Modern Physics for Scientists and Engineers Solutions

8/22/2019 Modern Physics for Scientists and Engineers Solutions

1/96

Modern Physics for Scientists and Engineers

Solutions

Joseph N. Burchett after consultations with John C. Morrison

August 4, 2010

Copyright 2010 by Elsevier, Inc. All rights reserved.


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2



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Introduction - Solutions

1. We are given a spherical distribution of charge and asked to calculate thepotential energy of a point charge due to this distribution. There are twosituations to consider which deliver quite different results. The problem does

not specify whether the point charge of interest is inside or outside the chargedistribution so we must take both possibilities into consideration. Before weget started, lets set up an appropriate description of our charge distributionso that we may proceed with each of the above situations. It is most usefulto express the charge distribution in terms of a charge density :

=Q

43R

3

This is the charge per unit volume as found by merely dividing the totalcharge Q by the total volume of the sphere. We now handle the outside the

sphere configuration. All of the charge inside the sphere may be consideredas though it is concentrated at a single point. We may then use the CoulombsLaw result for the force acting on the point charge q:

F =1

40

Qq

r2

We integrate to find the potential energy:

V = Qq40

r

1

r2dx =

Qq

40

1

r

This result is also identical to the one obtained in the text in the case of thetwo-point charge configuration. We see that this should in fact be the casesince we started with the same quantity of force.

Now, to handle the inside the sphere situation, we shall use the abovestated charge density. We easily see that the distance r from the center of the

3



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4

distribution of charge to the point charge is less than the radius R. Imagine

now a sphere of radius r that is inside the larger spherical charge distributionbut centered at the same point. Only the charge within this inner sphere willact on the point charge. We may now use our charge density to find thecharge contained in the smaller sphere:

Qin =4

3r3

The force on a charge q due to the inner sphere is then given by CoulombsLaw:

F =q

40

4r 33r2

=qr

30

Before we integrate to find the potential energy, we must make an importantdistinction from the outside the sphere procedure above. We were able tointegrate directly from infinity to the location of our point charge because theamount of source charge taken into account did not change (Q). However, aswe enter the sphere, the amount of source charge begins to decrease until wereach our point at distance r from the center. We split the integration intotwo parts corresponding to both regions to find the potential energy:

V = [R

Qq

40

1

r2dr +

rR

qr

30dr]

= Qq40

1R q

60(r2 R2)

Note that the first integral above was identical to the one for the outsideconfiguration, just evaluated at r = R. Now, we just substitute our chargedensity and simplify:

V =Qq

40

1

R Qqr

2

80R3+

Qq

80R

=3Qq

80R Qqr

2

80R3

3. Looking at the definition of kinetic energy:

KE =1

2mv2



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5

Lets solve for velocity:

v =

2(KE)m

Let KE = 4KE.

v =

2(KE)

m=

2(4KE)

m= 2

2(KE)

m

= 2v

So, the speed will increase by a factor of 2 if the kinetic energy is increasedby a factor of 4. By the definition of momentum:

p = mv

m(2v) = 2p

We should also expect the momentum of the particle above to increase bya factor of 2. In fact, as long as the mass does not change, the momentumshould increase or decrease by the same factor as the changing speed.

5. The frequency and wavelength of light are related by equation I.23:

f =c

The constant crefers to the speed of light which we know to be 3.00108 m/s.Taking care to convert our wavelength, which is given in nanometers, to

meters:f =

3.00 108 ms500 109m = 6 10

14 Hz

7. Equation I.25 relates the photon energy to the wavelength:

E =hc

Substituting the value hc = 1240 eVnm:

E =1240 eVnm

500nm= 2.48 eV

9. The energy of a quantum of light (the photon) must be equal to thedifference in energies of the two states:

Ephoton = E2 E1



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6

According to equation I-25, the energy also obeys the following:

Ephoton =hc

Equating the two and solving for , we get:

=hc

E2 E1



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1

The Wave-Particle Duality -Solutions

1. The energy of photons in terms of the wavelength of light is given by eq.1.5. Substituting the given wavelength into eq. 1.5:

Ephoton =hc

=

1240 eV nm200nm

= 6.2 eV

3. Considering that we are given the power of the laser in units of milli-

watts, the power maybe expressed as: 1 mW = 1 103

J/s. This gives usa strong hint how to proceed. We are hoping to find the rate of emissionof photons, so that if we can find the energy of a single photon, we can usethe power as stated above to calculate the number of photons. The energyof a single photon can be calculated as in Problem 1 by substituting thewavelength of the light into equation 1.5:

Ephoton =hc

=

1240 eV nm632.8 nm

= 1.960 eV

We now convert to SI units:

1.960 eV

1 J

1.6 1019 eV= 3.14

1019 J

Now, using the given laser power:

Rate of emission =power

energy per photon=

1 103 Jsphoton

3.14 1019J = 3.181015 photons

s

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8 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS

5. This problem concerns the photoelectric effect. Given the work functionof the material and the emitted electron kinetic energy, we wish to calculatethe wavelength of the light incident to the surface. Equation 1.6 providesthe following:

(KE)max =hc

W where W is the work f unction of the material.

The hc/ term describes the energy supplied by the incoming photons. Byviewing the work function as an energy threshold for producing photoelectriccurrent, we see that the amount by which the photon energy (hc/) exceedsthe work function is the resultant maximum kinetic energy of the emittedelectrons. We may thus write:

hc

= (KE)max + W

= 2.3 eV + 0.9 eV

= 3.2 eV

Using hc = 1240 eV nm,

= 1240 eV nm3.2 eV

= 387.5 nm

6. Here, we are given the stopping potential of a photoelectric experimentand wish to determine the work function of the metal. Since 0.72 eV is thenecessary potential energy to cease the flow of electrons, the maximum ki-netic energy of the electrons being emitted must equal 0.72 eV. Our problemis then reduced to solving eq. 1.6 for the work function:

W =hc

(KE)max =1240 eV

nm

460 nm 0.72 eV = 1.98 eV

11. For this atomic transition, the energy of the emitted photon must equal



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9

the difference in energy of the two states of hydrogen (n = 2 and n = 5).

Equation 1.22 gives us those energies, thus:

Ephoton = E5 E2 = 13.6 eV52

(13.6 eV22

) = 2.86 eV

From eq. 1.12:

=hc

E=

1240 eV nm2.86 eV

= 434.2 nm

15. We must first find the energy of the UV photons, which may be done

via eq. 1.5:

Ephoton =hc

=

1240 eV nm45 nm

= 27.6 eV

If electrons are to be emitted, we need to overcome the energy that is keepingthem in a bound state. For hydrogen atoms, equation 1.22 gives us thatenergy. We need only supply the n quantum number for the state of theatom (for ground state, n = 1):

E = 13.6 eVn2

= 13.6 eV12

= 13.6 eV

The kinetic energy of these emitted electrons should then be equal to thedifference between the energy provided by the incident light and the groundstate electron energy: KE = 27.6 eV 13.6 eV = 14.0 eV For the velocitycalculation, it will be useful to convert from eV to J:

14.0 eV 1.6 1019J

1 eV= 2.24 1018J (1.1)

We can then find the velocity using the definition of kinetic energy andthe mass of the electron:

KE =1

2mV2 v = 2(KE)m = 2.24 10

18J9.11 1031kg = 2.21 106

m

s (1.2)

17. It should be first noted that the wavelength of the emitted light increases



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as the photon energy decreases. Thus, the longest acceptable wavelength

will correspond to a transition to the nearest excited state. Transitions tohigher-lying states will require more energy and thus correspond to a shorterwavelength. We then examine the energy of the transition from the n = 3state of the hydrogen atom. Equation 1.22 gives:

E3 = 13.6 eVn2

= 13.6 eV32

= 1.51 eV

The next available state is the state corresponding to n=4 whose energy maybe found by eq. 1.22 again:

E4 =

13.6 eV

n2

=

13.6 eV

42

=

0.85 eV

So, in order for the absorption to take place, the incident photons mustcontain at least as much energy as the difference in these energies (0.85 eV(1.51 eV) = .66 eV). We may now solve eq. 1.5 for the wavelength:

=hc

E=

1240 eV nm.66 eV

= 1878 nm

Thus, any light of wavelength greater than 1878 nm would not possess suffi-cient energy to be absorbed by the hydrogen atom.

