Foundations for Nanoscience and Nanotechnology 1st Edition by Nils Petersen ISBN 9781482259100

Nils O. Petersen

Solutions Manual for Foundations for Nanoscience and Nanotechnology

v

Dedicated to those instructors who want to use problems as one of the pedagogical tools in education.

vi This Solution Manual is created for instructors and provides the answers to all the questions posed in the various Chapters in the Foundations for Nanoscience and Nanotechnology.

The questions are deliberately of varying levels of difficulty and structure.

This is in the spirit of the text, which is aimed at students who want to understand better the world of nanoscience and nanotechnology, but do not necessarily have the mathematics or physics or chemistry background to delve into great detail. Hence, some of the questions are designed to test the memory of concepts and are fairly simple with access to the text, but could be a test question. Others questions are designed to test the understanding of the subject matter, so they may be appropriate as a question to address collectively, or to incorporate into an assignment or a test.

Recognizing that some students may have stronger mathematical skills and want to understand or explore some of the details of derivations, a final category of questions (annotated with a star) are designed to explore the derivation of key equations in the text.

CHAPTER

1

Chapter 1 Problems

Problem 1.1 What are some common features of the various definitions of of nanomaterials?

Answer: All the definitions reference a range of sizes - typically 1 to 100 nm. Most definitions reference that there could be substructures within the material at that scale. Most definitions reference that there must be special properties of the material.

Problem 1.2 Consult the web-site for the Emerging Nanotechnologies project (www.nanotechproject.org) and discuss some of the most frequent applications of nanotechnology in consumer products and speculate on why these are so common.

Answer: The answer will depend on when the web-site is consulted as it is frequently updated. Nevertheless, some of the most common applications use nano-silver in various forms, particularly for its anti-bacterial properties. Many other applications use common, and previously known materials, such as oxides of titanium or other metals, partly for their catalytic properties and partly for their sun-blocking activities, as in cosmetics. A common feature of many of the early adopters is that these are relatively cheap materials already in commercial use in some other form.

1

CHAPTER

2

Chapter 2 Problems

Problem 2.1 Without consulting Table 2.1, define each of the terms in the Time-Dependent Schrödinger Equation [ ] ~2 2 ~ ∂ ĤΨ(x, y, z; t) = − ∇ + V̂ (x, y, z) Ψ(x, y, z; t) = − Ψ(x, y, z; t).

2m i ∂t Answer: Ĥ is the Hamiltonian operator, which is the operator for the total energy of the system. Ψ(x, y, z; t) is the total wave function that describes the states of the system. ~2 − 2m ∇2 is the kinetic energy operator of the system.

V̂ (x, y, z) is the potential energy operator of the system. ∂ − ~i ∂t Ψ(x, y, z; t) is the time-derivative of the wave function.

This form of the Hamiltonian does not include many terms that may arise from other interactions such as spin, magnetism, etc.

Problem 2.2 Using the operators in Table 2.1, calculate the result of the following operations: [ ( )] 1. x̂ ax2 + b = ax3 + bx. [ ( )] ( 2 ) ∂ 2. p̂ ax2 + b = ~i ∂x ax + b = ~i 2ax.

3. [x̂ (e−ax )] = xe−ax . ∂ 4. [p̂ (e−ax )] = ~i ∂x (e−ax ) = − ~i ae−ax . [ ] ∂2 2 2 5. p̂2 (cos(ax)) = −~2 ∂x 2 (cos(ax)) = ~ a cos(ax).

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4 Solutions Manual for Foundations for Nanoscience and Nanotechnology

Problem 2.3 Use any test function, such as one of those in Problem 2.2, to show that the position operator and the momentum operator do not commute.

Answer: For the position and momentum operators to commute we require that [x̂, p̂] = 0, or that (x̂p̂ − p̂x̂) f (x) = 0.

Selecting f (x) = ax2 + b we can evaluate each pair of operations to get 2 x̂p̂(ax2 + b) = x ~i 2ax = 2a~ i x ,

and ∂ ~ 2 p̂x̂(ax2 + b) = ~i ∂x (ax3 + bx) = 3a~ i x + i x.

CHAPTER

3

Chapter 3 Problems

Problem 3.1 Show that for conservative forces, the wave function in Equation 3.3 satisfies the TDSE leading to Equation 3.5, the TISE. Answer: Given the wave function Ψ(x, y, z; t) = ψ(x, y, z)e−iωt then ~ ∂ ψ(x, y, z)e−iωt i ∂t ( ) ~ ∂ −iωt = ψ(x, y, z) − e i ∂t ( ) ) ~ ( = ψ(x, y, z) − −iωe−iωt i

Ĥψ(x, y, z)e−iωt = −

= ψ(x, y, z)~ωe−iωt .

Eliminating the common term e−iωt and recognizing that E = ~ω we get Ĥψ(x, y, z) = Eψ(x, y, z) q.e.d.

*Problem 3.2 Show that for conservative forces, Equation 3.1 can be separated into two differential equations (Equation 3.2 and 3.5) where the constant is the energy, E. Answer: Assume Ψ(x, y, z; t) = f (x, y, z) · g(t), then [ 2 ] ~ ~ ∂ ∇2 + V (r) f (x, y, z) · g(t) = − f (x, y, z) · g(t), 2m i ∂t or [ 2 ] ~ ~ ∂ ∇2 + V (r) f (x, y, z) · g(t) = − f (x, y, z) g(t).

2m i ∂t Dividing by Ψ = f (x, y, z)g(t) gives [ 2 ] 1 ~ ~ 1 ∂ 2 ∇ + V (r) f (x, y, z) = − g(t). f (x, y, z) 2m i g(t) ∂t 5

6 Solutions Manual for Foundations for Nanoscience and Nanotechnology We see here that the left-hand side of the equation is only dependent on spatial coordinates while the right-hand side is only dependent of time. Since the coordinates and time are independent variables, the equation must hold for all values of both variables, which can only be true if both sides are equal to the same constant. Let us call this constant C. We get from the left-hand side that [ 2 ] ~ ∇2 + V (r) f (x, y, z) = Ĥf (x, y, z) = C · f (x, y, z).

2m Correspondingly, we get from the right-hand side that 1 ∂ iC g(t) = − , g(t) ∂t ~ which, when solved gives lng(t) = −

iCt ~

⇒

g(t) = e−iCt/~ .

It follows that if the constant is the energy, E, then C/~ = E/~ = ω, and

Ĥf (x, y, z) =Ef (x, y, z) g(t) =e−iωt Ψ(x, y, z; t) =f (x, y, z)e−iωt , which is what we wanted to show.

(3.1)

CHAPTER

4

Chapter 4 Problems

Problem 4.1 Show that ψ(x) in Equation 4.3 satisfies the TISE in Equation 4.2.

Answer: Given that ψ(x) = (A+B)coskx+i(A−B)sinkx we get, when substituting into Equation 4.2, that ∂2 [(A + B)coskx + i(A − B)sinkx] +k 2 [(A + B)coskx + i(A − B)sinkx] = ∂x2 ∂ [−(A + B)ksinkx + i(A − B)kcoskx] +k 2 [(A + B)coskx + i(A − B)sinkx] = ∂x[ ] −(A + B)k 2 coskx − i(A − B)k 2 sink +k 2 [(A + B)coskx + i(A − B)sinkx] = 0, which demonstrates that this wave function is a solution to the TISE.

Problem 4.2 Show that ψ(x) in Equation 4.3 is NOT an eigenfunction of the momentum operator.

Answer: ∂ The momentum operator is p̂ = − ~i ∂x , which gives, when it operates on the wave function in Equation 4.3, ~ ∂ [(A + B)coskx + i(A − B)sinkx] i ∂x ~ = − [−(A + B)ksinkx + i(A − B)kcoskx] i ~ = k [(A + B)sinkx − i(A − B)coskx] i ̸= c [(A + B)coskx + i(A − B)sinkx] ,

p̂ψ(x) = −

which shows that the wave function is not an eigenfunction of the momentum operator.

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8 Solutions Manual for Foundations for Nanoscience and Nanotechnology

Problem 4.3 Solve the TISE for V (x) = V0 for all values of x to show that √ the momentum for a particle moving along the positive x-axis is p = 2mp (E − V0 ) with E > V0 .

Answer: When V (x) = V0 the TISE becomes

[

Ĥψ(x) = Eψ(x) ⇒ ] ~ ∂ − + V0 ψ(x) = Eψ(x) ⇒ 2mp ∂x2 ∂2 2mp (E − V0 ) ψ(x) = ψ(x) 2 ∂x ~2 ∂2 ψ(x) = k 2 ψ(x). ∂x2 2

2

⇒

This is the√same wave equation we solved before, only with a different value for k = 2mp (E − V0 )/~. Thus we get the same solutions for the wave functions, namely ψ(x) = (A + B)coskx + i(A − B)sinkx.

The momentum along the positive direction is p = ~k = when (E > V0 ).

√ 2mp (E − V0 ),

*Problem 4.4 Use a spreadsheet or computer programming language (e.g.

Excel, Origin, Sigmaplot, or MATLAB) to reproduce Figure 4.3.

Answer: Follow these steps if you use Excel or a similar spreadsheet: 1. Create a column of position values ranging from -25 to +25 in increments of 0.1 in Column A. This is easily done by entering the increment value at the top of the column in Cell A1, entering the value of -25 in the next cell at A2, then entering the value of A2+$A$1 in cell A3, finally copying Cell A3 and pasting it into the cells below it till you get the value of 25.

2. Create a set of values in the top row (cells B1 to L1) ranging from 0.5 to 1.5 in increments of 0.1. These will be used to modify the wavelength of the cosine function and represent the value of ”k” in the wave function.

3. Enter in the second row (cells B2 to L2) the functions ”= cos(A2*$B$1)”, ”= cos(A2*$C$1)”, ”= cos(A2*$D$1)” etc. This will calculate the cosine of the x-value in Cell A2 modified by the value in the top row to give cosine functions of different wavelengths.

4. Copy and paste the second row cells all the way through all the rows with values in Column 1.

Chapter 4 Problems 9 5. Create a column for the sum of each of the rows from B to L, this will be the wave packet function.

6. Plot these functions (to reproduce Figure 4.3 you will have to add an offset to each of the columns from B to L to ensure that they are not plotted on top of each other).

*Problem 4.5 Use the program in Problem 4.4 to show the effect of changing the uncertainty in k values.

Answer: To do this, vary the range of values in the top row in columns B through L. For example decreasing the range to 0.75 to 1.25 in steps of 0.05 will be like decreasing the uncertainty in k, and conversely, increasing the range to 0 to 2 in steps of 0.2 will be like increasing the uncertainty in k.

*Problem 4.6 Show that Equation 4.21 follows from Equations 4.17 and 4.20.

Answer: Equation 4.17 is the general solution to the wave equation ψ(x) = Aeikx + Be−ikx , and Equation 4.20 simply states that A = −B. Thus [ ] ψ(x) = Aeikx − Ae−ikx = A eikx − e−ikx , but

e±ikx = cos(kx) ± isin(kx),

so, ψ(x) = A [(cos(kx) + isin(kx)) − (cos(kx) − isin(kx))] = 2Aisin(kx). √

Problem 4.7 Use Equations 4.21, 4.22 and 4.23 to show that A = 1/(i 2b).

Answer: Equation 4.21 shows us that ψ(x) = 2Aisin(kx), and Equation 4.21 tells us that the wave function is zero when b = 0.

In order to normalize, we require from Postulate I that ∫ ∞ ψ ∗ (x)ψ(x)dx = 1. −∞

We substitute the wave function for the box, and recognize that the wave

10 Solutions Manual for Foundations for Nanoscience and Nanotechnology function is zero for all values of x outside the box. Hence, we only need to integrate from 0 to b, and ∫ b

(2A∗ (−i)sin(kx)) (2Aisin(kx)) dx = 1

0

⇒

∫ b

∗

sin2 (kx)dx =1.

4A A 0

nπ We substitute k = nπ b and use y = b x, such that y = 0 when x = 0, b y = nπ when x = b, and dx = nπ dy. Then the integral becomes

4A∗ A ∗

∫ b

∫ b

sin2 (kx)dx =1

⇒

nπ x)dx =1 b

⇒

0

sin2 (

4A A

∫ nπ

0

b sin2 (y)dy =1 ⇒ 4A∗ A nπ 0 [ ]nπ b 1 1 4A∗ A y − sin(2y) =1 ⇒ nπ 2 4 0 [ ] 1 1 b nπ − 0 − sin(2nπ) + sin(0) =1 ⇒ 4A∗ A nπ 2 4 4 b 4A∗ A =1 ⇒ 2 1 A∗ A = , 2b which is satisfied when A = i√12b .

Problem 4.8 Show that substituting ψ(x) given by Equation 4.24 into the TISE yields the energy given in Equation 4.25.

Answer: √ Given the wave function ψ(x) = 2b sin( nπ b x) we can substitute into the left-hand side of the TISE to get:

Chapter 4 Problems 11

~2 ∂ 2 ψ(x) = 2mp ∂x2 ] [√ ( nπ ) ~2 ∂ 2 2 − sin x = 2mp ∂x2 b b [√ ] ( nπ ) ~2 nπ ∂ 2 − cos x = 2mp b ∂x b b [√ ] ( nπ )) ~ 2 n2 π 2 2( − −sin x = 2mp b2 b b −

+

~2 n2 π 2 ψ(x) = Eψ(x).

2mp b2

It follows that E=

~2 n2 π 2 h2 n2 = , 2mp b2 8mp b2

since h = 2π~.

Problem 4.9 Show that ∆En = (2n + 1)h2 /(8mp b2 ).

