Pong ball is E1 = 5.50 × 10 −64 J . (b) The speed is calculated as follows: 2

E1 = 5.50 × 10 −64 J =

1 2

mv 2 =

1 2

(10 × 10 −3 kg)v 2 ⇒ v =

2(5.50 × 10 −64 J ) 10 × 10 −3 kg

= 3.32 × 10 −31 m / s


24.35. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:

Using Equation 24.14 for n = 1 and 2,

(6.63 × 10 −34 Js) (3) = 1.809 × 10 −37 Jm 2 h2 −19 2 2 ⇒ 1 . 0 × 10 J = E2 − E1 = 2 − 1 ( ) L2 8mL2 8(9.11 × 10 −31 kg) L2 2

⇒L=

1.809 × 10 −37 Jm 2 = 1.35 × 10 −9 m = 1.35 nm 1.0 × 10 −19 J


24.36. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:

From Equation 24.14,

En =

h2 2 n 8mL2

h2 2 n + 1) 2 ( 8mL

En+1 =

Taking the ratio,

En+1  n + 1 =  n  En

2

Thus

n +1 = n

6.4 4 = 1.333 = 3.6 3

The two allowed energies have n = 3 and n = 4. Using n = 3 in the first energy equation,

En = 3.6 × 10

−19

J =

(6.63 × 10 8(9.11 × 10

−34

−31

Js)

2

kg) L2

(3)2 ⇒ L = 1.23 nm


24.37. Model: A particle confined in a one-dimensional box of length L has the discrete energy levels given by Equation 24.14. Solve: (a) Since the energy is entirely kinetic energy,

En =

h2 2 p2 h n = = 12 mvn2 ⇒ vn = n 8mL2 2m 2 mL

n = 1, 2, 3,K

(b) The first allowed velocity is

v1 =

6.63 × 10 −34 Js = 1.819 × 10 6 m / s 2(9.11 × 10 −31 kg)(0.20 × 10 −9 m )

For n = 2 and n = 3, v2 = 3.64 × 106 m/s and v3 = 5.46 × 106 m/s.


24.38. Model: The allowed energies of a particle of mass m in a two-dimensional square box of side L are Enm = Solve:

h2 (n 2 + m 2 ) 8mL2

(a) The minimum energy for a particle is for n = m = 1:

Emin = E11 =

h2 h2 2 2 + = 1 1 ( ) 8mL2 4 mL2

(b) The five lowest allowed energies are Emin , 25 Emin (for n = 1, m = 2 and n = 2, m = 1), 4Emin (for n = 2, m = 2), 5Emin (for n = 1, m = 3 and n =3, m = 1), and 132 Emin (for n = 2, m = 3 and n = 3, m = 2).


24.39. Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure 24.7. Solve: The Bragg diffraction condition is 2d cosθ m = mλ . The plane spacing is dA = 0.20 nm and the x-ray wavelength is λ = 0.12 nm. Thus

cosθ m =

m(0.12 × 10 −9 m ) mλ = (0.3)m ⇒ θ A1 = cos −1 (0.3) = 72.5° = 2 dA 2(0.20 × 10 −9 m )

Likewise for m = 2 and m = 3, θ A2 = cos −1 (0.6) = 53.1° and θ A3 = cos −1 (0.9) = 25.8° . These three angles for the x-ray diffraction peaks match the peaks shown in Figure 24.7c. (b) The new interplaner spacing is dB = d A 2 = 0.141 nm (see Figure 24.7b). The Bragg condition for the tilted atomic planes becomes mλ = 0.4243m cosθ m = 2 dB For m = 1, θ B1 = cos −1 (0.4243) = 64.9° . For m = 2, θ B2 = cos −1 (0.8486) = 31.9° . (c) The crystal is already tipped by 45° to get the tilted planes (see Figure 24.7b). So, for m = 1, θ1 = 64.9° − 45° = 19.9° . θ = 64.9° + 45° = 109.9° also, but we can’t see beyond 90°. For m = 2, θ 2 = 31.9° + 45° = 76.9° . These two angles match the angles in the diffraction peaks of the tilted planes.


24.40. Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure CP24.40. Solve: The Bragg diffraction condition is 2d cosθ m = mλ , where d is the interplanar separation. Because smaller m values correspond to higher angles of incidence, the diffraction angle of 71.3° in the x-ray intensity plot must correspond to m = 1. This means

2 d cos 71.3° = 1(0.10 × 10 −9 m ) ⇒ d =

0.10 × 10 −9 m = 1.56 × 10 −10 m 2(cos 71.3°)

The cosines of the three angles in the X-ray intensity plot are cos 71.3° = 0.321, cos 50.1° = 0.642, and cos15.8° = 0.962. These are in the ratio 1:2:3, which tells us that these are the m = 1, 2, and 3 diffraction peaks from a single set of planes with d = 0.156 nm. We can see from the figure that the atomic spacing D of this crystal is related to the interplanar separation d by

D=

d 0.156 nm = = 0.180 nm sin 60° sin60°


24.41. Model: Electrons have a de Broglie wavelength given by λ = h p . Trapped electrons in the confinement layer behave like a de Broglie wave in a closed-closed tube or like a string fixed at both ends. Solve: (a) The four longest standing-wave wavelengths in the layer are λ = 2L , L, 23 L , and 12 L . This follows from the general relation for closed-closed tubes: λ = 2L / n . Thus, λ = 10.0 nm, 5.0 nm, 3.33 nm, and 2.50 nm. (b) We have

p = mv =

h h 6.63 × 10 −34 Js 0.7278 × 10 −3 m 2 / s ⇒v= = = −31 mλ (9.11 × 10 kg)λ λ λ

Using the above four longest values of λ we get the four smallest values of v. Thus,

v1 =

0.7278 × 10 −3 m 2 / s = 7.28 × 10 4 m / s 10.0 × 10 −9 m

v2 = 1.46 × 10 5 m / s , v3 = 2.18 × 10 5 m / s , and v4 = 2.91 × 10 5 m / s .


24.42. Model: As light is diffracted by matter, matter can also be diffracted by light. Visualize: Please refer to Figure CP24.42. Solve: The de Broglie wavelength of the sodium atoms is

λ=

h h 6.63 × 10 −34 Js = = = 3.45 × 10 −10 m p mv (3.84 × 10 −26 kg)(50 m / s)

The slit spacing of the “diffraction grating” is d = 12 λ laser = 12 600 nm = 300 nm . Using the diffraction grating equation with m = 1, we have

λ = 1.151 × 10 −3 rad d In the small-angle approximation, sin θ ≅ tan θ = y L . We get d sin θ = (1)λ ⇒ sin θ =

y = L sin θ = (1.0 m )(1.151 × 10 −3 ) = 1.15 mm

Chapter 24  

where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, λ =389.0 nm and λ =379.9 nm . 1 6 24.1. Model: Balmer’s...

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