19. Using a procedure such as that of Example 1.5, we can relate the kineticenergy to the de Broglie wavelength by the following equation:

=h

2m(KE)

We need only convert the kinetic energy from eV to J and substitute theappropriate masses.

KE = 20 eV 1.6 1019 J

1 eV= 3.2 1019J

Electron : = 6.626 1034Js2(9.11 1031kg)(3.2 1018J) = 2.74 10

10m

Proton : =6.626 1034Js

2(1.67 1027kg)(3.2 1018J) = 6.41 1012m



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11

An -particle is the most commonly occurring isotope of the helium atom

consisting of two protons and two neutrons. We can thus approximate itsmass: m = 2(1.672 1027) + 2(1.674 1027) = 6.692 1027kg.

particle : = 6.626 1034Js

2(6.692 1027kg)(3.2 1018J) = 3.20 1012m

24. The Davisson-Germer experiment measured the scattering of electronsby a crystal during which the interference patterns characteristic of light wereshown to also occur with particles. We can solve Braggs Law (eq. 1.24) forthe inter-atom spacing of the crystal in terms of the de Broglie wavelength

and the scattering angle:

2d sin() = n d = n2 sin()

We may use the relation derived in example 1.5 to find the de Broglie wave-length from kinetic energy after first converting our kinetic energy to SIunits:

KE = 54 eV 1.6 1019

1 eV

=

h2m(KE) =

6.626

1010m

2(9.11 1031kg)(8.64 1018 J) = 1.66 1010

m

We let n=1 and solve Braggs Law for d, the spacing in the crystal:

d =

2 sin()=

1.66 1010m2 sin(50)

= 1.09 1010m



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The Schrodinger WaveEquation - Solutions

1. Equation 2.17 gives the energies for the particles in an infinite potentialwell:

E =n2h2

2mL2

Part (a) asks for the size of the region, which we can find by solving eq. 2.17for L:

L =

n2h2

8mE

We need only supply some information such as the energy (1.0 eV as given),the mass of the electron (9.11 1031kg from Appendix A), and the appro-priate value of n. The lowest energy will be found when the electron is atthe ground state (n=1). We must also convert 1.0 ev to 1.6 1019J. Nowwe are ready:

L =

12(6.63 1034 Js)2

8(9.11 1031 kg)(1.6 1019 J) = 6.14 1010 m

While the derivation of eq. 2.17 is correct, lets do a little dimensional anal-ysis to better see through our length calculation above. The quantum

number n is dimensionless as is of course the constant 8, so lets examine theoperations of units of the rest of the quantities:

n2h2

8me

J2 s2kg J =

J s2

kg

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14 2. THE SCHRODINGER WAVE EQUATION - SOLUTIONS

Now substitute 1 J = 1 kg m2

s2 s2:

J s2

kg=

kg m

2

s2 s2

kg=

m2 = m

It should be emphasized that the m here is the abbreviation for the unit oflength meters, not the quantity of mass.

Part (b) can be approached in a couple of different ways. The essentialquantity we must obtain now is the energy of the first level above groundstate (also called the first excited state). The energy required to excite theelectron must be:

E = En2

En1 = E2

E1

The quantum numbers n=2 and n=1 correspond to the first excited stateand ground state respectively. So, for E2 from equation 2.17:

E2 =22h2

8mL2=

(6.63 1034 Js)22(9.11 1031 kg)(6.14 1010 m)2

= 2.14 1018 J 1 eV1.6 1019 J

= 4.00 eV

Here we used the result of the size of the well from part (a). The othermethod would involve solving eq. 2.17 for everything except n and using ourgiven quantities as follows:

E =n2h2

2mL2 h

2

8mL2=

E

n2

h2

8mL2= (1.0 eV)(12) = 1.0 eV

En = n2(1.0 eV) E2 = 4.0 eV

Now for our desired result:

E = 4.0eV 1.0eV = 3.0eV



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3. As with the atomic transitions we dealt with in chapter 1, the emit-

ted photon energy will equal the difference of the energies of the two statescorresponding to n = 3 and n = 2. So our task becomes finding each of theseenergies via equation 2.17:

E3 =32h2

8mL2=

9(6.63 1034 Js)28(9.11 1031 kg)(10 109 m)2

= 5.43 1021 J 1 eV1.6 1019 J = .034 eV

E2 =22h2

8mL2=

4(6.63 1034 Js)28(9.11

1031 kg)(10

109 m)2

= 2.41 1021 J 1 eV1.6 1019 J = .015 eV

E = .034 eV .015 eV = .019 eVThis value of E will then be equal to the energy of the photon and we mayuse eq. 1.5 to calculate the wavelength of the light:

KE =hc

= hc

KE

=

1240 eV

nm

.019 eV = 65263 nm5. Note the typo, Draw the wave function... should be Find the wavefunction...

We are given the following information: finite well of depth 0.3 eV andthickness L = 10 nm, we are dealing with conduction electrons if GaASin their lowest state, and these electrons have an effective mass of 0.067times the electron mass. From section 2.3, we see the process of solvingthe finite square well which culminates in the numerical solution of tan =

(20/2) 1, for the even solutions, and cot =

(20/

2) 1 for the oddcase. Our variables and 0 correspond to the following expressions:

=kL

2, 0

2 =mV0L2

22(from eq. 2.30)

Notice that the latter of the two is a squared quantity where the former isnot, but well deal with that a little later on. We are interested in the ground



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state of the electron, which will correspond to n=1, leading us to an even

solution which may be found by equation 2.29:

tan =

0

2

2 1

In the text near the end of section 2.3, the value of 02 is given to be 13.2.

Our equation becomes:

tan =

13.22

2 1

By using a computer algebra system or calculator to graph the left-hand and

right-hand sides, we find the first intersection at 1.22. Thus, we maysolve = kL/2 for k:

k =2

L=

2(1.22)

10 109 m = 2.44 108 m1

For the equations outside the well, we must solve for as well. Lets firstsolve equation 2.24 for E:

k =2mE

2

2

E = k22

2m=

(2.44 108 m1)2(1.054 1034 J s)22(6.103 1032) kg

E = 5.419/times1021 J 1 eV1.6 1019 J = .034 eV

Now for (from eq. 2.26):

=

2(6.103 1032 kg)(.3 eV .034 eV)(1.6 1019 JeV )

(1.054 1034 J s)2 = 6.84108m1

We know have our necessary information for the wave equations:

(x) = B exp(6.84 10

8

m1

)x : x L

2A cos(2.44 108 m1)x : L2 x L2B exp(6.84 108 m1)x : x L2



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9. Outside the finite potential well, V = V0, thus our time-independent

Schrodinger equation is:

2

2m

d2

dx2+ V0 = E

If we multiply both sides by 2m/2:

d2

dx2 2mV0

2 = 2mE

2

Moving 2mE/2 to the LHS:

d2dx2

+ 2m(E V0)2

= 0

We make the following substitution:

k =

2m(E V0)

2

12

Since E is greater than V0, k is in fact a real number and our Schrodingerequation becomes:

d2

dx2 + k

2

= 0And, we may readily confirm by substitution that the general form of thesolution of this equation is a linear combination of the functions, A cos(kx)and B sin(kx), and is oscillatory in nature.

11. The normalized wave function must satisfy the normalization condition(equation 2.18):

|(x)|2 dx = 1

Ae

mx2

\2

Aemx2

\2

dx = A

2emx2

\

Therefore,1

A2=

emx2\dx = 2

0

emx2\dx



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This looks very similar to the integral we are given if we let a = m/. Then:

1

A2=

m

=

m

A =m

14

13. a) From 0 x L, V = 0, and our Schrodinger equation becomes:

2

2m

d2

dx2= E or

d2

dx2+

2mE

2 = 0

For x L, V = V0 and:

2

2m

d2

dx2+ V0 = E or

d2

dx2 2(V0 E)m

2 = 0

b) If we let:

k =

2mE

2and =

2m(V0 E)

2

Then,d2

dx2+ k2 = 0 ; 0 x L

d2

dx2 k2 = 0 ; x

L

Similar to the situation of the finite well discussed in the text, the aboveequations are thus satisfied by the following:

(x) = A cos(kx) and (x) = A sin(kx) ; 0 x L(x) = Bex ; x L

Once again, the negative argument in the exponential is to insure our wavefunction decays as x .

c) If the potential is to be infinite at x = 0, then the wave function mustgo to zero at x = 0. Therefore, we must impose new boundary conditions

to insure the continuity of the wave function. Above we had both even andodd solutions for the wave function, so lets examine their behavior at x = 0given the new conditions:

Even : (0) = A cos(k0) = 0



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Odd : (0) = A sin(k0) = 0

We see immediately that the fist of the equations is not physically acceptable,so we must reject the even solutions, thus we are left with: (x) = A sin(kx)for 0 x L. We may now proceed with our boundary conditions at x = L:

(L) = A sin(kL) = BeL

Imposing the continuity of the first derivative:

Ak cos(kL) = BeL

Divide eq. 2 by eq. 1:k cot(kL) = cot(kL) =

k

We insert our as defined in part (b):

cot(kL) =

2mV02k2

2mE2k2

Substitute k into the second term of the RHS:

cot(kL) =

2mV02k2

1

Let = kL: cot() =

2mV0L2

22 1

We can transform this into a form similar to that of eq. 2.31 if:

02 =

2mV0L2

2

Our equation to solve then becomes:

cot() =

02

2 1

This can be solved graphically for the values of for which the LHS and RHSexpressions intersect. From here we obtain the corresponding values of k:

k =

L



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And solve eq. 2.24 for E:

E =k22

2m

Note that the value for 02 which we derived above is slightly different from

that of eq. 2.30. This occurs because our potential well for this problem isactually oriented from x = 0 to x = L, as opposed to x = L/2 to x = L/2in the example of section 2.3. This in turn gave us a different argument of thecotangent function and thus necessitated we choose a different 0

2 in orderto keep the same clean form of eq. 2.31 to solve numerically. Procedurally,the process we just followed is identical to that of section 2.3.

15. The infinite potential well represents a situation where the potential

energy does not evolve with time. The time-dependent solutions must thensatisfy eq. 2.45:

(x, t) = uE(x)eit where w =

E

From eq. 2.20, the normalized spatial wave functions of a particle in theinfinte well are given by:

(x) =

2

Lsin

nxL

The corresponding energies are given by eq. 2.17:

E =n2h2

8mL2

We may then calculate from eqs. 2.39 and 2.17:

=E

=

n2h

4mL2

Our total wave function is then:

(x, t) = 2L

sinnxL

expin2ht

4mL2



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3

Operators and Waves -Solutions

1. We may obtain the momentum by multiplying the momentum operatoras given in equation 3.2

p = i ddx

times the wave function given in the problem

p(x) = i ddx

Aeix = i2Aeix

p(x) = Aeix

Our wavefunction is therefore seen to be an eigenfunction of the momentumoperator corresponding to the eigenvalue . Thus, a measurement of themomentum of the electron in this state would yield a value of .

The kinetic energy may be found via the relation:

KE =p2

2m

The square of the momentum operator that appears here in the numeratorwill result in the operator begin applied twice, therefore:

p2 = (i)2d2

dx2 = 2d2

dx2

We substitute this expression into the expression of the kinetic energy:

KE =p2

2m=

2

2m

d2

dx2

21



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22 3. OPERATORS AND WAVES - SOLUTIONS

and operate on the wavefunction:

22m

d2dx2

= 22m

d2

dx2Aeix

=2

2mA2eix KE =

22

2m

3. Our task is to find = A1(x) + B2(x) such that:

p = p

Where p is the eigenvalue of the momentum operator. So, lets try operatingon the following wavefunction:

= A cos(kx) + B sin(kx)

p = iddx

= iA ddx

cos(kx) iB ddx

sin(kx)

= ik Asin(kx) ik cos(kx)This choice of is not generally an eigenfunction of p, as we can not factorout our original = A cos(kx) + B sin(kx). Let us now try the following:

= cos(kx) + i sin(kx)

p = ik(sin(kx)) i2k(cos(kx)) = k cos(kx) + ik sin(kx)= k(cos(kx) + i sin(kx))

Thus, we have found an eigenfunction of the momentum operator whichcorresponds to the eigenvalue k. Note, that using the Euler formula:

eikx = cos(kx) + i sin(kx)

Our wavefunction may be written:

= eikx

Using this form of the wavefunction and the result of Problem 1, we may



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23

thus see that the wavefunction is an eigenfunction of the momentum opera-

tor corresponding to the eigenvalue

k.

5. Notice that if the energy is less than the minimum value of V(x), the[V(x)E] portion of the RHS is always positive. Therefore, if this equationholds, the second derivative of the wavefunction will have the same sign asthe wavefunction itself, and the function will increase or decrease in valuemonotonically.

7. Sections 3.3.1 and 3.3.2 provide the initial setup for this problem. Wemust solve the appropriate Schrodinger equation for each region, then useboundary conditions to find our desired ratios. In the first region, where

V = 0, we have the following from section 3.3.1:

d21dx2

+ k121 = 0

and given by eq. 3.32:

k1 =

2mE

2

This yields solutions:1 = Ae

ik1x + Beik1x

Region 2 produces equations 3.39 through 3.40:

d22dx2

k222 = 0 where k2 =

2m(V0 E)2

With physically acceptable solutions:

2(x) = Dek2x

Now, we impose continuity of the wave equations at x = 0 and of their firstderivatives at x = 0:

1(0) = 2(0)

A + B = D

1(0) = 2

(0) ik1(AB) = k2DDivide both side of the second equation by ik1 and our two equations become:

A + B = D (1)



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AB = k2ik1

D = ik2k1

D (2)

Here, the fact that 1/i = i was used to bring i out of the denominator andcancel the negative. If we add these two equations, we solve for A in termsof D:

2A = D(1 + ik2k1

)

A =1

2D(1 + i

k2k1

) (3)

If we subtract (1) from (2), we can solve for B in terms of D:

2B = D(1

ik2

k1)

B =1

2D(1 i k2

k1) (4)

From here, we divide each (3) and (4) by D:

A

D=

1

2(1 + i

k2k1

)

B

D=

1

2(1 ik2

k1)

The reflection coefficient R is found by

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

=|B|2|A|2

Using our ratios above:

R =( B

D)( B

D)

( AD

)( AD

)=

14

(1 i k2k1

)(1 + i k2k1

)14

(1 + i k2k1 )(1 i k2k1 )= 1

9. Following figure 3.6, we are given the general solutions for each region:

1(x) = Aeik1x + Beik1x, x 0

2(x) = Cek2x + Dek2x, 0 x L

3(x) = Eeik1x, x 0



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25

We impose the boundary conditions at x = 0 and x = L:

1(0) = 2(0)

1(0) = 2

(0)

2(L) = 3(L)

2(L) = 3

(L)

The first two yield the following system of equations:

A + B = C+ D

ik1(AB) = k2(C D)We divide the second of these by ik1 and add the result to the first to obtainan expression for A in terms of C and D:

2A = C+ D ik2k1

(CD) (1)

Now, lets turn to the interface between regions 2 and 3. The boundaryconditions produce the following system:

Ce

k2L

+ Dek2L

= Ee

k1L

k2Cek2L k2Dek2L = ik1Eek1L

Dividing the second of these equations by k2 reduces to the system to a moremanageable one:

Cek2L + Dek2L = Eek1L

Cek2L + Dek2L = ik1k2

Eek1L

Add the two equations and we can solve for C in terms of E:

2Cek2L = Eek1L + i k1k2

Eek1L

C =E

2

1 + i

k1k2

e(ik1k2)L (2)



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Subtract those same equations to solve for D in terms of E:

2Dek2L = Eek1L ik1k2

Eek1L

D =E

2

1 ik1

k2

e(ik1+k2)L (3)

Insert (2) and (3) into (1):

2A =E

2

1 + i

k1k2

e(ik1k2)L +

E

2

1 ik1

k2

e(ik1+k2)L

i

k2

k1

E

21 + ik1

k2 e(ik1k2)L + 1 i

k1

k2 e(ik1+k2)L

Divide both sides by A and simplify:

4 =E

A

1 + i

k1k2

ik1k2 1

eik1Lk2L +

E

A

1 i k1

k2

+

i

k1k2

+ 1

eik1L+k2L

=E

A

2 + i

k21 + k22

k1k2

eik1Lk2L +

2 + i

k22 k21k1k2

eik1L+k2L

Solve for E/A:

EA

= 4

2 + ik21 + k

22

k1k2

eik1Lk2L +

2 + ik

22 k

21

k1k2

eik1L+k2L

1

11. The Heisenberg uncertainty principle states:

Et 2

Therefore we solve for E:

E =

2t

=6.626 1034 J s

2(4.0 1010 s)= 8.28 1025 J



27/96

27

Now, converting to electron volts:

E = 8.28 1025 J 1 eV1.6 1019 J = 5.18 10

6 eV

13. Equation 3.48 gives the average value of an observable:

< Q >=

(x)Q(x) dx

Where Q is the corresponding operator, in this case the kinetic energy oper-

ator, which is given by: p2

2m=

2

2m

d2

dx2

The integral becomes:

< KE > =

10

Bex

2

2m

d2

dx2

Bex dx

= 2

2mB2

10

e2xdx

=

2

4m

B2 e2 1

where B is the normalization constant found in problem 12.