Answer: We define ∆En = En+1 − En so, 2

(n + 1) h2 n2 h2 − 8mp b2 8mp b2 ( ) h2 = n2 + 2n + 1 − n2 8mp b2 2 h = (2n + 1) .

8mp b2

∆En =

q.e.d.

Problem 4.10 Solve Equation 4.26 using Equation 4.27 to get Equations 4.28 and 4.29.

Answer: Given the wave function, ψ(x, y) can be expressed as the product of two independent wave functions, ϕ(x) and ϕ(y), i.e, ψ(x, y) = ϕ(x)ϕ(y), then the TISE becomes: [ 2 ] ∂ ∂2 ~2 + 2 ϕ(x)ϕ(y) = Eϕ(x)ϕ(y), − 2mp ∂x2 ∂y

12 Solutions Manual for Foundations for Nanoscience and Nanotechnology or − or −

[ 2 ] ~2 ∂ ∂2 ϕ(x)ϕ(y) + ϕ(x)ϕ(y) = Eϕ(x)ϕ(y), 2mp ∂x2 ∂y 2

~2 ∂2 ~2 ∂2 ϕ(y) 2 ϕ(x) − ϕ(x) 2 ϕ(y) = Eϕ(x)ϕ(y).

2mp ∂x 2mp ∂y

Dividing by the wave function on both sides yields −

~2 1 ∂ 2 ~2 1 ∂ 2 ϕ(x) − ϕ(y) = E. 2mp ϕ(x) ∂x2 2mp ϕ(y) ∂y 2

We recognize that the first term only depends on the variable x and that the second term only depends on the variable y and that the sum is equal to a constant. Since this must be true for all values of x and y, each term on the left-hand side must equal a constant, and each become a representation of the TISE in the corresponding variable. We therefore conclude that ~2 ∂ 2 ϕ(x) =Ex ϕ(x) 2mp ∂x2 ~2 ∂ 2 − ϕ(y) =Ey ϕ(y).

2mp ∂y 2

−

Each of these can be solved independently as a one-dimensional particlein-a-box problem and the wave functions will become those in Equations 4.28 and 4.29, respectively.

*Problem 4.11 Show that when the wave functions multiply the energies add.

Answer: From Problem 4.10 we saw that [ 2 ] ~2 ∂ ∂2 − ϕ(x)ϕ(y) + 2 ϕ(x)ϕ(y) = Eϕ(x)ϕ(y), 2mp ∂x2 ∂y and that ~2 ∂ 2 ϕ(x) =Ex ϕ(x) 2mp ∂x2 ~2 ∂ 2 − ϕ(y) =Ey ϕ(y).

2mp ∂y 2

−

Substituting the latter two equations into the first, gives [Ex ϕ(x)ϕ(y) + Ey ϕ(x)ϕ(y)] = Eϕ(x)ϕ(y), which shows that E = Ex + Ey , i.e, the energies add when the wave functions multiply.

Chapter 4 Problems 13

Problem 4.12 Show that E3,4 = E4,3 for a two-dimensional square box with infinite walls.

Answer: From Equation 4.33 we know that for a two-dimensional square box with infinite walls, Em,n = (m2 + n2 )

h2 , 8mp b2 2

thus in general Em,n = En,m , and specifically E3,4 = E4,3 = 25 8mhp b2 .

Problem 4.13 Describe the concept of degeneracy and give an example for a 2-D box with infinite walls.

Answer: Degeneracy arises when two states characterized by diffent quantum numbers have the same energy. For a 2-D box with infinite walls, that can occur whenever the box is square, since then we know that Em,n = (m2 + n2 )

h2 .

8mp b2

This means that Em,n = En,m .

Problem 4.14 What is the maximum number of degenerate states in a 3-D cubic box?

Answer: The energy for the three-dimensional cubic box is ( ) Emnp = m2 + n2 + p2

h2 , 8mp b2

which shows that there is a maximum of six possible degenerate states for a particular set of quantum numbers in which m ̸= n ̸= p, that is, Emnp = Empn = Enmp = Enpm = Epmn = Epnm .

*Problem 4.15 Show in detail that for 1. cylindrical coordinates, ∂ ∂ sinϕ ∂ = cosϕ − ; ∂x ∂r r ∂ϕ 2. spherical coordinates, ∂ ∂ cosθcosϕ ∂ sinϕ ∂ = sinθcosϕ + − . ∂x ∂r r ∂θ rsinθ ∂ϕ

14 Solutions Manual for Foundations for Nanoscience and Nanotechnology Answer: For cylindrical coordinates we know that x = rcosϕ,

y = rsinϕ,

and r2 = x2 + y 2 ,

ϕ = arccot

z = z,

( ) (y) x = arctan y x

We also know from the chain rule that ∂f ∂ϕ ∂f ∂z ∂f ∂f ∂r = + + ∂x ∂r ∂x ∂ϕ ∂x ∂z ∂x ∂f ∂f ∂r ∂f ∂ϕ ∂f ∂z = + + ∂y ∂r ∂y ∂ϕ ∂y ∂z ∂y ∂f ∂f ∂r ∂f ∂ϕ ∂f ∂z = + + , ∂z ∂r ∂z ∂ϕ ∂z ∂z ∂z which can be expressed in matrix format as  ∂f 

 ∂r

∂x

∂x

    ∂f   ∂r  =  ∂y   ∂y    ∂f ∂z

∂r ∂z

∂ϕ ∂x

∂z   ∂f  ∂x ∂r

∂ϕ ∂y

∂z   ∂f  ∂y   ∂ϕ  .

∂ϕ ∂z

∂z ∂z

 

 





∂f ∂z

The task now is to find the components of the matrix.

We calculate ) ∂r ∂r ∂ ( 2 ∂r2 x + y 2 = 2x = 2rcosϕ ⇒ = 2r = = cosϕ ∂x ∂x ∂x ∂x ) ∂r2 ∂r ∂ ( 2 ∂r = 2r = x + y 2 = 2y = 2rsinϕ ⇒ = sinϕ ∂y ∂y ∂y ∂y ∂r = 0. ∂z Similarly, we calculate

Chapter 4 Problems 15 ( ) ∂ x 1 ∂ 1 −1 ∂ϕ = arccot = arccot (v) = ∂x ∂x y y ∂v y 1 + v2 1 −1 −y −rsinϕ −sinϕ = = 2 = = y 1 + xy22 y + x2 r2 r (y) 1 ∂ ∂ 1 1 ∂ϕ = arctan = arctan (w) = ∂y ∂y x x ∂w x 1 + w2 1 1 x rcosϕ cosϕ = = = 2 = 2 2 2 y x1+ 2 x +y r r x

∂ϕ = 0. ∂z The matrix then becomes  ∂f 



∂x

−sinϕ r

cosϕ

    ∂f     = sinϕ  ∂y      ∂f 0 ∂z

cosϕ r

0

  ∂f  0 ∂r     ∂f    0   ∂ϕ  .   ∂f 1 ∂z

It follows that ∂ ∂ sinϕ ∂ = cosϕ − ∂x ∂r r ∂ϕ ∂ ∂ cosϕ ∂ = sinϕ + ∂y ∂r r ∂ϕ ∂ ∂ = . ∂z ∂z Correspondingly for spherical coordinates: x = rsinθcosϕ, and 2

2

2

y = rsinθsinϕ,

( ) (y) x = arctan , ϕ = arccot y x

2

r =x +y +z ,

The matrix we need to find is now  ∂f 

 ∂r

∂x

∂x

    ∂f   ∂r  =  ∂y   ∂y    ∂f ∂z

∂r ∂z

z = rcosθ,

∂θ ∂x

∂ϕ   ∂f  ∂x ∂r

∂θ ∂y

∂ϕ   ∂f  ∂y   ∂θ  .

∂θ ∂z

∂ϕ ∂z

 

 





∂f ∂ϕ

θ = arccos

(z ) r

.

16 Solutions Manual for Foundations for Nanoscience and Nanotechnology

We calculate ) ∂r ∂ ( 2 ∂r ∂r2 = 2r = x + y 2 + z 2 = 2x = 2rsinθcosϕ ⇒ = sinθcosϕ ∂x ∂x ∂x ∂x 2 ) ∂r ∂r ∂ ( 2 ∂r = 2r = x + y 2 + z 2 = 2y = 2rsinθsinϕ ⇒ = sinθsinϕ ∂y ∂y ∂y ∂y ) ∂r ∂ ( 2 ∂r ∂r2 = 2r = x + y 2 + z 2 = 2z = 2rcosθ ⇒ = cosθ. ∂z ∂z ∂y ∂z Similarly, we calculate ( ) ∂ϕ ∂ x 1 ∂ 1 −1 = arccot = arccot (v) = ∂x ∂x y y ∂v y 1 + v2 1 −1 −y −rsinθsinϕ = = 2 = 2 2 y 1 + xy22 y + x2 r sin θsin2 ϕ + r2 sin2 θcos2 ϕ −rsinθsinϕ −sinϕ = r2 sin2 θ (sin2 ϕ + cos2 ϕ) rsinθ (y) 1 ∂ ∂ϕ ∂ 1 1 = arctan = arctan (w) = ∂y ∂y x x ∂w x 1 + w2 1 1 x +rsinθcosϕ = = 2 = 2 2 x 1 + y22 x + y2 r sin θcos2 ϕ + r2 sin2 θsin2 ϕ =

x

=

rsinθcosϕ r2 sin2 θ (cos2 ϕ + sin2 ϕ)

(y) ∂ϕ ∂ = arctan = 0. ∂z ∂z x Moreover, since cosθ = zr

=

cosϕ rsinθ

Chapter 4 Problems 17

∂cosθ ∂θ ∂ z ∂ 1 −1 ∂r (−rcosθ) (sinθcosϕ) = −sinθ = =z =z 2 = ⇒ ∂x ∂x ∂x r ∂x r r ∂x r2 ∂θ cosθcosϕ = ∂x r ∂cosθ ∂θ ∂ z ∂ 1 −1 ∂r (−rcosθ) (sinθsinϕ) = −sinθ = =z =z 2 = ⇒ ∂y ∂y ∂y r ∂y r r ∂y r2 cosθsinϕ ∂θ = ∂y r ∂cosθ ∂θ ∂ z 1 ∂z ∂ 1 1 −1 ∂r = −sinθ = = +z = +z 2 ∂z ∂z ∂z r r ∂z ∂z r r r ∂z ) sin2 θ 1 (rcosθ) (cosθ) 1( 2 = − = 1 − cos θ = ⇒ r r2 r r ∂θ sinθ =− . ∂z r The matrix then becomes  ∂f  ∂x



sinθcosϕ

    ∂f     = sinθsinϕ  ∂y      ∂f ∂z

cosθ

cosθcosϕ r

−sinϕ   ∂f  ∂r rsinθ

cosθsinϕ r

cosϕ   ∂f  . rsinθ   ∂θ 

−sinθ r

 

 



0



∂f ∂ϕ

It follows that ∂ cosθcosϕ ∂ sinϕ ∂ ∂ = sinθcosϕ + − ∂x ∂r r ∂θ rsinθ ∂ϕ ∂ ∂ cosθsinϕ ∂ cosϕ ∂ = sinθsinϕ + + ∂y ∂r r ∂θ rsinθ ∂ϕ ∂ ∂ sinθ ∂f = cosθ − . ∂z ∂r r ∂θ

*Problem 4.16 Show in detail that for spherical coordinates ( ∇2 =

1 ∂ r2 ∂r

( ) ( ) ) ∂ 1 ∂ ∂ 1 ∂2 r2 + 2 sinθ + 2 . ∂r r sinθ ∂θ ∂θ r sinθ ∂ϕ2

Answer: This problem is truly only for the most enthusiastic. It demands great attention to detail. The full solution is available in several places. One useful approach is provided by planetmath.org at planetmath.org/DerivationOfTheLaplacianFromRectangularToSphericalCoordinates.

18 Solutions Manual for Foundations for Nanoscience and Nanotechnology The basic steps include defining the Laplacian in rectangular coordinates ∇2 f =

∂2f ∂2f ∂2f + 2 + 2.

2 ∂x ∂y ∂z

Using the solutions in Problem 4.15 for the spherical coordinates we write ∂f cosθcosϕ ∂f sinϕ ∂f ∂f = sinθcosϕ + − ∂x ∂r r ∂θ rsinθ ∂ϕ ∂f ∂f cosθsinϕ ∂f cosϕ ∂f = sinθsinϕ + + ∂y ∂r r ∂θ rsinθ ∂ϕ ∂f ∂f sinθ ∂f = cosθ − . ∂z ∂r r ∂θ ∂f Substituting ∂f ∂x for f in the first of these, ∂y for f in the second of these,

and ∂f ∂z for f in the third of these, leads to ∂2f ∂ ∂f cosθcosϕ ∂ ∂f sinϕ ∂ ∂f = sinθcosϕ + − 2 ∂x ∂r ∂x r ∂θ ∂x rsinθ ∂ϕ ∂x ∂2f ∂ ∂f cosθsinϕ ∂ ∂f cosϕ ∂ ∂f = sinθsinϕ + + 2 ∂y ∂r ∂y r ∂θ ∂y rsinθ ∂ϕ ∂y ∂2f ∂ ∂f sinθ ∂ ∂f = cosθ − . ∂z 2 ∂r ∂z r ∂θ ∂z ∂f ∂f Substituting in turn the expressions for ∂f ∂x , ∂y , and ∂z and carrying out the appropriate derivatives, gives three [horrendous] expressions of first and second order derivatives, but when the three terms are added, a number of these will cancel (as long as you keep track of the signs!) or lead to sums of cos2 θ and sin2 θ that add to unity, to give a simpler expression for the Laplacian:

∇2 f =

∂2f 1 ∂2f 1 ∂2f cosθ ∂f 2 ∂f + + + 2 + .

2 2 2 2 2 2 ∂r r ∂θ r sin θ ∂ϕ r sinθ ∂θ r ∂r

This can in turn be rearranged to give the desired expression: ( ∇ = 2

1 ∂ r2 ∂r

( ) ( ) ) 1 ∂ ∂ 1 ∂2 2 ∂ r + 2 sinθ + 2 , ∂r r sinθ ∂θ ∂θ r sinθ ∂ϕ2

which may be most easily seen by working backward.

Problem 4.17 Normalize ψ(r0 , ϕ, 0) in Equation 4.47 for one direction.

Answer: To normalize we set the integral equal to unity:

Chapter 4 Problems 19 ∫ ∞

ψ ∗ (q)ψ(q) = 1,

−∞

but in this wave function, ψ(r0 , ϕ, 0) = Ae−imϕ , the only variable is the angle ϕ, which in turn has values from 0 to 2π and hence, we get ∫ 2π

Aeimϕ Ae−imϕ dϕ = 1 ⇒

0

∫ 2π dϕ = A2 2π = 1 ⇒

A2 0

1 A= √ .

2π q.e.d.

Problem 4.18 Show that the energy for a particle on a ring bounded by an infinite potential is given by Equation 4.49.

Answer: The quantum number, m, is given by √ 2mp r02 E m= , ~2 so m2 ~2 2mp r02 E m2 ~2 = m2 = ⇒ E= .

2 2 ~ 2mp r0 2I

Problem 4.19 Show that Ering = Ebox when b = 2πr0 , whenever n=2m.

Answer: The energy of the particle-in-a-box is Ebox =

n2 h2 .

8mp b2

If the box has a length equal to the circumference of the ring with radius, r0 , then b = 2πr0 and 2

Ebox =

n2 h2 n2 ~2 (n/2) ~2 = = = Ering , 8mp (4π 2 r02 ) 8mp r02 2mp r02

whenever n = 2m, which corresponds to all the even states in the particle-inthe-box. This makes sense since these are the only ones where the derivative is continuous at the edges if they fold back on themselves.

20 Solutions Manual for Foundations for Nanoscience and Nanotechnology

Problem 4.20 Show that the probability distribution for the spherical harmonic Y11 is independent of the angle ϕ.

Answer: Given the wave function √ 3 sinθeiϕ , Y11 = − 8π the probability of finding the particle is given by (see Postulate I) ψ ∗ (q)ψ(q)dq = Y11∗ Y11 dθdϕ =

3 3 sin2 θe−iϕ eiϕ dθdϕ = sin2 θdθ, 8π 8π

which is independent of the angle ϕ.

Problem 4.21 Show that the energy in Equation 4.68 results from Equation 4.66 when r = r0 and V (r0 ) = 0.

Answer: The starting point is Equation 4.66: [

~2 − 2mp

(

1 ∂ r2 ∂r

( r

2 ∂

)

∂r

l (l + 1) − r2

)

] − (E − V (r)) R(r) = 0.

Substituting r = r0 and V (r) = 0, yields zero for the first term since the derivative of a constant is zero, which leaves [ ( ) ] ~2 l (l + 1) − − − (E) R(r0 ) = 0, 2mp r02 which rearranges to Equation 4.69 E=

~2 2mp

(

l (l + 1) r02

) .

Problem 4.22 Confirm Equation 4.69 using Equation eq:4.68.

Answer: Equation 4.68 [

~2 − 2mp

(

l (l + 1) − r02

)

] − (E) R(r0 ) = 0

rearranges to Equation 4.69 when dividing by the wave function, R(r0 ),: ( ) ~2 l (l + 1) E= , 2mp r02 which is possible since R(r0 ) is a constant.

Chapter 4 Problems 21

Problem 4.23 Use a spreadsheet or computer program to calculate R1,0 (r) and R2,1 (r) with Z = 1 and d0 = 1 and plot these as a function of r, in units of a0 , from 0 to 3a0 .

Answer: the By substituting the values of n, l, Z, and d0 , we find that ( R1,0 (r) =

r a0

)0 ( ) ( )0 −r r e a0 = e−r/a0 . a0

This will be an exponential decay function as shown in Figure 4.22.

Correspondingly, ( R2,1 (r) =

r a0

)1 ( ) ( )0 ( ) −r r r 2a0 e = e−r/2a0 .

2a0 a0

which will be a rising function at small values of r and a decaying function as r increases as shown in Figure 4.22.

Problem 4.24 For the Hydrogen Atom, 1. Evaluate E1 , E2 , E3 , and E4 and plot these.

2. Calculate ∆E1,2 , ∆E1,3 , and ∆E1,4 .

3. Calculate ∆E1,∞ , ∆E2,∞ , and ∆E3,∞ [this corresponds to the H-lines in the emission spectrum of the H-atom].

Answer: Given that, for the Hydrogen Atom, µ ≈ me and Z = 1, we get me e4 1 µZ 2 e4 =− .

2 2 2~ n 2~2 n2 ( )2 ( ) With me = 9.109×10−31 kg, e2 = q 2 /4πϵ0 = 1.602 × 10−19 C / 1.112 × 10−10 C 2 J −1 m−1 = 2.308 × 10− = −28 J m, and ~ = 1.054 × 10−34 J s, we get En = −

En = −

9.109 · 2.3082 × 10−87 kg J 2 m2 1 1 = −21.84 × 10−19 J 2 .

2 −68 2 2 2 2 · 1.054 × 10 J s n n

Note that in our solution, we used the potential V (r) = e2 /r. Many prefer to use the potential V (r) = q 2 /(4πϵ0 r), which leads to the expression for the energy that includes permittivity of vacuum, ϵ0 and the charge of the electron, q. The difference becomes the definition of e2 = q 2 /(4πϵ0 ).

Thus

22 Solutions Manual for Foundations for Nanoscience and Nanotechnology

E1 = −21.84 × 10−19 J

∆E12 = 16.38 × 10−19 J

∆E1∞ = 21.84 × 10−19 J

E2 = − 5.46 × 10−19 J

∆E13 = 19.41 × 10−19 J

∆E2∞ = 5.46 × 10−19 J

E3 = − 2.43 × 10−19 J

∆E14 = 20.48 × 10−19 J

∆E3∞ = 2.43 × 10−19 J

E4 = − 1.36 × 10−19 J.

Problem 4.25 Use a spreadsheet program or a computer program to calculate and plot the Hermite polynomials and the wave functions for the simple harmonic oscillator i.e., 1. H0 (q), H1 (q), and H2 (q) 2. ψ0 (q), ψ1 (q), and ψ2 (q).

Answer: Recall that ψn (q) =Nn Hn (q)e−q /2 ( mω )1/2 q= x ~ ( mω )1/4 ( 1 )1/2 Nn = ~π (2n n!) n 2 ∂ 2 Hn (q) =(−1)n eq e−q ∂q n 1 En =(n + )~ω.

2 2

We therefore get that ∂ 0 −q2 e =1 ∂q 0 1 2 ∂ 2 2 2 H1 (q) =(−1)1 eq e−q = (−1)eq (−2q)e−q = 2q ∂q 1 2 1 2 ∂ 2 2 ∂ 2 H2 (q) =(−1)2 eq e−q = eq (−2q)e−q 2 1 ∂q ∂q ) ( 2 −q 2 q2 (−2q)(−2q)e − 2e−q = 4q 2 − 2. =e H0 (q) =(−1)0 eq

2

Plots of the Hermite polynomials will then show a horizontal line at unity, a line with a slope of two, and a quadratic centred at the origin on the q-axis.

Further,

Chapter 4 Problems 23

N0 =

( mω )1/4 ( ~π ( mω )1/4 (

1 (20 0!)

)1/2 =

( mω )1/4

~π ( mω )1/4 1 1 √ = N1 = ~π (21 1!) ~π 2 ( mω )1/4 ( 1 )1/2 ( mω )1/4 1 √ .

N2 = = ~π (22 2!) ~π 2 2 )1/2

This leads to ψ0 (q) =N0 H0 (q)e−q /2 =

( mω )1/4

e−q /2 ~π ( mω )1/4 √ 2 2 ψ1 (q) =N1 H1 (q)e−q /2 = 2qe−q /2 ~π ) ( mω )1/4 √ ( 1 −q2 /2 −q 2 /2 ψ2 (q) =N2 H2 (q)e = 2 q2 − e . ~π 2 2

2

To show the principles, it is acceptable to plot these as a function of q without the numeric term in front. In each case the wave functions are the Gaussian multiplied by a polynomial (a constant, a linear function, and a quadratic in these cases).

Problem 4.26 Show that ∆En = ~ω for the simple harmonic oscillator.

Answer: ( ) Given that En = n + 12 ~ω, it follows that ( ) ( ) 1 1 n+ ~ω − n − 1 + ~ω 2 2 ( ) 1 1 = n+ −n+1− ~ω = ~ω.

2 2

∆En = En − En−1 =

Problem 4.27 Calculate the energy differences for the rigid rotator: ∆E1 , ∆E2 , ∆E3 , and ∆E4 .

Answer: ~2 Given that El = l(l + 1) 2µr 2 , then it follows that 0

∆El = (l(l + 1) − (l − 1)l) and hence

~2 ~2 = 2l , 2µr02 2µr02

24 Solutions Manual for Foundations for Nanoscience and Nanotechnology

~2 2µr02 ~2 ∆E2 =4 2µr02 ~2 ∆E3 =6 2µr02 ~2 ∆E4 =8 , 2µr02 ∆E1 =2

~ which gives a series of equally spaced lines, with spacing 2 2µr 2.

2

0

CHAPTER

5

Chapter 5 Problems

Problem 5.1 Derive Equation 5.3.

Answer: Given that ∆En = (2n + 1)

h2 c = hν = h , 8me b2 λ

it follows that (2n + 1) h2 λ (2n + 1) hλ b = = ⇒b= 8me hc 8me c 2

√ (2n + 1) hλ .

8me c

Problem 5.2 Find the structure of beta-carotene, determine the number of conjugated double bonds, and estimate the length of the box and the number of electrons in them. Then use the particle in the box to estimate the wavelength of light expected to be absorbed by this molecule.

Answer: Beta-carotene is a polyene with two six-member rings attached at each end. There are a total of 9 trans-conjugated double bonds in the linear chain and one cis-conjugated double bond in each of the rings. This leads to a total of 11 conjugated double bonds. Assuming each double bond is on average 134 pm and the single bond is on average 148 pm, then the total length of the conjugated system would be about 2854 pm, which would be the length of the box that the electrons can move in. Since there are 11 double bonds, there are 22 electrons, and hence we expect the HOMO to correspond to n = 11 and the LUMO to n = 12. We therefore want to calculate the energy corresponding to the transition from the 11th state to the 12th state.

Given that the mass of the electron is me = 9.109 × 10−31 kg, Planck’s constant is h = 6.626 × 10−34 J s, the size of the box is b = 2.854 × 10−9 m, and the speed of light is c = 3.000 × 108 m s− 1, we get

25

26 Solutions Manual for Foundations for Nanoscience and Nanotechnology

( ) ( ) hc 8me b2 c 8me b2 hc λ= = = ∆En (2n + 1) h2 (2n + 1) h 3 · 8 · 9.109 · 2.8542 × 108 × 10−31 × 10−18 m s−1 kg m2 = (2 · 12 + 1) · 6.626 × 10−34 J s −41 1780 × 10 = m = 1079 nm.

165 × 10−34 The actual absorption maximum for beta-carotene is around 470 nm, so the estimate is off by about a factor of two.

Problem 5.3 Find the maximum wavelength for the absorption spectrum of trans-retinal and estimate, using Equation 5.3, the effective size of the box occupied by the electrons and compare this to the estimated length of the conjugated system in the molecule.

Answer: Given that the maximum wavelength of absorption for trans-retinal is about 370 nm = 3.70 × 10−7 m, we can use Equation 5.3: √ (2n + 1) hλ b= 8me c to calculate the effective length of the box. Trans-retinal has a total of six double bonds, so the transition from the HOMO to the LUMO is for n = 7.

Further, the mass of the electron is me = 9.109 × 10−31 kg, Planck’s constant is h = 6.626 × 10−34 J s,and the speed of light is c = 3.000 × 108 m s− 1, so we obtain √ b= √ =

(2 · 7 + 1) · 6.626 · 3.70 × 10−34 × 10−7 J s m 8 · 9.109 · 3 × 10−31 × 108 kg m s−1 347.4 × 10−41 (kg m2 s−2 ) s m √ = 1.589 × 10−18 m2 = 1.26 nm.

218.61 × 10−23 kg m s−1

This compares with an estimate of 6 · 0.134 + 5 · 0.148 nm = 1.54 nm, for six double bonds and five single bonds.

Problem 5.4 What are the assumptions of the Fermi gas model?

Answer: There are four assumptions: • There are N positively charged nuclei and N negatively charged electrons.

Chapter 5 Problems 27 • There is no lattice structure.

• The Coulomb interactions are negligible.

• There are two electrons per energy level.

Problem 5.5 Using the Fermi–Dirac statistics, calculate the fraction of electrons at the energy of the chemical potential.

Answer: The fraction of electrons at energy E is given, according to the Fermi–Dirac statistics, by f (E) =

1 , exp ([E − µ] /kT + 1)

where µ is the chemical potential.

When the energy is equal to the chemical potential, we get f (E = µ) =

1 1 1 = = . exp ([E − µ] /kT + 1) exp(0) + 1 2

*Problem 5.6 Using the properties of a wave function defined by Postulate I, set up the boundary conditions of the periodic potential illustrated in Figure 5.5 and write out the four equations that need to be solved to get to Equation 5.14. You do not need to solve them.