28/96




29/96

4

The Hydrogen Atom -Solutions

2. Per equation 4.9, the radial probability density for the 1s state hydrogenis found by:

|P1s(r)|2 =

2

1

a0

r

a0er/a0

2=

2r2

a30e2r/a0

The 1s radial wave function used here is listed in table 4.2 in the row corre-sponding to n=1 and l=0. Since hydrogen is the atom of interest, we used theappropriate atomic number Z=1. We shall maximize the above expression

by taking the first derivative and equating to zero:d

dr|P1s(r)|2 = 4r

a30e2r/a0 +

2r2

a30

2a0

e2r/a0 = 0

=4r

a30e2r/a0 4r

2

a40e2r/a0 = 0

Divide both sides by e2r/a0 and solve for r:

4r

a30 4r

2

a40= 0

4r

a30

1 r

a0

= 0

Therefore we have one solution at r = 0 and one at r = a0. Lets test theintervals (0, a0) and (a0,) to reveal the nature of these points. Let r = a0/2

29



30/96

30 4. THE HYDROGEN ATOM - SOLUTIONS

for the former and r = 2a0 for the latter:

d

dr|P1s(r)|2

x=a0/2

=4

a30

a02

e1 4

a30

a204

e1

= e1

2

a20 1

a20

> 0

d

dr|P1s(r)|2

x=2a0

=4

a30(2a0) e

4 4a30

4a20

e4

=8

a20e4(1 2) < 0

Since the first derivative of the radial probability density takes a positive val-ues for (0, a0) and negative values for (a0,), we may conclude that r = a0is in fact the maximum value. This is consistent with the description of theBohr radius as the innermost orbital radius of the electron in the hydrogenatom.

4. Starting with equations 4.7 and 4.8,

dP = |(r)|2 r2 sin() dr d d

We construct our wavefunction by equation 4.4:

(r,,) =Pnl(r)

rlm() m(, )

Pnl(r), lm(), and m(, ) for the 1s state may be found in tables 4.1 and4.2 in rows corresponding to n=1, l=0, and ml=0 (Note that ml=0 is the onlypossibility for l=0 thus only one row for l=0 appears in table 4.1). Carryingon,

(r,,) = 2r

1

a0

r

a0

er/a0 1

212

=1

a3/20

er/a01



31/96

31

We now integrate to find the probability:

P = |(r,,)|2 r2 sin() dr d d

=1

1

a30

a00

r2e2r/a0 dr0

sin() d

20

d

=

a00

4r2

a30e2r/a0 dr

Grouping 1/a0 with r and dr:

P = 2a0

0 2r

a02

e2r/a0 dr

a0Substitute = r

a0:

P = 2

10

2e2 d

This integral may be solved analytically using integration by parts:10

2e2 d = 12

e221

0

10

2

1

2e2

d

= 12

e2 +1

2

e2

1

0

10

1

2

e2 d

= 14

1 5e2

5. For a given n-state, the possible values of l are n-1, n-2, . . . 0. Therefore,n=5 allows l=0,1,2,3,4. For any given l value, the possiblities for the mag-netic quantum number are ml = -l, -l+1, . . . 0, 1, . . . l. The results are asfollows:

l = 0 : m = 0

l = 1 : m =

1, 0, 1

l = 2 : m = 2,1, 0, 1, 2l = 3 : m = 3,2,1, 0, 1, 2, 3l = 4 : m = 4,3,2,1, 0, 1, 2, 3, 4



32/96


7. Recall from equation 4.4 our construction of a wave function as a productof radial and angular parts:

(r,,) =Pnl(r)

rlml()()

Our strategy will be to express the given wave function in terms of radial andangular parts, then use their properties to check whether the wavefunctionis an eigenfunction of the Schrodinger equation and find the correspondingenergy. For our given wavefunction, we see that we may separate it in termsof dependence on and dependence on r as cos() and r2eZr/2a0. Note

that the presence of r2 is due to fact the radial part is to be divided by r ineq. 4.4 to construct the total wave function. We need not incorporate theconstant C as it will not affect whether the wave function is a solution of theSchrodinger equation. For the angular part, note that in table 4.1, cos()appears in the spherical harmonic corresponding to l = 1 and ml = 0.When we distribute the left-hand side of eq. 4.4, we get:

22m

d2

dr2Pnl(r) +

2l(l + 1)

2mr2Pnl(r) 1

40

Ze2

rPnl(r)

Taking the second derivative of the radial wave function:

d2dr2

Pnl(r) = 2eZr/2a0 2rZ

a0eZr/2a0 + Z2r2

4a0eZr/2a0

Upon substitution and using l = 1, the left-hand side of eq. 4.4 becomes:

22m

d2

dr2Pnl(r) +

2l(l + 1)

2mr2Pnl(r) 1

40

Ze2

rPnl(r)

= 2

meZr /2a0 +

2rZ

ma0eZr/2a0 +

2Z2r2

8ma0eZr /2a0 +

2

meZr/2a0 Ze

2r

40eZr/2a0

=

2rZ

ma0 eZr/2a0

Ze2r

40 eZr/2a0

+

2Z2r2

8ma0 eZr /2a0

Recall from equation 1.20 that the Bohr radius a0 may expressed as follows:

a0 =40

2

me2



33/96

33

If we substitute this expression for a0 into the first term of the previous

equation, the first two terms cancel and equation 4.5 becomes:2Z2

8ma0r2eZr/2a0 = EPnl(r)

The radial wave function is in fact an eigenfunction of the Schrodinger equa-tion with corresponding eigenvalue (energy):

E =2Z2

8ma0

9. We shall start with eq. 4.5:22m

d2

dr2+

2l(l + 1)

2mr2 1

40

Ze2

r

Pnl(r) = EPnl(r)

And 1.20:

a0 =40

2

me2

The substitution as directed is = r/a0, so we can directly substitute for F:r = a0. We let Z=1 as is the case hydrogen and carry out the substitution:

22m

d2

d(a0)2+

2

l(l + 1)2m(a0)2

140

Ze2

a0

Pnl(r) = EPnl(r)

Since a0 is a constant, we may bring it outside the derivative in the first termon the left-hand side and substitute eq. 1.20 for one a0 in each of the firsttwo terms.

1a0

2

2m

me2

402d2

d2+

1

a0

2l(l + 1)

2m(a0)2me2

402 1

a0

Ze2

a0

Pnl(r) = EPnl(r)

After making the appropriate cancellations and factoring we have:

e2a0

140

1

2d2

d2+ 1

2l(l + 1)

2 1

Pnl(r) = EPnl(r)

1

2

d2

d2+

l(l + 1)

22 1

Pnl(r) =

E

(1/40)(e2/a0)Pnl(r)



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12. The transition integral for the general case is given byn1l1m1l

z

a0

n2l2m2l dV

We are interested in the transition 3d1 2p1, so the transition integralbecomes:

3,2,1

z

a0

2,1,1 dV

From here on, we assume that the radial distance will be given in terms

relative to a0, so we take a0 = 1 and constructs the wavefunctions by tables4.1 and 4.2.

3,2,1 =P3,2(r)

rY2,1(, )

=2

2

81

15r2er/3

15

8sin()cos()ei

2,1,1 =P2,1(r)

rY1,1(, )

=

1

26rer/2

3

8 sin()e

iNow, we take the complex conjugate of 3,2,1, substitute z = rcos(), andproceed with the integration:

3,2,1z2,1,1 dV =1

648

r4e5r/6sin2()cos2() dV

=1

648

0

0

20

r6e5r/6sin3()cos2() dr d d

The integration may be separated as follows:

1

648

0

r6e5r/6 dr0

sin3()cos2() d

20

d =1

648

40310784

15625

4

152

= 2.123



35/96

35

13. In order to find the transition coefficient for hydrogen, we will use the

results found in Appendix FF (online) along with eqs. 4.20 and 4.21. WeAccording to the selection rules, summarized in table 4.3, ml = 1 for x-and y-polarized light. Therefore, we must calculate the transition integralsfor both the 2p1 1s0 and 2p1 1s0 cases. As directed in Appendix FF,we use the following in place of the z operator in eq. 4.21:

x =1

2(r+ + r)

y =1

2i(r+ r)

We first solve the case for x-polarized light.

I12 = 2|1s0x2p1|2 + |1s0x2p1 |2

Which gives us the following when we substitute the above expression for x:

1s0x2p1 =1

2(1s0r+2p1 +

1s0r2p1)

1s0x2p1 =1

2(1s0r+2p1 +

1s0r2p1)

Since 1s0r+2p1 and 1s0r2p1 will equal zero, we may use the results givenin Appendix FF for the remaining terms, giving us:

I12 = 2

1

6R2i +

1

6R2i

=

2

3R2i

Using the results of eqs. 4.23 and 4.26 into eq. 4.22:

A21 =6.078 1015

(121.6)31.109

6= 6.25 108 per atom per second

We may follow the same procedure for y-polarized light. First, the transitionintegral:

I12 = 2|1s0y2p1|2 + |1s0y2p1|2



36/96


Using the operator substitution:

1s0y2p1 = 12i(1s0r+2p1 1s0r2p1)

1s0y2p1 =1

2i(1s0r+2p1 1s0r2p1)

Once again, the nonzero terms lead to:

I12 = 2

16

R2i +16

R2i

=23

R2i

The result for y-polarized light is then:

A21 = 6.078 1015

(121.6)31.109

6= 6.25 108 per atom per second

15. a) Lets imagine that the angular momentum vector l points from theorigin and lies on the surface of a cone, then the radius of the circle at thebase of the cone is:

r = |l|sinAnd, for a change of the azimuthal angle d, the distance that the tip of theangular momentum vector moves is given by:

d|l| = |l|sin d

b) We start with the relationship of the torque to angular to momentum:

d|l|dt

= ||

This equation together with eq. 4.40 immediately leads to the equation:

d

|l

|dt =e

2m |lB|The definition of the vector product then gives:

d|l|dt

=e

2m|l|B sin()



37/96

37

Substituting d|l| = |l|sin d, we may then find our Larmor frequencies:

|l|sin()ddt

=e

2m|l|B sin()

L =d

dt=

eB

2m

17. The total angular momentum of the 4f electron will contain contribu-tion form both the electrons spin and its orbital angular momentum. Section4.3.4 shows that the total angular momentum may take on the values:

J = j1 +j2, j1 +j2 1, . . . |j2 j1|

We let j1 be our orbital angular momentum and j2 the spin. The electronoccupies an f-state, therefore j1 = 3, and the intrinsic spin of the electronmeans j2 = 1/2. We may now list the possible values for J:

J =7

2,

5

2

The spin-orbit coupling energy is then found by:

Eso = 2

2l = 3

2

2for j = l + 1

2

Eso =2

2(l + 1) = 22 for j = l 1

2

19. Using the two equations following eq. 4.50, the separation in energiesdue to spin-orbit coupling is:

E =2

2

l

2

2 (l + 1) =

2

2 (2l + 1)

Since we are using the atomic system of units, = 1 and

=

2(2l + 1) (1)



38/96


We then let l=1 (due to the occupation of a p-orbital):

E = 32

= 7.9 106

= 5.27 106

The spin-orbit constant for the levels neon may be calculated directly byeq. 4.48, however we may use equation 4.51 to set up a relationship betweenHe and N e. Since He

+ and Ne9+ are both hydrogenic atoms (containingonly one electron), the quantity

a0

Zr3

should be equal in both cases as the grouping of a0 and z with r will accountfor the scaling of the radial distance with changing nuclear charge. We nowuse the appropriate atomic numbers of helium and neon to generate a ratiobetween He and Ne .:

N eHe

=104

24= 625 N e = 625He = 3.29 103

For the separation between the 3p3/2 and 3p1/2 levels, we use equation (1)above:

E = 3.29 1032

(2l + 1) = 4.94 103

A comparison with the separation of the corresponding helium states revealsthe vastly increased impact of spin-orbit coupling in the case of neon.



39/96

5

Many-Electron Atoms -Solutions

1. An electron has an intrinsic spin s = 1/2, therefore the possible val-ues of ms are 1/2. The 4f state of the electron has an orbital angularmomentum quantum number l = 3 and therefore may have values ml =3,2,1, 0, 1, 2, 3.

In the case of two electrons occupying the 4f state, we first see from ourresults above, that there are 7 possible ml quantum numbers 2 ms quantumnumbers. There are thus 14 possible states in which to put the first electron.Once we assign first electron, there are now 13 possible states. Because the

order in which placed them does not matter in our configuration, we thendivide the product by two:

# of distinct states =14 13

2= 91

3. Using the form of eq. 5.6, we attribute each row of the determinantto a particular state and each column to a particular electron occupying acorresponding state (row). Our system contains these electrons, thereforeN=3 in our coefficient and

=

1

3

10(1) 10(2) 10(3)10(1) 10(2) 10(3)20(1) 20(2) 20(3)

39



40/96

40 5. MANY-ELECTRON ATOMS - SOLUTIONS

4. For each of these elements, we shall fill their shells in the following order:

1s, 2s, 2p, 3s, 3p, 4s, 4p, and so on. The number of electrons in a neutralatom of any element will equal the number of protons in its nucleus, thereforewe use the atomic number of each element as the number of electrons we useto fill these shells:

Flourine(Z = 9) : 1s2 2s2 2p5

Magnesium(Z = 12) : 1s2 2s2 2p6 3s2 [Ne] 3s2

Silicon(Z = 14) : 1s2 2s2 2p6 3s2 3p2 [Ne] 3s2 3p2

Potassium(Z = 19) : 1s2 2s2 2p6 3s2 3p6 4s1 [Ar] 4s1

Cobalt(Z = 27) : 1s2 2s2 2p6 3s2 3p6 4s2 3d7 [Ar] 4s2 3d7

An alternate (shorter) way to notate these, as Ive done above to the right,is to use the symbol of the inert gas that occurs closest before a particularelement to denote the filled shells through complete rows of the periodic tableand merely denote the filling of the remaining shells.

7. The nitrogen atom will have the following ground configuration: 1s2 2s2 2p3.Since the 1s and 2s orbitals have less energy than the 2p orbital, the nexthigher configurations are 1s2 2s2 2p2 3s and 1s2 2s2 2p2 3p.

8. ns2: Here are two electrons with l=0, therefore, L=0. Their spins pro-duce S=0,1. Since L+S must be even for two equivalent electrons, the onlypossibility is 1S (S=0, L=0).

nd2: Two electrons with l=2 may produce L=4,3,2,1,0. Again, the addi-tion of their spin angular momenta produce S=0,1. Imposing the conditionthat L+S must be even, we have the following configurations: 1G (L=4,S=0), 3F(L=3, S=1), 1D (L=2, S=0), 3P(L=1, S=1), and 1S (S=0, L=0).

4f2: These two electrons occupy f-states (l=3), thus may contribute toL=6,5,4,3,2,1,0. As has been the case above, S=0,1. Ipmosing the condi-

tion that L+S must be even, we have the following LS terms:1

I (L=6,S=0), 3H (L=5, S=1), 1G (L=4, S=0), 3F (L=3, S=1), 1D (L=2, S=0),3P(L=1, S=1), and 1S (S=0, L=0).

Lets apply Hunds rules to the 4f configuration. Of these terms, the ones



41/96

41

corresponding to the maximum value of S are 3H, 3F, and 3P. Of these, the

one with maximum value of L (

3

H) should lie lowest.

11. It is important to take note here that we have a case of non-equvalentelectrons for the [Xe] 4f 5d configuration and one of equivalent electronswith [Xe] 4f2. For [Xe] 4f 5d, one electron has l=3 and the other l=2.Since there is no additional constraint on the values of L and S except forthe rules of addition of angular momentum, L may take L=5,4,3,2, or1 andS=0,1.

For 4f2, we must impose the condition that L+S be even. The two l=3electrons could form L=6,5,4,3,2,1,0 and S=0,1. The allowed LS terms are

then (L=6, S=0), (L=5, S=1), (L=4, S=0), (L=3, S=1), (L=2, S=2), (L=1,S=1), and (L=0,S=0).