Answer: The wave function, ψ(x), must be continuous and have continuous derivatives at all edges. The diagram in Figure 5.5 shows the edges of the boxes at positions x = 0, x = c, x = c+b. Let the wave function be ψ0 (x) in the regions where the potential is zero (between x = 0 and x = c) and let it be ψV (x) in the regions where the potential is V , then the wave functions are given by

ik0 x

ψ0 (x) = Ae

+ Be

−ik0 x

ψV (x) = CekV x + De−kV x

√ 2mp E & k0 = √~ 2mp (V − E) & kV = . ~

The boundary conditions at the edge where x = 0 now require that ψ0 (x = 0) = ψV (x = 0) ⇒ A + B = C + D ∂ ∂ ψ0 (x = 0) = ψV (x = 0) ⇒ ik0 (A − B) = kV (C − D) . ∂x ∂x The wave function must be periodic, which means that for all x we have:

28 Solutions Manual for Foundations for Nanoscience and Nanotechnology

ψ0 (x + a) = ψ0 (x)eika ψV (x + a) = ψV (x)eika , where k is the wave vector for the nearly free electron.

Using this, we can equate the wave functions and their derivatives at their periodic boundaries, i.e., they must be equal at x = c and x = c + a. This gives ψV (c) = ψ0 (c) = ψ0 (c + a)e−ika ∂ ∂ ψV (c) = ψ0 (c + a)e−ika , ∂x ∂x which, when substituting the wave functions, gives ( ) CekV c + De−kV c = Aeik0 (c+a) + Be−ik0 (c+a) e−ika ( ) ( ) kV CekV c − De−kV c = ik0 Aeik0 (c+a) − Be−ik0 (c+a) e−ika .

In the end there are four equations with four unknowns (A, B, C, and D): A+B−C −D =0 ik0 A − ik0 B − kV C + kV D = 0 e

ik0 (c+a) −ika

e

−ik0 (c+a) −ika

A+e

e

B − ekV c C − e−kV c D = 0

ik0 eik0 (c+a) e−ka A − e−ik0 (c+a) e−ika B − kV ekV c C + kV e−kV D = 0.

These four equations with four unknowns can be solved by evaluating the determinant (coming from the coefficients of the unknowns): 1 ik0

eik0 (c+a) e−ika ik0 eik0 (c+a) e−ka

1 −ik0 e−ik0 (c+a) e−ika −e−ik0 (c+a) e−ika

−1 −kV −ekV c −kV ekV c

−1 = kV = 0. −e−kV c kV e−kV

Problem 5.7 Describe qualitatively, the difference between a semiconductor and an insulator.

Answer: A semiconductor has a band gap with values around a few electron volts or less so that an applied voltage can cause electrons to be promoted to the conduction band where they can move freely and conduct electricity.

An insulator has a band gap much greater than a few electron volts so that an applied voltage will not cause electrons to flow.

Chapter 5 Problems 29

Problem 5.8 Describe qualitatively, the difference between a direct and an indirect band gap semiconductor and explain why it is important.

Answer: A direct band gap semiconductor has the minimum of the energy curve in the conduction band aligned in k-space with the maximum of the energy curve in the valence band. There is no significant distortion of the lattice structure when electrons are promoted to the conduction band and therefore transitions can readily occur in excitation and emission — allowing for luminescence.

An indirect band gap semiconductor has the minimum of the energy curve in the conduction band shifted in k-space relative to the maximum of the energy curve in the valence band. There is a significant distortion of the lattice structure when electrons are promoted to the conduction band. This makes emissions much more difficult and hence luminescence is low or absent.

Problem 5.9 Describe qualitatively, the difference between an n-doped and a p-doped semiconductor.

Answer: An n-doped semiconductor has an excess of electrons over holes so the Fermi level is closer to the conduction band.

A p=doped semiconductor had an excess of holes over electrons so the Fermi level is closer to the valence band.

Problem 5.10 Describe zero-, one-, two-, and three-dimensional structures and give examples of each.

Answer: Zero-dimensional structures have boundaries or barriers to movement of electrons in all three dimensions such as in quantum dots.

One-dimensional structures have boundaries or barriers to movement of electrons in two dimensions, but not in the third, such as in nano wires.

Two-dimensional structures have boundaries or barriers to movement of electrons in one dimension only, such as in thin plates.

Three-dimensional structures have no boundaries or barriers to movement of electrons and represent bulk materials.

Problem 5.11 Provide a rationalization for the trends observed in Figure 5.16.

Answer: Figure 5.16 show that the energy varies inversely with the dimension to some power greater than 1, i.e., E ∝ d−n where n > 1. Thus energy decreases as the size of the gap in the junction increases. This is consistent with the particle-in-the-box formulation, although these data show that n ̸= 2 for all dimensions.

30 Solutions Manual for Foundations for Nanoscience and Nanotechnology

Problem 5.12 Using the data in Figure 5.22, discuss whether there is a difference between a small cluster of molecules and a small nanoparticle of the same material.

Answer: Based on the comparison of the data for the three small clusters of defined size and the extrapolation of the data for the nanoparticles of decreasing size, there appears to be a nearly continuous change. Thus, it appears that these clusters mimic very small nanoparticles very well.

CHAPTER

6

Chapter 6 Problems

Problem 6.1 Describe the differences between isolated, closed, and open systems and give an example of each.

Answer: An isolated system is one in which neither energy nor mass can be exchanged with the surroundings. An example would be a closed thermos bottle.

A closed system is one in which energy, but not mass, can be exchanged with the surroundings. An example would be a stoppered flask on a table.

An open system is one in which both energy and mass may be exchanged with the surroundings. An example would be an open beaker on a table.

Problem 6.2 What are the key characteristics of isochoric and adiabatic processes?

Answer: An isochoric process is one in which the volume remains constant.

An adiabatic process is one in which there is no exchange of heat with the surroundings

Problem 6.3 State the first, second, and third laws of thermodynamics.

Answer: First Law: In an isolated system the energy is constant.

Second Law: In an isolated system the entropy of a spontaneous process must always increase.

Third Law: It is impossible to attain the temperature of absolute zero.

Problem 6.4 Define the heat capacities at constant volume and at constant pressure in terms of thermodynamic functions.

Answer: The heat capacity at constant volume is the change in internal energy per

31

32 Solutions Manual for Foundations for Nanoscience and Nanotechnology unit temperature:

( CV =

∂U ∂T

) .

V

The heat capacity at constant pressure is the change in enthalpy per unit temperature: ( ) ∂H CP = . ∂T P

*Problem 6.5 Show that the equalities in Equations 6.34 and 6.35 follow from the definitions of the Helmholtz and Gibbs free energy, respectively, and expressions for the Clausius Inequality.

Answer: The Clausius Inequality states that T sys dS sys ≥ dq sys ⇒ dq sys − T sys dS sys ≤ 0 T sys dS sys ≥ (dU )V ⇒ (dU )V − T sys dS sys ≤ 0 T sys dS sys ≥ (dH)P ⇒ (dH)P − T sys dS sys ≤ 0.

The definitions of the Helmholtz and Gibbs free energy are, respectively A ≡ U − T S and G ≡ H − T S, which give the differentials dA = dU − T dS − SdT ⇒ (dA)V,T = (dU )V − T dS ≤ 0 dG = dH − T dS − SdT ⇒ (dG)P,T = (dH)P − T dS ≤ 0. q.e.d.

Problem 6.6 Provide four different criteria for spontaneous change at different conditions.

Answer: (dU )V,S ≤ 0 (dH)P,S ≤ 0 (dA)V,T ≤ 0 (dG)P,T ≤ 0 (dS)V,U ≥ 0 (dS)H,T ≥ 0.

Chapter 6 Problems 33

Problem 6.7 Show that equations 6.36 to 6.39 are correct.

Answer: From the definitions of U , H, A, and G, we can write

dU = dq + dw dH = d(U + P V ) dA = d(U − T S) dG = d(H − T S), which leads to dU = T dS − P dV dH = dU + P dV + V dP = T dS + V dP dA = dU − T dS − SdT = −P dV − SdT dG = dH − T dS − SdT = V dP − SdT. q.e.d.

*Problem 6.8 Given that the property of an exact differential df (x, y) =

∂f ∂f dx + dy ∂x ∂y

requires that ∂2f ∂2f = . ∂x∂y ∂y∂x Derive Equations 6.40 to 6.43 from Equations 6.36 to 6.39.

Answer: 1: From Equation 6.36 to 6.40: Given dU = T dS − P dV, then from the rules of the exact differential, ( ) ( ) ∂U ∂U =T & = −P. ∂S V ∂V S Taking the second derivatives of both terms and equating them, gives us (

∂T ∂V

)

( = S

∂ ∂V

(

∂U ∂S

) )

( =

V

S

∂ ∂S

(

∂U ∂V

) )

( =−

S

V

∂P ∂S

) .

V

34 Solutions Manual for Foundations for Nanoscience and Nanotechnology 2: From Equation 6.37 to 6.41: Given dH = T dS + V dP, then from the rules of the exact differential, ( ( ) ) ∂H ∂H =T & = V. ∂S P ∂P S Taking the second derivatives of both terms and equating them, gives us (

∂T ∂P

(

) = S

∂ ∂P

(

∂H ∂S

(

) ) = P

S

∂ ∂S

(

∂H ∂P

(

) ) = S

P

∂V ∂S

) .

P

3: From Equation 6.38 to 6.42: Given dA = −SdT − P dV, then from the rules of the exact differential, ( ) ) ( ∂A ∂A = −S & = −P. ∂T V ∂V T Taking the second derivatives of both terms and equating them, gives us ( −

∂S ∂V

)

( =

T

∂ ∂V

(

∂A ∂T

) )

( =

V

T

∂ ∂T

(

∂A ∂V

) )

( =−

T

V

∂P ∂T

) .

V

4: From Equation 6.39 to 6.43: Given dG = −SdT + V dP, then from the rules of the exact differential, ) ( ) ( ∂G ∂G = −S & = V. ∂T P ∂P T Taking the second derivatives of both terms and equating them, gives us ( −

∂S ∂P

)

( = T

∂ ∂P

(

∂G ∂T

) )

( =

P

T

∂ ∂T

(

∂U ∂P

) )

( =

T

P

∂V ∂T

) .

P

Chapter 6 Problems 35

Problem 6.9 Define the chemical potential in terms of the Helmholtz free energy.

Answer: The chemical potential for species i, µi is the molar Helmholtz free energy when the temperature, volume, and the number of all other species are kept constant, thus ( ) ∂A . µi = ∂ni T,V,nj̸=i

CHAPTER

7

Chapter 7 Problems

Answer:

Problem 7.1 Explain qualitatively why the chemical potential of a gas decreases more rapidly with temperature than the chemical potential of a solid and why the chemical potential of a gas increases more rapidly with pressure than the chemical potential of a liquid.

Answer: The rate of change in the chemical potential with temperature at constant pressure equals the negative of the molar entropy, i.e., ( ) ∂µ = −Sm . ∂T P Since the molar entropy of a gas is much larger than than that of the solid (being more disordered), then the change of the chemical potential of the gas is more sensitive to temperature than the solid. The gradient of the chemical potential of the gas with temperature is greater, but negative, so the slope has a large negative value, so it decreases more rapidly than does the solid.

The rate of change in the chemical potential with pressure at constant temperature equals the molar volume, i.e., ( ) ∂µ = Vm . ∂P T Since the molar volume of a gas is much larger than that of the liquid (except at the critical point), then the change of the chemical potential of the gas is more sensitive to pressure than the liquid. The gradient of chemical potential of the gas with pressure is greater, and positive, so the slope is larger, so it increases more rapidly than does the liquid.

Problem 7.2 Draw schematically the chemical potential as a function of temperature at a pressure corresponding to the triple point.

37

38 Solutions Manual for Foundations for Nanoscience and Nanotechnology Answer: The graph would be similar to Figure 7.3 but all three lines would intersect at one single point - the temperature of the triple point.

Problem 7.3 Explain why water does not freeze at the high pressures at the bottom of a deep lake.

Answer: The molar volume of ice is larger than that of liquid water at temperatures below about four degrees centigrade. Hence, the chemical potential of the liquid water will change less with pressure than will that of the ice, and as a consequence the chemical potential of the liquid water will be lower than that of the ice at these temperatures. The consequence is that the liquid is the thermodynamically more stable phase even as the pressure increases. If the temperature decreases sufficiently and the pressure increases sufficiently, then the solid phase will eventually appear.

*Problem 7.4 Derive Equation 7.4 from 7.2 assuming ideal gas behavior.

Answer: Given (

∂µ ∂P

) = Vm ⇒ dµ = Vm dP, T

we can substitute Vm = nRT /P for an ideal gas and integrate from the initial to the final pressures to get ( ) ∫ f nRT Pf ∆µ = µf − µi = dP = nRT ln .

P Pi i

Problem 7.5 Consider a hollow tube with a bubble at each end with the bubble at the left being bigger than that at the right and with a continuous connection of air between them through the tube. This is an unstable situation, so what will happen next?

Answer: Since the pressure inside the small bubble has to be greater in order to sustain the smaller diameter than does the pressure inside the large bubble, the surface tension of the small bubble will try to decrease its size more than the large bubble. Since the air can travel along the length of the tube, the molecules will move from the region of the small bubble, causing the small bubble to get smaller and the large bubble to get larger.

*Problem 7.6 Derive the Kelvin Equation 7.10 from the Laplace Equation 7.8 and Equation 7.9.