13. As described in section 5.5.2, the nuclear charge increases as we goacross a row in the periodic table, thus decreasing the average value of theelectrons distance from the nucleus. This causes an increase in the Coulombinteraction, thus we expect sulfur (S) to possess the greatest Coulomb inter-action of these three elements.

15. Recall from equation 4.51 that the strength of the spin-orbit interac-tion increases with atomic number as Z4. Since the spin-orbit interaction ismainly responsible for the breakdown of LS coupling, the O atom has thesmallest spin-orbit interaction and the states of O corresponds most nearlycorrespond to pure LS terms.

17. To find the total energy, we simply click on neon in the periodic ta-ble interface, then the red arrow on the bottom row of controls to run thecalculation. Then, click on the Averages tab to display the kinetic, po-tential, and total energies. We see a value of -128.55 for the total energy.We now remove a 2p electron by selecting the 2p shell via the incrementalobjects in the bottom row of controls, then clicking the (-e) button. Verify

you have removed the electron from the correct shell as the configuration inthe second to last row of controls should now read 1s2 2s2 2p5. Now, run theapplet again for the neon ion and finding the energy under the Averagestab, you should find the energy to be approximately 127.82. The binding



42/96

42 5. MANY-ELECTRON ATOMS - SOLUTIONS

energy is the difference of these two values:

|128.55 (127.82)| .73Going back to the neutral neon atom, the third column to the left on the pageobtained after striking the Averages tab gives a 2p single electron energy of0.85. This is very close to the difference in total energies of the neutral Neatom and the ion.

19. We can calculate the Slater integrals for Si, P, and S by first select-ing the appropriate element in the periodic table of the Hartree-Fock applet,clicking the red arrow to run the calculation, then clicking the Other tab.Now, the second box should be changed to 2, then continuing across: 3,p, 3, p. The red arrow on the right runs the calculation. We obtainthe following values.

F2(3p, 3p)Element au cm1

Si .166 36418P .197 43180S .220 48323

We see a definite increase along the row, constitent with the qualitativeresult obtained in problem 13.

21. Using fig. 5.5, the energy of 1S of the 1s 3s configuration is approxi-mately 185,000 cm1 while 1P has an energy of approx. 170,000 cm1. Theenergy of the emitted photon should equal the difference of the energy levels,thus:

E = 25, 000 cm1

According to equation 5.12, 1 eV of photon energy corresponds to light ofwave number 8065.54 cm1. The photon energy emitted in this transitionmust equal:

25, 000 cm1

8065.54 cm1/eV= 3.10 eV

Using the relation E = hc/,

=hc

E=

1240 eV nm3.10 eV

= 400 nm



43/96

6

The Emergence of Masers andLasers - Solutions

1. As given in figure 6.7, the range of the 4F2 band is approximately17, 000 cm1 to 19, 000 cm1 . As these quantities are given in terms of1/, we may quickly obtain the corresponding frequencies in Hz by using therelation I.22:

c = f f = c

Therefore, we may convert c to cm/s and simply multiply:

17 103 cm1 3 1010 cms = 5.1 1014 Hz19 103 cm1

3 1010 cm

s

= 5.7 1014 Hz

Similarly, for 4F1:

23 103 cm1

3 1010 cms

= 6.9 1014 Hz

27 103 cm1

3 1010 cm

s = 8.1 1014 Hz

3. The 2p core is given to be in the state 2P3/2. Using the following schemefor spectroscopic notation:

2s+1LJ

43



44/96

44 6. THE EMERGENCE OF MASERS AND LASERS - SOLUTIONS

we take J = 3/2. Since the excited electron (3p) has orbital angular momen-

tum l=1, the possible values for K are:

K = J + l, J + l 1, . . . |J l| = 52

,3

2,

1

2

The spin orbit interaction spits each level with a value of K into two levelscorresponding to the total angular momentum having the values K 12 .

5. The situation described in problem 4 gives I = 1/2 and the fact thatwe are dealing with hydrogen which means we have one electron, for whichS = 1/2. The possible values for F are then F = 0, for which MF = 0;and F = 1, for which MF =

1, 0, 1. Since the F=0 level is not split by a

magnetic field, we only calculate gF for F = 1.Using eq. 6.17,

For F=1, gF =(1 + 1) 12(12 + 12) + 12(12 + 1)

1 + 1=

9

8

Therefore, we consider the F=1 states via equation 6.5. Since gF is positive,the MF = 1 state will have a negative z value and be drawn to regions oflow magnetic field. Consequently, the MF = 1 will have a positive z valueand be drawn to regions of high magnetic field outside the magnetic trap.These atoms will be lost, decreasing the kinetic energy and temperature ofthe atoms caught in the trap.

7. Given S = 1/2 and I = 5/2, our possible values for F are F = 3 andF = 2. These levels split as follows:

F = 3 : MF = 3,2,1, 0, 1, 2, 3F = 2 : MF = 2,1, 0, 1, 2

As we see, the MF = 3 and MF = 3 states are not mixed with any stateswhere F = 2, thus will be pure states with straight lines. All other statesof F = 3 and all states where F = 2 will be mixed, thus will have curvedlines.



45/96

7

Statistical Physics - Solutions

1. As described in the text, there are sixteen possible outcomes of flippinga coin four times. The possible outcomes can be classified according to thenumber of heads. We may use equation 7.2 (or 7.3) to calculate the statisti-cal weight associated with each outcome by letting N = 4 and n equal thenumber of heads obtained in each distribution:

HHHH :

4

4

=

4!

4!0!= 1

HHHT :43 = 4!

3!1!= 4

HHTT :

4

2

=

4!

2!2!= 6

H T T T :

4

1

=

4!

1!3!= 4

T T T T :

4

3

=

4!

0!4!= 1

3. We may set up equation 7.7 for each energy level:

n2g2

=n22

=N

Ze2/kBT (1)

45



46/96

46 7. STATISTICAL PHYSICS - SOLUTIONS

n1g1

=n11

=N

Ze1/kBT (2)

Divide (1) by (2):n2n1

= 2e(21)/kBT

We are given 2 1 = 0.025 eV, T = 298K, and we use kB = 8.617 105eV/K:

n2n1

= 2 exp

0.025 eV(8.617 105 eV/K)(298 K)

= .755

5. F1(u) is defined in conjunction with Fig. 7.2:

F1(u) =4

u2eu2

We maximize F1(u) by setting the derivative of F1(u) with respect to u tozero:

dF1du

=4

2ueu

2 2u3eu2

= 0

2ueu2

(1

u2) = 0

u = 0,1We test the intervals (0, 1) and (1,):

F1

1

2

> 0 and F1(2) < 0

Thus, F1(u) has a maximum at u = 1.

7. To find the average value of v2, we calculate the integral:

(v2)av =

0

v2P(v) dv

Use equation 7.22 for P(v)dv:

(v2)av =

0

4v4

m

2kBT

3/2emv

2/2kBT dv



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47

We may use eq. F.2 in Appendix F to find I4(1) = 3

/8, thus (v2)av =

3kBT /m.