Answer:

Chapter 7 Problems 39 The Laplace equation is 2γ . r Equation 7.9 shows the change in the chemical potential for a liquid and its vapor when in equilibrium as the liquid is changed from a large drop with ’infinitely’ large radius to a smaller drop of radius r. Substituting the ∆P of the Laplace equation gives, ( ∗) Pr 2γ Vm (l)∆P = RT ln = Vm (l) . ∗ P∞ r ∆P =

Rearranging gives ( ln

Pr∗ ∗ P∞

) =

2γVm (l) , rRT

or the vapor pressure of the liquid of a small drop becomes ∗ Pr∗ = P∞ e

2γVm (l) rRT

,

which will increase as the radius of the drop decreases.

Problem 7.7 Explain the observation in Figure 7.7 that the vapor pressure of mercury increases more rapidly than water as the size of the drop decreases.

Answer: The density of liquid mercury is 13.5 g cm−3 , the molar mass is 200 g mol−1 , and the surface tension is 485 mN m−1 . Correspondingly, the density of water is 1 g cm−3 , the molar mass is 18 g mol−1 , and the surface tension is 72.8 mN m−1 . Thus, 200 · 485 cm3 mol−1 mN m−1 13.5 = 7185 cm3 mol−1 mN m−1 ; 18 · 72.8 V m(H2 O, l)γH2 O = cm3 mol−1 mN m−1 1 = 1310 cm3 mol−1 mN m−1 .

V m(Hg, l)γHg =

For a given size of drop, the vapor pressure of mercury will be larger than that of water, and as a consequence it will also be more sensitive to the radius and rise more rapidly as the radius decreases.

Problem 7.8 Calculate the critical radius for a water droplet at room temperature when the pressure is 10% above the vapor pressure.

Answer: The density of water is 1 g cm−3 , the molar mass is 18 g mol−1 , and the

40 Solutions Manual for Foundations for Nanoscience and Nanotechnology surface tension is 72.8 mN m−1 . At room temperature (T = 298 K) and a pressure 10 % above the vapor pressure, the critical radius is

rc =

2γVm 2 · 18 · 72.8 g mol−1 mN m−1 (P )= 1 · 8.314 · 298 ln(1.1) g cm−3 KJ K −1 mol−1 RT ln P ∗ =

2620 × 10−3 N m−1 = 1.05 × 10−9 m = 1.05 nm 2477 × 106 m−3 N m

Problem 7.9 Use the vector diagram in Figure 7.11 to derive Young’s equation (Equation 7.18).

Answer: When the surface tension is at equilibrium, the projection of the surface tension vectors must balance. In the diagram on the left, it follows that γsv must balance γsl and the projection of γlv , which is γlv cosθ. In the diagram on the right, it follows that γsl must balance γsv and the projection of γlv , which is -γlv cosθ. In either case, we get the same answer, namely γsl = γsv − γsv cosθ

Problem 7.10 Show that Equation 7.19 follows directly from the balance of pressures as shown in Figure 7.13.

Answer: The pressure difference is on, the one hand, given by the Laplace Equation, and, on the other hand, the pressure due to gravity on a column of height h and density ρ, so ∆P =

2γ = hρg ⇒ r

h=

2γ . rρg

Problem 7.11 The xylem in trees are small tubes created from dead cells and can have diameters as small as about 25 microns. Calculate the theoretical height of the water column that such a tube could support. You should find that this is not likely enough to support the water rising to the top of a tall tree. Explore alternative explanations for the rise of water to the top of a very tall tree.

Answer: The height of the water column in the xylem would be

h=

2γ 2 · 72.8 mN m−1 = rρg 12.5 · 1 · 9.80 µm g cm−3 m s−2 145.6 × 10−3 kg m s−2 m−1 = = 1.188 m 122.50 × 10−6 × 103 m kg m−3 m s−2

Chapter 7 Problems 41 Clearly, the capillary action can only account for part of the rise of water in the trees. It is expected that evaporation from the leaves creates an additional negative pressure which accounts for the bulk of the upwards driving force.

*Problem 7.12 Derive Equation 7.43.

Answer: The definition of G = H − T S can be written as G/T = H/T − S. This can be differentiated at constant pressure to give (

∂ ∂T

)(

G T

(

( )( ) ) ∂ H ∂S − ∂T T P ∂T P P ( ( ) ) −H 1 ∂H ∂S −H = 2 + − = 2 .

T T ∂T P ∂T P T

)

=

The last part follows from the facts that ( ) ( ) ( ) ∂G ∂H ∂S = −T − S, ∂T P ∂T P ∂T P (

and

(

so

∂H ∂T

∂G ∂T

) = −S, P

)

( =T P

∂S ∂T

) .

P

Problem 7.13 Provide a possible rationale for the difference in the variation of binding energy with the size of spherical particles of gold (Au) and silicon (Si) illustrated in Figures 7.16 and 7.18.

Answer: The argument presented by the authors is that, in contrast to the metals, such as gold, the group IV elements, such as carbon and silicon, can re-hybridize so that at the surface they may form multiple bonds between the elements, akin to forming sp2 rather than sp3 bonds. This will make the surface interactions stronger for these elements, leading to the greater variation with size.

Problem 7.14 Consider the following chemical reactions. For each, write an expression for the reaction coefficient, Q. 1. CO2 (g) + H2 O(g)

H2 CO3 (s).

2. 2CO(g) + O2 )(g)

CO2 (g).

3. N2 (g) + 3H2 (g)

2N H3 (g).

42 Solutions Manual for Foundations for Nanoscience and Nanotechnology 4. CuO(s) + 2N aHSO4 (aq)

N a2 SO4 (aq) + H2 O(l).

Answer: The general form for the reaction coefficient is given in Equation 7.34 ( )v ∏ ∏ Pj j vi Q= ci × .

Pjo i j We therefore get (noting the for solids we use activities and frequently these are unity for ideal systems): 1: ( )−1 ( )−1 ( )−1 P P P P H O CO H O CO 2 2 2 2 = aH2 CO3 .

Q = a1H2 CO3 · 2 Po Po (P o ) 2:

( Q=

PCO2 Po

)−1 )+1 ( 2 2 PO2 PCO PCO2 (P o ) · = .

3 2 PO2 PCO (P o )

3:

2

Q= 4: Q=

PN2 H3 (P o ) .

3 PN PH 2

[N a2 SO4 ] [H2 O] . aCuO [N aHSO4 ]

CHAPTER

8

Chapter 8 Problems

Problem 8.1 Explain how the density of states in the valence band is observed in the photoelectron spectrum.

Answer The density of states in the valence band is reflected in electrons that emerge from the surface with kinetic energy corresponding to the energy difference between the energy in the excited light and the binding energy. The number of electrons observed at a particular energy is interpreted as proportional to the density of states (i.e., the number of bound electrons) at that particular energy level in the valence band.

Problem 8.2 What is the origin of inelastically scattered electrons in the photoelectron spectrum?

Answer The inelastically scattered electrons arise from electrons emitted from below the immediate region of the surface. The deeper the exciting photons can penetrate, the more electrons will be excited from the region below the surface.

As these electrons travel through the material to the surface, the may scatter off the nuclei in the lattice, and hence loose energy. They will then emerge with less kinetic energy that those arising from the surface itself.

43

CHAPTER

9

Chapter 9 Problems

Problem 9.1 Describe some important attractive forces between particles.

Answer The most important attractive force is the electrostatic force between oppositely charged particles. This operates at long distances and usually have a 1/D dependence, where D is the separation.

Others forces arise from dipole-dipole interactions (Keesom interactions) and dipole-induced dipole (Debye interactions) both of which involve a permanent dipole.

The weakest forces arise from the London dispersion forces, that involve transient dipoles in each particle. The important part is that these are additive forces, so the overall attractive force between particles can be quite large.

These last three attractive forces are quite sensitive to the dielectric constants and the refractive indexes of the materials and the medium between them. These are also long-range interactions with a 1/Dm , where m can be between 1 and two depending on the geometry of the objects.

At very close distances, the exponent can can approach 6, as in the famous Lennard-Jones 6-12 potential.

Problem 9.2 Describe some important repulsive forces between particles.

Answer The most important repulsive force is the electrostatic force between particles of the same charge. These are long range forces with the effect of the separation typically being exponential with an exponent that is sensitive to the ionic medium separating them.

At very small separations, the physical contact of the orbitals of the surface atoms becomes the dominant repulsive interaction. This is reflected in the 1/D12 dependence in the famous Lennard-Jones 6-12 potential.

Problem 9.3 Using the expressions for the Hamaker constant in Equations 9.6 and 9.7, calculate its value for a system of two drops of water in air.

45

46 Solutions Manual for Foundations for Nanoscience and Nanotechnology Further, calculate the attractive force between them assuming they both have a radius of 100 nm.

Answer For water at room temperature, the refractive index is 1.330 and the dielectric constant is 80. The refractive index and dielectric constant for air at room temperature is nearly that same as those in vacuum (defined as unity) with values of 1.0003 and 1.0006 respectively. Thus we get from Equation 9.6 that ( )( ) 3kT ϵ1 − ϵ3 ϵ2 − ϵ3 4 ϵ1 + ϵ3 ϵ2 + ϵ3 ( )( ) −23 3 · 1.38 × 10 J K −1 298 K 80 − 1 80 − 1 = 4 80 + 1 80 + 1 −21 2 3 · 4.11 × 10 79 = J = 2.93 × 10−21 J. 4 812

A1 =

From Equation 9.7 we get ) )( ( 2 n1 − n23 n22 − n23 3hν A2 = √ √ [√ ]1/2 √ √ 8 2 (n21 + n23 ) (n22 + n23 ) (n21 + n23 ) + (n22 + n23 ) 3 · 6.63 × 10−34 J s · 3.30 × 1015 s−1 √ 8 2 ( )( ) 1.332 − 12 1.332 − 12 ×√ [√ ]1/2 √ √ (1.332 + 12 ) (1.332 + 12 ) (1.332 + 12 ) + (1.332 + 12 ) =

= 5.8 × 10−19 · 0.117 J = 6.786 × 10−20 J, so that AH = 7.08 × 10−20 J. Assuming further that they are spheres of radius r = 100 nm, we get AH r2 7.08 × 10−20 J 100 × 10−9 m =− 2 2 6D 2r 6D 2 ( ) 1 = 5.90 × 10−28 N m2 , D2

Fattractive = −

which for a separation of D = 100 nm, as an example, would give F = 5.90 × 10−14 N

Chapter 9 Problems 47

Problem 9.4 Calculate the Debye length 1. in a 60-mM solution of N aCl. 2. in a 20-mM solution of CaCl2 .

3. in a 10-mM solution of AlCl3 .

Answer The Debye length, 1/κ, is given by Equation 9.19 ( 2 ) ( 2 ) e NA ∑ 2 e NA κ2 = zi Mi = · I. ϵr ϵ0 kT ϵr ϵ0 kT i The pre-factor is independent of the solution concentration and can be calculated separately at 298 K as ( 2

κ =

e2 N A ϵr ϵ0 kT

) ·I

( )2 1.6 × 10−19 C · 6.02 × 1023 mol−1 = ·I 80 · 8.85 × 10−12 J −1 C 2 m−1 · 1.38 × 10−23 J K −1 · 298 K = 5.29 × 10−4 × 10+20 m mol−1 · I = 5.29 × 1016 m mol−1 · I The ionic strength is I=

∑

zi2 Mi

i

Thus we get 1. in a 60-mM solution of N aCl, where z1 = z2 = 1 and M1 = M2 = 60 × 10−5 mol m−3 so, ( ) κ2 = 5.29 × 1016 m mol−1 12 · 6 × 10−4 mol m3 + 12 · 6 × 10−4 mol m3 ( ) = 5.29 × 1016 12 × 10−4 m−2 .

2. in a 20-mM solution of CaCl2 where z1 = 2, z2 = 1, M1 = 20 × 10−5 mol m−3 , andM2 = 40 × 10−5 mol m−3 so, ( ) κ2 = 5.29 × 1016 m mol−1 22 · 2 × 10−4 mol m3 + 12 · 4 × 10−4 mol m3 ( ) = 5.29 × 1016 12 × 10−4 m−2 .

3. in a 10-mM solution of AlCl3 : where z1 = 3, z2 = 1 and M1 = M2 = 10 × 10−5 mol m−3 , andM2 = 30 × 10−5 mol m−3 so, ( ) κ2 = 5.29 × 1016 m mol−1 32 · 1 × 10−4 mol m−3 + 12 · 3 × 10−4 mol m−3 ( ) = 5.29 × 1016 12 × 10−4 m−2 .

48 Solutions Manual for Foundations for Nanoscience and Nanotechnology In three all cases the Debye length is the same and is equal to 1 1 1 =√ = × 10−6 m = 0.125 µm = 125 nm.

12 −2 κ 7.97 5.29 · 12 × 10 m

Problem 9.5 Describe the concept of the diffuse double layer.

Answer The diffuse double layer arises in an ionic solution close to a charged surface and describes the region in the solution next to the surface where the ion concentration differs from that in the bulk.

Problem 9.6 How does the zeta-potential relate to the Stern Layer?

Answer The zeta-potential is an experimentally determined potential attributed to the potential near a surface where the hydrated layer of ions move with the particle (or stays with the surface if the liquid moves across the surface). This potential is different from the true surface potential because of the shielding provided by the ions in this hydrated layer.

The Stern layer is a theoretical construct associated with that region close to a surface where there are ions bound to the surface and hence shield the solution from the true surface potential.

The zeta-potential is interpreted as the potential at the outer edge of the Stern layer.

Problem 9.7 Use a spreadsheet to calculate the overall interaction energy between two 5-nm-diameter gold spheres in an aqueous solution with 10 mM of N aCl. Compare this to the interaction energy between one of these spheres and a flat surface of gold.

Answer The calculation will use the same approach as in Problems 9.3 and 9.4, first calculating the attractive term by evaluating the Hamaker constant, then calculating the repulsive term.