9. To find rms, we need to first find (2)av, then take the square root.

Analogous to example 7.3,

(2)av =

0

2P() d

Using equation 7.26:

(2)av =2

0

5/2

(kBT)3/2e/kBT d

Per equation 7.27, u = /kBT and we may write the above equation in termsof u:

(2)av =2

(kBT)2

0

u5/2eu du

At this stage, we can work the integral according to the procedure expressedin eqs. F.8-F.14:

(2)av =2

(kBT)2(7/2)

=2 (kBT)

2

(7/2)

=2

(kBT)2 5

2(5/2)

=2

(kBT)2 5

2

3

2(3/2)

=2

(kBT)2 5

2

3

2

1

2(1/2)

=15

4(kBT)

2

We now take the square root of this last quantity:

rms =

15

2kBT



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11. According to the equipartition of energy (end of section 7.3.1), eachparticle has an average kinetic energy of kBT /2 for each degree of freedom.For three degrees of freedom:

av =3

2kBT

Therefore, total kinetic energy of a mole of gas should be:

=3

2kBT NA =

3

2(1.3811023J/K)(298 K)(6.0221023particles) = 3717 J

13. Per equation 7.28,

u(f) =8f2

c3

hf

ehf/kBT 1

We apply the product rule and quotient rule of differentiation:

du(f)

df=

16f

c3

hf

ehf/kBT 1

+8f2

c3

h

ehf/kBT 1 h2fkBTehf/kBT(ehf/kBT 1)2

Lets set this expression equal to zero and multiply both sides by

ehf/kBT 12:16f2

c3

ehf/kBT 1 + 8f2c3

ehf/kBT 1 8f2

c3

hf

kBTehf/kBT

= 0

Multiply both sides by c3/8hr2 and let x = hf/kBT, then simplify:

2 (ex 1) + (ex 1) xex = 03ex xex 3 = 0 (3 x)ex = 3

15. Equation 7.28 gives the energy density as a function of frequency:

u(f) =8f2

c3

hf

ehf/kBT 1



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49

We may derive an expression as a function of wavelength by using relation

I.23 from the introductory chaper: f = c/.

u() =8(c/)2

c3

h(c/)

ehc/kBT 1

=8

3

h

ehc/kBT 1

19. According to eq. 7.50, the work done on the gas is given by:

dW = P dV

Per the ideal gas law, we solve for P:

P = nRTV

We then integrate to find W and change sign to reflect work done by the gas:

W = nRT

V3V2

dV

V= nRT ln

V3V2

Eq. 7.48 states dE = dQto + dWon. As found above, the work done by thegas is equal to nRT ln(V3/V2). If the energy remains constant, the amountof heat added to the gas must be equal to the work done by the gas and

dQto = nRT ln

V3V2

21. We first solve eq. 7.66 for Tc:

1

2.612

N

V

=

2mkBTc

h2

3/2

Tc = 1

2.612

N

V2/3 h2

2mkB

Substitute our given density:

N

V= 5.0 1014 atoms/cm3



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Tc = 1.16 109h2

mkB

=7.31 1017

m

We may supply the appropriate atomic mass to calculate the critical tem-perature. This equation may appear a bit peculiar due to a quantity oftemperature appearing on the left-hand side and one of mass on right. Wecan take a closer look at the units to more clearly see this relationship. Forclarity, lets use SI units of J, K, m, and kg for the necessary quantities:

Tc =

1

2.612

N

V

2/3h2

2mkB [Tc] = 1

m2J2 s2

kg J/K=

1

m

kg m/s2 s2 K

kg= K

23. Equation 7.76 gives the probability that an energy will fall between and + d:

P() d =3

23/2F

1/2 d1

e(F)/kBT + 1

Thus, we may substitute our values for Fand T then carry out the integration

numerically:

At T = 295K :3

2

4.13.9

33/21/21

e(F)/295kB) + 1d = 5.9 1017

At T = 3000K :3

2

4.13.9

33/21/21

e(F)/3000kB) + 1d = 5.91017 = 2.41103

It is seen here that the electron is much more likely to be found with anenergy between 3.9 ecV and 4.1 eV at 3000K.

25. We substitute our given quantities into the given formula, along withnc = 1 since gold is a monovalent metal:

N

V=

19.32 g/cm3(6.022 1023atoms/mole)197 g/mole



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51

N

V= 5.90 1022 electrons/cm3

27. (a) Using the formula from problem 24 and the fact that nc = 1 sincesodium is a monovalent metal:

N

V=

ncNAM

=(.971 g/cm3)(6.022 1023atoms/mole)

23.0 g/mole

= 2.54 1022 electrons/cm3

(b) To find the Fermi energy, we use equation 7.73 as well as the quantitywe just calculated in part (a):

F =h2

2m

3

8

N

V

2/3

=(4.13 1015)22(5.48 104)

3

8(2.54 1022)

2/3= 3.26 1012 eV

(c) The equation after example 7.6 relates the Fermi temperature to theFermi energy:

F

= kB

TF

TF

=F

kB

Now, we substitute the Boltzmann constant and the Fermi energy found inpart (c):

TF =3 eV

8.617 10 =5 = 35, 418K



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53/96

8

Electronic Structure of Solids -Solutions

1. The cesium choride lattice is depicted in Figure 8.5. If we take a lighter-shaded sphere to represent a cesium ion, we see that there are 8 chloride ionsthat serve as the nearest neighbors. Or, if the cesium ion serves as the centerpoint in a body-centered cubic lattice, the eight nearest neighbors are thecholoride ions at the corners.

Returning to Figure 8.5, recall that the designation of which shade cor-responds to which ion is arbitrary. Therefore, now alllow the darker spheresto be the cesium ions. By extrapolating the figure in one more direction, it

can be seen that cesium would have 6 next-nearest neighbors.

3. We refer to Fig. 8.8(b) where we will use vectors to denote the loca-tions of the carbon atoms from a fixed origin, then use vector relationshipsto determine the bond angle. First, we select our origin to be located at pointA in the lattice. The other two point of interest are point B and the atomlocated in the center of the top face shown. Our primitive i and j will beoriented along the edges of the top face (starting at point A) and the k vectorwill point directly upward from point A. We may now construct the positionvectors to each point. Let a be the position vector from the origin to the top

face-centered atom, while b points from point B to the origin (point A). It isour intent to construct a vector c pointing from the top face-centered atomto point B. We will then find the angle between b and c Thus:

a =1

2i +

1

2j, b = 1

4i 1

4j +

1

4k

53



54/96

54 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS

Our strategy wil be to find the angle between b and c, so lets first find c.

By inspection we see that: c = a + b

We substitute the expressions for a and b:

c = a + b =

1

2 1

4

i +

1

2 1

4

j 1

4k

=1

4i +

1

4j 1

4k

Now we use the dot (or scalar) product of b and c to find the angle betweenthe two:

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

b c = 116 1

16+

1

16= 1

16

For the right-hand side:

|b||c| cos() =

3

16

3

16cos() =

3

16cos()

Equating the two:

1

16 =3

16cos()

cos() = 13

= cos11

3

= 10928

5. For the simple cubic lattice, referring to figure 8.2, the nearest neigh-bor of any point will be located along the length of a side of the cube. Thus,the nearest neighbor distance is simply a. In the case of the face-centered

cubic lattice, Figure 8.6, for example, we may find the nearest neighbor dis-tance by finding the midpoint distance along the diagonal of a face. We maysimply find the diagonal distance via the Pythagorean theorem:

a2 + a2 =

2a



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55

The midpoint, thus nearest neighbor distance, is then (

2/2)a.

7. Our strategy to find the center-to-center distance of the copper ions willbe to find the volume of one cell, from which we will find the length of anedge of the cube, then use this to arrive at the face-centered nearest-neighbordistance. First, we calculate the number of atoms per unit volume:

8.96 g/cm3

63.5 g/mole= .141

mole

cm3

.141mole

cm3

6.022 1023 atoms

mole

= 8.494 1022 atoms

cm3

Now, we wish to know the volume per unit cell. A face-centered cubic latticehas four atoms per cell, therefore:

4 atoms/cell

8.494 1022 atoms/cm3 = 4.709 1023 cm

3

cell

One edge of the cube should now have length:

1.648 10221/3 = 3.611 108cm

As given in problem 5, the nearest-neighbor distance is then (

2/2)a, where

a is the length of an edge of the cube.2

2a = 2.553 108 cm

9. We shall begin by constructing the position vectors of each of the corners,starting with the corner directly above the origin indicated in Fig. 8.6 (b)and continuing clockwise:

l1 = ak

l2 = aj + ak

l3 = ai + aj + ak

l4 = ai + ak



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The primitive vectors from eq. 8.4 are:

a1 = a2

j + k

, a2 = a2

k + i

, a3 = a2

i + j

Our strategy will be to solve these primitive vectors for i, j, and k andsubstitute into the lattice vectors. We first add a1 and a2:

a1 + a2 =a

2(i + j) + ak

Then subtract a3:a1 + a2 a3 = ak

We now solve for k:

k =a1 + a2 a3

aWe continue in such manner to find:

i =a2 + a3 a1

a

j =a1 a2 + a3

aWe finally carry out the substitution:

l1 = a1 + a2 a3l2 = a1 a2 + a3 + a1 + a2 a3

= 2a1l3 = a2 + a3 a1 + a1 a2 + a3 + a1 + a2 a3= a1 + a2 + a3

l4 = a2 + a3 a1 + a1 + a2 a3= 2a2

11. The results from problems 4 & 5 are as follows:

simple cubic : d = a

R =

a

2

bcc : d =3

2a R =

34

a

fcc : d =2

2a R =

2

4a



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57

We now substitute these values of R into the packing fraction equation:

simple cubic : F = 143

a3

3

a3=

6

bcc : F = 2

43

3a4

3a3

=

3

8

fcc : F = 4

43

2a4

3a3

=

2

6

13. Equation 8.2 gives the following bcc primitive vectors:

a1 =a

2

j + k i

, a2 =

a

2

k + i j

, a3 =

a

2

i + j k

We must use eqs. 8.18-8.20 to determine the primitive vectors of the corre-sponding reciprocal lattice. However, it would be of great benefit to calculatethe necessary cross-products first. We shall use the determinant method here:

a2

a3

=a2

4

i j k

1 1 11 1 1 =

a2

2

j +a2

2k

a3 a1 = a2

4

i j k1 1 11 1 1

=a2

2i +

a2

2k

a1 a2 = a2

4

i j k1 1 11 1 1

=a2

2i +

a2

2j

Now we are ready to use eqs. 8.18-8.20 to calculate the primitive vectors of

the reciprocal lattice:

b1 = 2

a2

3

j + k

a2

4

j + k i

j + k = 4

a

j + k



58/96


b2 = 2

a2

3 i + ka24

j + k i j + k =

4

a

i +

k

b3 = 2

a2

3

i + j

a2

4

j + k i

j + k = 4

a

i + j

Compared to eq. 8.4, we see that this reciprocal lattice is nothing but anfcc-lattice with sides of length 8/a.

15. We shall once again use eqs. 8.18-8.20 to calculate the primitive vectorsof the reciprocal lattice. But, again we carry out the cross-products first:

a2 a3 = a2

4

i j k1 0 11 1 0

=a2

4

i + j + k

a3 a1 = a2

4

i j k1 1 00 1 1

=a2

4

i j + k

a1

a2 =a2

4

i j k0 1 11 0 1

=a2

4i + j k

Since the denominator of each of eqs. 8.18-8.20 is the same, we may go aheadand calculate it to substitute into each equation:

a1 (a2 a3) = a2

j + k

a

2

4

i + j + k

=

a3

4

Now our reciprocal lattice primitive vectors are:

b1 =2

a i + j + k =4

a

1

2 i + j + kb2 =

2

a

i j + k

=

4

a

1

2

i j + k

b3 =2

a

i + j k

=

4

a

1

2

i + j k



59/96

59

We factored out 1/2 so as to get us closer to the form of eq. 8.2. Notice that

these reciprocal vectors are nothing but the primitive vectors of a bcc-latticewith sides of length 4/a.

18. Once again, we shall use eqs. 8.18-8.20 extensively as we want to showthat:

a1 = 2b2 b3

b1 (b2 b3) (1)

a2 = 2b3 b1

b1 (b2 b3) (2)

a1 = 2b1 b2

b1 (b2 b3)(3)

Where b1, b2, and b3 are given by eqs. 8.18-8.20. Let us calculate thecross-product b2 b3 first since it will be used in each equation.

b2 b3 = 42

a3 a1a1 (a2 a3)

a1 a2a1 (a2 a3)

We first make the important observation that the denominators of the termson each side of the cross-product results in a scalar quantity. Therefore, itcan be brought outside the vector product and we may carry out the cross-product. Let us proceed by writing just the operation in question:

(a3 a1) (a1 a2)

The following vector relation, which can be found in most books that discussvector calculus, can be of immense help here:

A (BC) = B (A C)C (A B)

We relate the vectors that occur in the relation above to the vectors involvedin our desired operation as follows:

A = a3

a1

B = a1

C = a2

After the substitution:



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(a3 a1) (a1 a2) = a1 [(a3 a1) a2] a2 [(a3 a1) a1]= a1 [a1 (a2 a3)] + a2 [(a1 a1) a3]

To arrive at this first term in the second expression of the right-hand side,we used the following vector relation:

A (BC) = (AB) CNotice that the part of this term in brackets is now identical to the expressionfor the denominator. We also see that the second term is equal to zero sincewe have a vector in a cross-product with itself. Our result for b2

b3 is now:

b2 b3 = 42a1We can use the same method for the other relevant cross-products to obtainthem as well:

b3 b1 = 42a2b1 b2 = 42a3

The denominators of (1), (2), and (3), which are identical may be also readilysolved using the result from the cross-products above:

b1 (b2 b3) = 2a2

a3a1 (a2 a3) 4

2

a1

= 83

By substituting the results we have obtained into the right-hand side of (1),we see that the equation holds true as can be shown for (2) and (3).

19. Three lattice planes for the simple cubic lattice are depicted in Fig8.15. Of these, figure 8.15(b) represents a member of the family with Millerindices (110). Therefore, the plane is perpendicular to the reciprocal latticevector:

g = b1 + b2

Equation 8.33 gives the distance between planes:

d =2

|g|



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61

We may find the magnitude of g using the primitive reciprocal vectors found

in example 8.1:g =

2

ai +

2

aj |g| = 2 2

a

Therefore our result:

d =a

2=

2a

2

21. a) First assume g = g:

a0 e

igx

e

igx

dx =a0 dx = a

On the other hand, if g = g our integral is as follows:a0

eigxeigxdx =

a0

eix(gg)dx

By using Euler formula, the integrand may be expressed as a combination ofsine and cosine functions:

a

0

cos(x(g

g)) + i sin(x(g

g))dx

Therefore, from equation 8.10 we see that the arguments of the cos and sinfunctions will be of the form:

2n

ax

Therefore, the integral from 0 to a is over a full period. Such integrals of cosand sin functions are equal to zero.

b) Eq. 8.14 gives:

f(x) = g

Fgeigx

Multiply by eigx:

eigxf(x) =

g

Fgeigxeigx



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Integrate from 0 to a:a0

eigxf(x) dx =

g

Fga0

eigxeigx dx

From part (a), we know that the integral on the right-hand side has a valueof zero unless g = g, in which case the value of the integral is a. Thus,the summation disappears as the only term that survives is g = g:a

0

eigxf(x) dx = aFg

25. We shall first carry out the substitutions then apply orthogonalityconditions as directed. Our terms as given in the text are:

Eq. 8.55 : k(r) =1

V1/2eikr

Eq. 8.58 : V(r) =

g

Vgeigr

Our other two needed quantities may be determined from these:

k(r) =1

V1/2eik

r and k(r) =1

V1/2eikr

Substituting into the integral:

Ik, k =

g

VgV

ei(g+k

k)r dV

We separate the exponential terms, so the the integral may be written:

Ik, k =

g

VgV

ei(g+k

)reikr dV

We may also think of this as:

gVgV

k(r)(k+g)(r) dV

Using the orthogonality condition of 8.52:

k = k + g

which is equation 8.62.



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10

Semiconductor Lasers -Solutions

1. As with other radiative transitions we have dealt with, the photons emit-ted must have the same energy as the difference in energies between the twostates. Here, the two states are the bottom of the conduction band and thetop of the valence band, with an energy difference = 1 .519 eV. Usingequation I.25:

E =hc

=

hc

E =

1240 eV

nm

1.519 eV = 816.3nm

3. As outlined in section 10.3.1, heterostructures may be formed using semi-conductors with similar lattice constants. Thus, the heterostructures grownon a substrate of InP must have the same lattice constant as InP. Well usethe linear interpolation procedure for each alloy:

a(InxGa1xAs) = a(InP) = xa(InAs) + (1 x)a(GaAs)

5.869 = 6.058x + 5.653(1

x)

x = .53

This leads to an ally of composition: In.53Al.48As

5. By examining figure 2.6, we see that the graph of

(20/2) 1 inter-

sects the graph of tan and cot three times as it decays. Therefore, if

63



64/96

64 10. SEMICONDUCTOR LASERS - SOLUTIONS

we insure that (20/

2) 1 reaches zero after it intersects once, but before

it does again, the well will have only one bound state. Per equation 2.30, it20 that includes dependence on V0, the depth of the well. So, we solve for 0where the whole expression equals zero:

202 1 = 0 20 = 2

We kepp only the positive solutions and impose the condition that < /2as that is where the graph of cot begins in the positive domain. If

(20/2) 1 decays completely before this point, then the electron will not

have more than one ground state.

< 2 20 < 2

2

We now solve eq. 2.30 for V0, using SI units at first:

20 =m0V0L

2

22

Modern Physics for Scientists and Engineers Solutions

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