In this case the Hamaker constant for gold in water is AH = 5.52 × 10−19 J m−1 .

To calculate the Z-term in the repulsive potential, you need to assume a surface potential (which for many gold nanoparticles is in the order of -40 mV because of the coating of citrate from the synthesis, but it can be instructive to vary the surface potential in the spread sheet to see the effect on the overall potential). With this assumption, Z = 1.292 × 10−11 J. For a 10 mM solution of a 1:1 electrolyte, κ = 1.33 × 10−6 m−1 .

With these numbers it is easy to calculate in a spreadsheet the overall potential for a given geometry using Equation 9.33 as a function of both particle size and separation.

The difference between the potential of interaction between two spheres of

Chapter 9 Problems 49 the same size (Equation 9.33) and that of a spherical particle of the same size with an infinite flat surface is given by Tables 9.1 and 9.3, and it is clear that the difference is a factor of two — with the sphere-surface having the large potential.

Problem 9.8 Show for the attractive potential in Equation 9.34 reduced to that of the DLVO theory when rr >> D. Answer The attractive potential as described by Equation 9.34 is AH Uattractive = − 6

(

8rr2 8rr2 D(8rr + h) + + ln D(8rr + D) (4rr + D) (4rr + h)2

)

When rr >> D, it follows 8rr2 rr → D(8rr + D) D and

8rr2 → 2rr (4rr + D)

and ln

D(8rr + h) →0 (4rr + h)2

where the first term will dominate, so Uattractive → −

A H rr AH r =− 6 D 6 2D

as expected if the two spheres are of the same size (rr = r1 r2 /(r1 + r2 )).

Problem 9.9 Describe the conditions that may lead to a dendritric aggregate rather than a compact aggregate.

The dendritic aggregate will arise when the aggregation reaction is diffusion limited, since this means that every encounter is successful in creating an irreversible association of a particle with another. This will arise whenever the activation energy for association is low, that is the barrier to approach is low. This in turn implies that the attractive forces are stronger than the repulsive forces for most of the approach. This will occur in solution whenever the concentration of salt is very high (Figure 9.8) or whenever the particles are very small.

CHAPTER

10

Chapter 10 Problems

Problem 10.1 Explain the differences between the mean, the median, and the mode.

Answer The mean is the arithmetic average, which means that it is the sum of all the elements divided by the total number of elements.

The median is the value of the element at which half the elements are below and half are above that value.

The mode is the value of the most frequently observed element.

Problem 10.2 Calculate and plot on the same graph the binomial distribution for n = 20 and p = 0.5; the Poisson √distribution with µ = 10, and the normal distribution with µ = 10 and σ = 10. Compare these in the context of the law of rare events and the central limit theorem.

Answer Use a spreadsheet, such as Excel, to create a column of numbers from 1 to 20. Then use a separate column to calculate each of the three distributions using the built-in functions of the spread sheet and the parameters above. Plot these. It can be instructive to normalize these to the maximum value (at n=10 in each case).

The plot shows that each of these functions have very similar shapes and most of the asymmetry of the Poisson distribution has disappeared. The binomial distribution is still narrower than the normal distributions, but the shape is close to.

It is evident that the law of rare events and the central limit theorem are close to being true, even with this relatively small number of events.

Problem 10.3 Assuming x can take on the values [7, 7, 8, 8, 8, 9, 10], calculate the inverse of the mean of x, 1/ ⟨x⟩, as well as the mean of the inverse of x, ⟨1/x⟩, and compare these numbers.

Answer The inverse of the mean is 51

52 Solutions Manual for Foundations for Nanoscience and Nanotechnology

1 7 = = 0.1228 ⟨x⟩ 2 · 7 + 3 · 8 + 9 + 10 The mean of the inverse is ⟨ ⟩ 2/7 + 3/8 + 1/9 + 1/10 1 = = 0.1245 x 7

(10.1)

Although these ⟨ ⟩ numbers are close in value, they are not the same, and in 1 general ⟨x⟩ ̸= x1

Problem 10.4 Consider that you want to determine the speed of the water in a stream. You decide to use the ”Poohsticks Game (from Winnie the Pooh books)” in which you infer the speed of the stream from the speed of an object moving on top of the stream of water. In other words, you measure the time that the object moves a fixed distance (from one side of a bridge to the other).

You repeat this experiment multiple times to get a set of measurements of time. You further want to do some statistical analysis of these data. a) When is it appropriate to average the time measurements and then divide this average into the known, fixed distance to get an average speed? b) When is it appropriate to calculate the individual speeds for each time measurement and then average these calculated speeds? (Similar questions arise in many experiments where the parameter of interest is inversely related to the parameter that can conveniently be measured. An example is the classical physics experiment in which we measure the viscosity of a fluid by measuring the time a small ball of radius, r, takes to descend a fixed distance in a column of the liquid. The viscosity, η, is then calculated from Stoke’s Law Fgravity = 6πηrv, where v is the calculated velocity).

Answer This is a matter of trying to determine where the greatest source of error may arise. a) It is appropriate to average the time measurements if the greatest source of error is in the time measurement itself, which can arise if the timing device cannot measure the time well relative to the appearance. This could occur if the timing was by counting seconds in the head and all the sticks move rapidly under the bridge. b) It is appropriate to average the speeds if the greatest source of error is the difference in speed between different measurements, which can arise if each stick encounters different turbulence or eddies as they move under the bridge and the time measurements are precise.

CHAPTER

11

Chapter 11 Problems

Problem 11.1 Explain the concept of an ensemble.

Answer An ensemble is a set of representative samples of the system of interest.

Members of the ensemble may or may not differ from each other, but the average properties of all the members of the ensemble will provide a best estimate of the properties of the system.

Problem 11.2 Explain qualitatively why the canonical partition function relative to the microcanonical partition function is different for distinguishable and indistinguishable particles as expressed in Equations 11.5 and 11.6.

Answer If particles are distinguishable (e.g. they all have different colors), then every combination of particles is unique. If, on the other hand, the particles are not distinguishable, then there will be many states that are identical and hence are being counted multiple times. To allow for this, one divides by the number of possible combinations that are identical, which will be N ! = N · (N − 1) · (N − 2) . . . 3 · 2 · 1.

53

CHAPTER

12

Chapter 12 Problems

Problem 12.1 Discuss the importance of the relative fluctuations being proportional to the inverse of the average number of molecules in a specific open volume.

Answer Since we have shown that ⟨ 2⟩ x 1 2 = ⟨N ⟩ , ⟨x⟩ then it is possible to create an experiment in which one can measure the variance and the mean of fluctuations in the systerm and from these measurements calculate an estimate of the average concentration of the species of interest, as long as the volume is known. In most cases, this is only practical when the concentration is low, so that the relative fluctuations are large.

It should be noted that both measurements will be sensitive to background and noise components, so we normally have to resort to an indirect calculation of the variance by using a correlation function that is extrapolated to the origin. ⟨

⟩

⟨

⟩

Problem 12.2 Show that (x − ⟨x⟩)2 = x2 − ⟨x⟩2 .

We simply multiply through and take the averages to get ⟨ ⟩ ⟨( )⟩ 2 2 (x − ⟨x⟩) = x2 + ⟨x⟩ − 2x ⟨x⟩ ⟨ ⟩ 2 = x2 + ⟨x⟩ − 2 ⟨x⟩ ⟨x⟩ ⟨ ⟩ 2 = x2 − ⟨x⟩ .

*Problem 12.3 Derive Equation 12.5.

Answer We can calculate the average energy from the partition function as

55

56 Solutions Manual for Foundations for Nanoscience and Nanotechnology 1 ∑ Ei e−Ei /kT .

Q If we differentiate the average energy with respect to temperature at constant volume and of particles, we get the heat capacity at constant ( number ) volume CV = ∂⟨E⟩ ∂T ⟨E⟩ =

VN

(

∂ ⟨E⟩ ∂T

) ( ∑ ∂ 1 −Ei /kT = Ei e ∂T Q VN ( ) [∑ ] 1 ∂ [∑ ] ∂ 1 = Ei e−Ei /kT + Ei e−Ei /kT ∂T Q Q ∂T [ ] ∂ ( 1 ) ∑ 1 ∂lnQ 2 −Ei /kT [⟨E⟩] + −Ei e =− ∂T Q ∂T kT ( ) ⟨ ⟩ −1 ∂lnQ =− [⟨E⟩] + E 2 . ∂T kT 2

)

Multiplying by kT 2 , and recognizing that kT 2 ∂lnQ ∂T = ⟨E⟩, we get ( kT 2

∂ ⟨E⟩ ∂T

)

⟨ ⟩ ∂lnQ ⟨E⟩ + E 2 ∂T ⟨ ⟩ 2 2 , = − ⟨E⟩ + E 2 = σE

= kT 2 CV = −kT 2 VN

which is what we wanted to show.

*Problem 12.4 Derive Equation 12.12.

Answer We know that the average number of molecules or particles is ⟨N ⟩ =

Nj µ Ei 1∑ Nj gi e− kT e kT Ξ i

We differentiate this with respect to the chemical potential at constant volume and temperature to get (

∂ ⟨N ⟩ ∂µ

)

Nj µ Ei ∂ 1∑ Nj gi e− kT e kT ∂µ Ξ i ] [ ( ) Nj µ Nj µ E E ∂ 1 ∑ 1 ∂ ∑ − kTi − kTi kT kT = e + e Nj gi e Nj gi e ∂µ Ξ Ξ ∂µ i i

=

=− =−

∂lnΞ 1 ∑ Nj2 − Ei Nj µ ⟨N ⟩ + gi e kT e kT ∂µ Ξ i kT

∂lnΞ 1 ⟨ 2⟩ ⟨N ⟩ + N . ∂µ kT

Chapter 12 Problems 57 Multiplying by kT and recognizing that kT ∂lnΞ ∂µ = ⟨N ⟩, we get ( ) ⟨ ⟩ ∂ ⟨N ⟩ 2 2 kT , = − ⟨N ⟩ + N 2 = σN ∂µ which is what we wanted to show.

*Problem 12.5 Show that Equation 12.13 is valid.

Answer We have shown in the Problem 12.4 that ( ) ∂ ⟨N ⟩ 2 σN = kT . ∂µ V T We can use the fact that the molar volume is defined as Vm = V /N in which we can use the average number of molecules, so that we can substitute ⟨N ⟩ = V /Vm to get ( ( ) ) ∂V /Vm −1 ∂Vm 2 σN = kT = V kT . ∂µ Vm2 ∂µ VT But we can use the Gibbs-Duhem relation N dµ = V dP − SdT to show that at constant temperature, dµ = Vm dP − Sm dT = Vm dP , which can be substituted in the above expression to yield ) dVm −1 2 V Vm dP ( m) −1 dVm = V kT V3 dP ( m ) 3 − ⟨N ⟩ d (V / ⟨N ⟩) = V kT V3 dP )( ( ) 2 −1 dV ⟨N ⟩ = kT V V dP (

2 σN = V kT

2

= since κT =

( −1 dV ) V dP .

kT ⟨N ⟩ κT , V

CHAPTER

13

Chapter 13 Problems

Problem 13.1 Consider a 1-L container with 1 mole of neon gas at 25 o C. Calculate the collision frequency, the average time between collisions, and the mean free path.

Answer: The frequency of collisions is zcoll = σρ ⟨v⟩ where the collision cross section σ = 4πr2 , r being the√radius of the molecule; the density ρ = ⟨N ⟩ /V ; and the mean speed ⟨v⟩ = 8kT /πµ, µ being the reduced mass of the molecule.

For 1 mol/L Neon at 25 o C ( )2 σ = 4πr2 = 4π × 1.54 × 10−10 m = 2.98 × 10−19 m2 ⟨N ⟩ = 6.02 × 1023 V = 10−3 m3 ρ = 6.02 × 1026 m−3 1 1 µ = · mN e = · 20.2 × 1.66 × 10−27 kg = 1.67 × 10−26 kg 2 2 k = 1.38 × 10−23 J K −1 T = 289.13 K so, √ ⟨v⟩ = =

8 · 1.38 · 298 × 10−23 kg m2 s−2 π · 1.67 × 10−26 kg

√ 62.7 × 104 m2 s−2 = 7.92 × 102 m s−1 .

Hence, the collision frequency becomes

59

60 Solutions Manual for Foundations for Nanoscience and Nanotechnology

zcoll = 2.98 × 10−19 m2

6.02 × 1023 7.92 × 102 m s−1 = 1.42 × 1011 s−1 , 10−3 m3

and the average time between collisions becomes tave =

1 1 = × 10−11 s = 7.03 × 10−12 s = 7 ps, zcoll 1.42

and the mean free path becomes ⟨v⟩ 1 = ⟨v⟩ · tave = zcoll σρ = 7.925 × 102 m s−1 · 7.03 × 10−12 s = 5.57 × 10−9 m = 5.57 nm.

λ=

As an aside, the collision frequency per unit volume for two like molecules is given by √ 2 2kT ⟨N ⟩ 1 1 ZAA = σ = σ ⟨v⟩ ρ2 = ρzcoll , πµ V 2 2 2 so ZAA = 2.98 × 10−19 m2 ·

1 · 7.92 × 102 m s−1 · 6.022 × 1052 m−6 2

= 4.27 × 1037 m−3 s−1 1 = · 6.02 × 1026 m−3 · 1.42 × 1011 s−1 2 = 4.27 × 1037 m−3 s−1 .

Problem 13.2 Derive Fick’s second law from Fick’s first law.

Answer: Fick’s First Law relates the flux to the concentration gradient with the diffusion coefficient being the proportionality constant, i.e., ∂N ∂x The flux depends on the position, so there will be a gradient in the flux with respect to position (∂J/∂x). The number of molecules that will move past a particular position per unit time, will depend on this gradient in flux.

When the gradient in the flux is negative, the number of molecules passing per unit time will be positive. Thus, ) ( ) ( ∂J ∂ ∂N ∂2N ∂N =− =− −D =D 2. ∂t x ∂x ∂x ∂x ∂x Jx = −D

Chapter 13 Problems 61

Problem 13.3 Compare the diffusion coefficient for spherical particles of radius 1 nm, 10 nm, and 100 nm moving in water at room temperature.

Answer: The diffusion coefficient is inversely proportional to the radius, r of a spherical object: D=

kT 6πηr

Thus d1nm = 10 · D10nm = 100 · D100nm 1.38 × 10−23 · 298 J 6π · 8.90 × 10−4 P a s · 1 × 10−9 m 411 × 10−23 kg m2 s−2 = 167 × 10−13 kg m−1 s−2 s m = 2.46 × 10−10 m2 s−1 .

=

Problem 13.4 Derive Equation 13.15 Answer: ( ) [B] = [A]0 − [A]t = [A]0 − [A]0 e−kt = [A]0 1 − e−kt .

Problem 13.5 Consider the first order decay of a molecule by four parallel processes in which the rate constants are k4 = 3k1 ; k3 = 2k1 ; and k2 = k1 .

Calculate the relative yield of each of the four products.

Answer: The yield of one product among a series of parallel reactions is given by Φj = ∑

kj . i (ki )

In this case, ∑ (ki ) = k1 + k2 + k3 + k4 = k1 + k1 + 2k1 + 3k1 = 7k1 . i

62 Solutions Manual for Foundations for Nanoscience and Nanotechnology Thus k1 1 = 7k1 7 k2 1 Φ2 = = 7k1 7 k3 2 Φ3 = = 7k1 7 3 k4 = . Φ4 = 7k1 7 Φ1 =

Problem 13.6 Under which circumstances will a pressure perturbation not allow you to measure the kinetics of a reaction?

Answer: Whenever there is no change in the molar volume of the reaction.

Problem 13.7 Consider a measuring volume with dimensions of 1 µm × 1 µm × 3 µm. At what concentration will there, on the average, be 9 molecules in that volume? What is the expected average relative fluctuation amplitude for this system?

Answer: The measuring volume is 1 µm × 1 µm × 3 µm× = 3 × 10−18 m3 = 3 × −15 10 L. The desired concentration is an average of ⟨N ⟩ = 9 molecules per measuring volume, so 9 3 × 1015 mol = = 5 nM 3 × 10−15 L 6.02 × 1023 L The expected average relative fluctuation will be ⟨ 2⟩ x 1 1 2 = ⟨N ⟩ = 9 . ⟨x⟩ C=

Problem 13.8 What conditions are required for a measurement technique to be a suitable candidate for fluctuation analysis to be useful in estimating concentrations?

Answer: The concentration has to be low enough that one can measure statistically meaningful fluctuations in concentration, or numbers of molecules, relative to the noise in the experiment.

Problem 13.9 Show that Equation 13.36 is correct (compare with Problem 12.2).

Answer:

Chapter 13 Problems 63 This answer follows the logic of Problem 12.2: 1 ∑ (N (ti ) − ⟨N ⟩) × (N (ti+j ) − ⟨N ⟩) k−j ] 1 ∑[ 2 N (ti ) · N (ti+j ) + ⟨N ⟩ − N (ti ) · ⟨N ⟩ − ⟨N ⟩ · N (ti+j ) = k−j ∑ ∑ 1 ∑ 2 = (N (ti )N (ti+j )) + ⟨N ⟩ − (N (ti ) ⟨N ⟩) − (N (ti+j ) ⟨N ⟩) k−j 1 ∑ 2 2 2 = (N (ti )N (ti+j )) + ⟨N ⟩ − ⟨N ⟩ − ⟨N ⟩ k−j 1 ∑ 2 = (N (ti )N (ti+j )) − ⟨N ⟩ . k−j q.e.d.

Problem 13.10 Derive Equation 13.45.

Answer: On the one hand we know that the absorbance is given by ( ) I(t) I(t) A(t) = ϵlc(t) = −log10 ⇒ = 10−ϵlc(t) .

I0 (t) I0 (t) On the other hand, ( ) I(t) IF (t) = qΦ(I0 (t) − I(t)) = qΦI0 1 − .

I0 (t) Substituting, gives, ( ) IF (t) = qΦI0 1 − 10−ϵlc(t) . q.e.d.

CHAPTER

14

Chapter 14 Problems

Problem 14.1 Describe the principle of orthonormality of wave functions and compare this to the orthonormal properties of unit vectors in a Cartesian coordinate system.

Answer: Orthonormal means simultaneously orthogonal and normalized.

In terms of wave functions this means that the integral of the product of one wave function and the complex conjugate of a different wavefunction is zero (orthogonal) while the integral of the product of the a wave function with its complex conjugate is unity (normal). This is embodied in the relations ∫ ∫

ψa∗ ψb dτ = 0 ψa∗ ψa dτ = 1

In terms of the unit vectors, for example in the Cartesian space, î, ĵ, k̂, this means that orthogonality requires î · ĵ = 0 î · k̂ = 0 ĵ · k̂ = 0 while normalization to unit vectors requires that î · î = 1 ĵ · ĵ = 1 k̂ · k̂ = 1 Because of the similarity between the properties of wave functions and unit vectors, the wave functions can be considered as basis functions, and there is a description of quantum mechanics that used matrix formulations for the wave functions.

65

66 Solutions Manual for Foundations for Nanoscience and Nanotechnology

Problem 14.2 Define the transition dipole moment. How does this compare to a dipole moment of a molecule?

Answer: The transition dipole moment is defined as ∫ µ ⃗ = ψn∗ (x, y, z; t)µ̂ψm (x, y, z; t)dτ.

In this definition, µ̂ is an operator characterized by the dipole moment of the molecules, so that the Hamiltonian is Ĥ = Ê · µ̂, where Ê is the electromagnetic field operator (Ê = E0 (exp (ihνt/~ + exp (−ihνt/~))). ⃗= ∑ The permanent dipole in a molecule (in whatever state) is defined as µ ei r⃗i .

*Problem 14.3 Prove that the integral of an odd function is zero.

Answer: The integral of a function from minus infinity to plus infinity can be separated into two terms: from minus infinity to zero and from zero to plus infinity: ∫ ∞

∫ 0

∫ ∞

f (x)dx =

f (x)dx +

−∞

f (x)dx.

−∞

0

Substituting x = −y in the first part of this relation gives ∫ ∞

∫ 0

∫ ∞

f (x)dx = or

f (−y)d(−y) +

−∞

−∞

∫ ∞

∫ ∞ f (x)dx =

−∞

f (x)dx, 0

∫ ∞ f (−y)dy +

0

f (x)dx, 0

but f (−y) = −f (y) whenever the function is odd, so ∫ ∞ ∫ ∞ ∫ ∞ f (x)dx = − f (y)dy + f (x)dx = 0, −∞

0

0

since the integrand is irrelevant and the two integrals cancel.

Problem 14.4 What is meant by stimulated emission?

Answer: Stimulated emission is a transition from an excited state of a molecule or particle to a lower state (or ground state) by the interaction with a photon of appropriate energy. In the process, the photon-molecule interaction leads to the emission of two photons, the incident photon as well as one arising from the transition. The stimulated emission gives off two photons for every photon arriving at the exited states. This is the basis for lasers.

Chapter 14 Problems 67

Problem 14.5 Describe the differences between homogeneously and inhomogeneously broadened lines. What is the source of homogeneous line broadening.

Answer: Homogeneously broadened absorption or emission lines arise from the damping of the transition rates of the molecules. For example, the relaxation of the excited state at a relaxation time of τ leads to a line of width proportional to 1/τ . It is a reflection of the fact that all molecules in the system behave the same.

Inhomogeneously broadened absorption or emission lines arise whenever different molecules are subjected to different environments in which their resonance energy vary. Thus different molecules experience different absorption maxima in their spectra. The overall, inhomogeneously broadened line shape is then a result of the superposition of all of these individual lines (each of which may be homogeneously broadened).

CHAPTER

15

Chapter 15 Problems

Problem 15.1 What is the Frank-Condon principle and why is it important in the concept of absorption and emission of electromagnetic radiation?

Answer: The Frank-Condon principle states that the induced electronic transitions occur must faster than any possible nuclear motions. It follows that the coordinates of the molecules are not changed during the transition which leads to ’vertical’ transitions in an energy-coordinate plot. The excited electronic state is then most likely also in an excited vibrational and rotational state.

Any nuclear motions will then follow as vibrational and rotational relaxations, also referred to as internal conversions.

Problem 15.2 Compare internal conversions to intersystem crossings.

Answer: Internal conversion describes the change in energy within an excited electronic state from a vibrationally and rotationally excited state to a lower vibrational and rotational state. This is usually accompanied by emission of radiation in the form of heat.

Intersystem crossing describes the adiabatic or isoenergetic transition from one electronic state to another accompanied by a change in spin state of an electron, for example from a singlet state to a triplet state or from a triplet state to a singlet state.

Problem 15.3 Sketch an example of a Jablonski diagram with one singlet excited state and one triplet state.

Answer: See Figure 15.6 and remove the S2 and T2 states.

Problem 15.4 Compare the properties of fluorescence and phosphorescence.

Answer: Fluorescence arises from spontaneous emission of a photon during an elec-

69

70 Solutions Manual for Foundations for Nanoscience and Nanotechnology tronic transition from one singlet excited state to the singlet ground state.

There is no spin change so it is a spin-allowed transition that occurs fairly soon after the excited state is formed — typically after a few nanoseconds.

Phosphorescence arises from spontaneous emission of a photon during an electronic transition from a triplet excited state to the singlet ground state.

There is a spin change so it is a spin-forbidden transition that occurs more slowly — typically after a few microseconds.

Problem 15.5 Show why the Stern-Volmer plot is linear and explain the significance of the intercept and the slope.

Answer: The Stern-Volmer plot is a plot of the ratio of the quantum yield in the absence of a quencher, Φ0F , to the quantum yield in the presence of a quencher, ΦF , as a function of the concentration of the quencher, [Q].

From Equation 15.10, it follows that the quantum yield in the absence and presence of the quencher differ only by one term in the denominator, so that kF Φ0F = ∑ ki ΦF = ∑

kF . ki + kQ [Q]

It follows that Φ0F = 1 + kQ [Q], ΦF which gives a linear plot with an intercept of one and a slope equal to the rate constant for quenching, kQ .

Problem 15.6 What are the key differences between the dynamic and static quenching processes?

Answer: Dynamic quenching refers to a process where all the molecules or particles in the system are affected by the quenching interaction, that is, there are many molecules that can interact with the quencher during the lifetime of the excited state. As a consequence, all the molecules experience some form of quenching and hence the lifetime of the fluorescence emission is shortened and the quantum yield of fluorescence is lowered.

Static quenching refers to a process where the interaction with the quencher is long-lived relative to the excited state lifetime, so the quenching is complete for those molecules that interact with the quencher. However, at any given time, only a fraction of the molecules experience contact with the quencher, the rest are unaffected. Those in contact with the quencher are not observed in the experiment, so the quantum yield is lower. However, those not

Chapter 15 Problems 71 in contact with quencher are unaffected and hence the lifetime of fluorescence is not changed.

Problem 15.7 Discuss the possible mechanisms whereby energy can be transferred from one molecule to another.

Answer: There are four principal mechanisms of energy transfer between excited states of molecules.

There is the through space energy transfer from a singlet excited state of a donor to a singlet ground state of an acceptor, creating the singlet ground state of the donor and the singlet excited state of the acceptor.

There is the contact based energy transfer through an electron exchange between a singlet excited state of a donor to a singlet ground state of an acceptor, creating the singlet ground state of the donor and the singlet excited state of the acceptor.

There is the contact based energy transfer through an electron exchange between a triplet excited state of a donor to a singlet ground state of an acceptor, creating the singlet ground state of the donor and the triplet excited state of the acceptor.

There is the contact based energy transfer through an electron exchange between a triplet excited state of a donor to a triplet ground state of an acceptor, creating the singlet ground state of the donor and the singlet excited state of the acceptor.

Problem 15.8 Why is triplet-triplet annihilation critical to photodynamic therapy?

Answer: The triplet-triplet annihilation describes the energy transfer from a triplet excited state of a donor to a triplet ground state of an acceptor, creating the singlet ground state of the donor and the singlet excited state of the acceptor. In the process of photodynamic therapy, the triplet donor of an organic molecules is created by application of light. This is sufficiently long lived that it can interact with dioxygen molecules in the tissue. Since the ground state of dioxygen is a triplet, the energy transfer through triplet-triplet annihilation, will create a singlet dioxygen molecule, which is highly reactive in its own right, but can also give rise to other reactive oxygen species, such as superoxide and hydroxide radicals. These reactive oxygen species react with critical intracellular molecular species, damaging the cells — leading to apoptosis (cell death). If the reactions are focused on cancer cells, this is highly desirable.

Problem 15.9 What are the three key factors that affect the efficiency of energy transfer by the Förster mechanism?

Answer:

72 Solutions Manual for Foundations for Nanoscience and Nanotechnology The three key factors are • Spectral overlap between the emission of the donor and absorbance of the acceptor.

• The separation of the donor and the acceptor.

• The relative orientation of the transition dipole moments of the donor and the acceptor.

Problem 15.10 What is the separation of two chromophores whose energy transfer efficiency is fifty percent?

Answer: The energy transfer efficiency is related to the separation, R, of the donor and the receptor by 1 (

E= 1+

R R0

)6 ,

where R0 is the separation at which the energy transfer efficiency is fifty percent. This is a constant for a given pair of donors and acceptors and is given by Equation 15.24.

CHAPTER

16

Chapter 16 Problems

Problem 16.1 What is the key difference between a normal epifluorescence microscopy and a confocal microscope?

Answer: The confocal microscope incorporates a pinhole in the emission plane of the microscope that restricts the amount of light arriving at the detector from the regions above and below the focal plane.

Problem 16.2 What is the point spread function for a confocal microscope?

Answer: The point spread function for a confocal microscope is the convolution of the excitation profile of the microscope (determined either by the focus of a laser or the diffraction limited volume of the objective) and the detection profile of the microscope (determined by the properties of the objective and the pinhole in the emission plane).

Problem 16.3 Discuss the diffraction limit and the impact on confocal microscopy.

Answer: The diffraction limit, d, is given by the Abbe relation λ λ = 2nsinθ 2N A where λ is the wavelength of the illumination light, θ is the angle subtended by the lens used for observation, and N A describes the numerical aperture of the lens.

This relationship tells us that two objects separated by distances less than the diffraction limit cannot be distinguished as separate objects through the lens. Thus, in a conventional confocal microscope, using objectives with numerical apertures as high as 1.5, it is impossible to distinguish objects that are less than about 200 nm apart, using visible light. d=

73

74 Solutions Manual for Foundations for Nanoscience and Nanotechnology Recently, there have been a number of techniques — super-resolution microscopy — developed to circumvent this limitation, either by performing some optical tricks or by ensuring that there are never any two objects observed simultaneously in a fluorescence microscope image that are close together than the diffraction limit. This has led to higher resolution images with resolutions down to 10 nm in some cases.

Problem 16.4 Describe two different methods that can be used to circumvent the diffraction limit in super-resolution microscopy.

Answer: There are optical approaches, which include the stimulated emission depletion, ground state depletion, or structured illumination techniques. In the first case, as an example, the molecules are excited with a laser beam of one shape and de-excited with a laser beam of a different shape to create an effective volume of emission that is smaller than the original excitation volume. The excitation beam is typically a Gaussian shape and the de-excitation beam is typically a donut-shape, leaving the emission volume determined by the ’hole’ in the donut.

There are also localization approaches, which include stochastic optical reconstruction, photoactivated localization, or variants of these. In these cases, the illumination is designed to create a low density of emitters, so that the probability of two emitters being within the diffraction limit of each other is nearly zero. With the assumption of a single emitter, and with the knowledge of the shape of the point spread function, it is possible to reconstruct where the molecule is most likely to be located (namely at the centre of the point spread function). By repeatedly measuring different molecule, it is possible to create an map of the location of all, or most, of the molecules with high resolution.

Problem 16.5 Compare and contrast the decay of the fluorescence correlation function for a system that is diffusing rapidly with that for a system flowing rapidly.

Answer: The correlation function for a diffusing species is g(τd ) =

g(0) , 1 + τtd

while that for a flowing species is g(τf ) = g(0)e−(t/tf ) .

2

The diffusion correlation function is a slowly decaying function where the correlation time can be estimated when g(τd ) = g(0)/2. The steepest part of the decay curve is at the origin and the tail of the correlation function is very long.

Chapter 16 Problems 75 The flow correlation function decays more rapidly, and the correlation time can be estimated when g(τf ) = g(0)/e2 . This function has zero slope at the origin and it decays most rapidly around its mid-point.

CHAPTER

17

Chapter 17 Problems

*Problem 17.1 Derive Equation 17.7 from the Maxwell Equations.

Answer: From the Maxwell equations we have ( ) ⃗ ∂ B ⃗ ×E ⃗ =− ∇ , ∂t ⃗ on both sides to get from which we can create the cross-product with ∇ ( ) ⃗ ∂ B ⃗ ×∇ ⃗ ×E ⃗ =∇ ⃗ × − ∇ ∂t ( ) ∂ ⃗ ⃗ =− ∇×B ∂t ( ) ⃗ ∂ ∂E =− µϵ ∂t ∂t = −µϵ

⃗ ∂2E .

2 ∂t

But the left-hand side is (this is shown separately below) ( ) ⃗ = −∇2 E ⃗ ⃗ ×∇ ⃗ ×E ⃗ =∇ ⃗ · ∇ ⃗ ·E ⃗ − ∇2 E ∇ ⃗ ·E ⃗ = ρ/ϵ = 0 when there are no charges. Hence, since ∇ ⃗ = −µϵ −∇2 E

⃗ ∂2E 2 ∂t

q.e.d. ⃗ ×∇ ⃗ ×E ⃗ =∇ ⃗ ·∇ ⃗ ·E ⃗ − ∇2 E: ⃗ Separate proof of ∇ Given ( ) ∂ ∂ ∂ ⃗ ∇= · î + · ĵ + · k̂ , ∂x ∂y ∂z 77

78 Solutions Manual for Foundations for Nanoscience and Nanotechnology we can write  ⃗ × E ⃗ ∇

=

(

)

∂Ey ∂Ex ∂x − ∂y

î

 ∂   ∂x  E⃗x



ĵ

k̂

∂ ∂y

∂  ∂z 

E⃗y

E⃗z

  

( =

∂Ey ∂Ez ∂y − ∂z

k̂, and hence  î ĵ   ∂ ∂ ⃗ ×∇ ⃗ ×E ⃗ = ∇  ∂x ∂y  ( ) ( ) ∂Ey ∂Ez −∂Ez ∂Ex − ∂y ∂z ∂x + ∂z

) î +

( −∂Ez

k̂ ∂ ∂z

(

∂Ey ∂Ex ∂x − ∂y

∂x

) x + ∂E ĵ + ∂z

    .  )

For simplicity we carry on only with the term in î, but the others are equivalent: [

( ) ( )] ∂ ∂Ey ∂Ex ∂ −∂Ez ∂Ex î − − + ∂y ∂x ∂y ∂z ∂x ∂z [( 2 ) ( 2 )] ∂ Ey ∂ 2 Ex −∂ Ez ∂ 2 Ex = î − − + ∂y∂x ∂y 2 ∂z∂x ∂z 2 [( 2 ) ( 2 )] ∂ Ey ∂ 2 Ez ∂ Ex ∂ 2 Ex î = + − + ∂y∂x ∂z∂x ∂y 2 ∂z 2 [( 2 ) ( 2 )] ∂ Ex ∂ 2 Ey ∂ 2 Ez ∂ Ex ∂ 2 Ex ∂ 2 Ex = î, + + − + + ∂x2 ∂y∂x ∂z∂x ∂x2 ∂y 2 ∂z 2 =

2

where we have added and subtracted the term ∂∂xE2x to get the last part of the equation.

But, ⃗ ⃗ ·E ⃗ − ∇2 E ⃗ ∇·∇ ( ) [( ) ( )] ∂ ∂ ∂ ∂ ∂ ∂ = · î + · ĵ + · k̂ · · î + · ĵ + · k̂ · Ex î + Ey ĵ + Ez k̂ ∂x ∂y ∂z ∂x ∂y ∂z ( 2 ) ( ) ∂ ∂2 ∂2 − + + · E î + E ĵ + E k̂ x y z ∂x2 ∂y 2 ∂z 2 ) ( ) ( ∂ ∂ ∂Ex ∂Ey ∂Ez ∂ · î + · ĵ + · k̂ · + + = ∂x ∂y ∂z ∂x ∂y ∂z ( 2 ) ( ) 2 2 ∂ ∂ ∂ − + 2 + 2 · Ex î + Ey ĵ + Ez k̂ . ∂x2 ∂y ∂z Focusing again on the î-term, we get [( 2 ) ( 2 )] ∂ Ex ∂ 2 Ey ∂ 2 Ez ∂ Ex ∂ 2 Ex ∂ 2 Ex + + − + + î ∂x2 ∂y∂x ∂z∂x ∂x2 ∂y 2 ∂z 2

Chapter 17 Problems 79 which shows the equivalence.

*Problem 17.2 Derive Equation 17.11 from the Maxwell Equations.

Answer: When the wave is propagating along the x-direction, then it is a function of x and t only. The derivatives with respect to y and z are zero. Hence (

) ∂Ex ρ = =0 ∂x ϵ ( ) ⃗ ·B ⃗ = ∂Bx = 0. ∇ ∂x ⃗ ·E ⃗ = ∇

*Problem 17.3 Derive Equation 17.12 from the Maxwell Equations.

Answer: Given that Ex = Ez = 0, we get that ) ∂Bx ∂By ∂Bz , , ∂t ∂t ∂t ( ) ∂E ∂Ey ∂Ex ∂Ez ∂Ey ∂Ex z ⃗ ⃗ =∇×E = − , − , − ∂y ∂z ∂z ∂x ∂x ∂y ( ) ∂Ey ∂Ey = − , 0, ∂z ∂x ( ) ∂Ey = 0, 0, , ∂x

⃗ ∂B − =− ∂t

(

∂E

where the term ∂zy = 0 because we are assuming that the electric field is propagating exclusively along the y direction, so there can be no gradient along the z-direction.

Thus ∂Bz ∂Ey = . ∂t ∂x q.e.d.

*Problem 17.4 Derive Equation 17.13.

Answer: Since

∂Ey ∂Bz = , ∂t ∂x we can integrate both sides with respect to time and recognize that Ey (x, t) = E0 ei(kx−ωt) to get

80 Solutions Manual for Foundations for Nanoscience and Nanotechnology

∫ t

∂ E0 ei(kx−ωt) dt 0 ∂x ∫ t −ik = −ik E0 ei(kx−wt) dt = E0 ei(kx−wt) −iω 0 k = Ey (x, t). ω

Bz (x, t) − Bz (x, 0) = −

If we let Bz (x, 0) = 0 and recognize that c = ω/k, then it follows that Bz (x, t) =

1 Ey (x, t). c

q.e.d.

Problem 17.5 Use a spreadsheet or other programming tool to calculate the functions in Equations 17.19 and 17.23 and to plot them to illustrate the subtle differences in the absorption curves.

Answer: The absorption part of Equation 17.19 corresponds to its imaginary part: ) ( )( e Γ xe (t) = .

2 2ω0 me (ω0 − ω) + Γ2 The absorption part of Equation 17.23 corresponds to α in Equation 17.23 ) ( )( N e2 ωΓ α=2 .

2 4ϵ0 me ω0 c0 (ω0 − ω) + Γ2 Normalization suggests that in both cases it is only the term within the brackets that need be plotted. In this case, the key difference is the factor of the frequency, ω, in the α term.

Create a spread sheet with a column with values of ω, set a value for each of ω0 and Γ, then calculate each of the two formulas.

Problem 17.6 Use Equations 17.16 and 17.25 to show that In = nI0 .

Answer: Equation 17.16 shows I0 =

1 c0 ϵ0 E02 .

2

Equation 17.25 shows c0 n ϵ n = ϵ 0 n2 .

cn =

Chapter 17 Problems 81

Figure 17.1: The refraction of a wave at an interface going from high to low refractive index.

It follows that 1 cn ϵn E02 2 1 c0 = ϵ0 n2 E02 2n 1 = n c0 ϵ0 E02 2 = nI0 .

In =

q.e.d.

Problem 17.7 Draw the wave-fronts corresponding to the refraction of a wave that enters a medium of lower refractive index and compare to Figure 17.9.

Answer: Figure 17.1 in this manual illustrates an example. The diagram is effectively the inverse of that in Figure 17.9 in the text.

Problem 17.8 Use the geometry of refracted beams to prove Snell’s Law. Answer: In the Figure 17.2 in this manual we see that the two triangles share a side, namely AC. The length of the side BC is related to the velocity of the wave in the incident medium, vi , and the length of the side AD is related to the velocity of the wave in transmitted medium, vt . Hence,

82 Solutions Manual for Foundations for Nanoscience and Nanotechnology

Figure 17.2: The refraction of a wave at an interface

c0 ∆t = ACsinθi ni c0 AD = vt · ∆t = ∆t = ACsinθt , nt BC = vi · ∆t =

which can be rearranged to give c0 ∆t = ni sinθi = nt sinθt .

AC q.e.d.

Problem 17.9 Describe the difference between s- and p-polarization.

Answer: The s-polarization refers to polarization where the E-vector is perpendicular to the plane of incidence, which is defined as the plane that contain both the incident and reflected wave vectors.

The p-polarization refers to polarization where the E-vector is parallel to the plane of incidence.

Problem 17.10 What is the Brewster Angle and why it is only relevant to p-polarizations?

Answer:

Chapter 17 Problems 83 The Brewster angle is the angle at which the p-polarization has zero reflection (or 100% refraction) at an interface. It is only relevant to the ppolarization since the reflection coefficient for s-polarization cannot take on the value of zero (see Figure 17.11).

The Brewster angle can be calculated by: ( ) nt −1 θB = tan . ni

Problem 17.11 What is the critical angle and under which circumstances is it important?

Answer: The critical angle is the angle above which both s- and p-polarization is completely reflected at an interface, which is only possible when the wave is going from a high refractive index medium to a low refractive index medium.

The critical angle can be calculated by: ( ) nt θc = sin−1 . ni

Problem 17.12 Describe the appearance of an evanescent wave in the low refractive index medium.

Answer: The evanescent wave is an exponentially decaying electromagnetic wave, decaying perpendicular to the plane of the interface.

Problem 17.13 Write the expression for the wave equation at the surface of a thin optical fiber.

Answer: Assuming that the optical fiber runs along the r-direction, then the electric field will oscillate along that direction, but will decay as evanescent waves in the radial directions. This would correspond to ⃗ = E⃗0 eikx ek′ (y+z)) e−iwt .

E