1 Problems Chapter 1: Introduction to Nanoelectronics

2 Problems Chapter 2: Classical Particles, Classical Waves, andQuantum Particles

2.1. What is the energy (in Joules and eV) of a photon having wavelength 650 nm? Repeat for an electronhaving the same wavelength and only kinetic energy.

Solution: For the photon,

p =hc

E; E =

hc

=

hc

650 109 = 3:058 1019 J (1)

EeV =EJjqej = 1:91 eV.

For the electron,

E =h2

2me2e

=h2

jqej 2me (650 109)2= 5:704 1025 J (2)

= 3:56 106 eV.

2.2. For light (photons), in classical physics the relation

c = f (3)

is often used, where c is the speed of light, f is the frequency, and is the wavelength. For photons,is the de Broglie wavelength the same as the wavelength in (3)? Explain your reasoning. Hint: useEinsteins formula

E = mc2 =pp2c2 +m0c4; (4)

where m0 is the particles rest mass (which, for a photon, is zero).

Solution: Yes, these wavelengths are the same. From Einsteins formula, E = pc for photons, andusing E = hf we have

c = f = E

h=

pc

h; (5)

so that we must have = h=p.

2.3. Common household electricity in the United States is 60 Hz, a typical microwave oven operates at2:4 109 Hz, and ultraviolet light occurs at 30 1015 Hz. In each case, determine the energy of theassociated photons in joules and eV.

Solution:

E = hf; (6)

Eelec = h60 = 3:98 1032 J = 2:48 1013 eVEoven = h

2:4 109 = 1:59 1024 J = 9:92 106 eV

Euv = h30 1015 = 1:99 1017 J = 124:2 eV.

2.4. Assume that a HeNe laser pointer outputs 1 mW of power at 632 nm.

(a) Determine the energy per photonSolution: Each photon carries

Ep = }! = }2c

= }

(2)3 108

632 109 = 3:145 1019 J = 1:963 eV. (7)

1

(b) Determine the number of photons per second, N .Solution: The sum of all N photons has power

P = NEp (1/s) J = 103 J/s (8)

! N = 103

3:145 1019 = 3:179 7 1015 photons/s.

2.5. Repeat 2.4 if the laser outputs 10 mW of power. How does the number of photons per second scalewith power?

Solution: The sum of all N photons has power

P = NEp (1/s) J = 10 103 J/s (9)

! N = 10 103

3:145 1019 = 3:179 7 1016 photons/s.

The number of photons scales linearly with power.

2.6. Calculate the de Broglie wavelength of

(a) a proton moving at 437; 000 m/s,

(b) a proton with kinetic energy 1; 100 eV,

(c) an electron travelling at 10; 000 m/s.

(d) a 800 kg car moving at 60 km/h.Solution: (a)

=h

p=

h

mpv=

h

(1:673 1027) (437000) = 9:065 1013 m (10)

(b)

E =1

2mpv

2 = 1100 jqej ! v =r(2) (1100 jqej)1:673 1027 = 4:59 10

5 m/s (11)

=h

p=

h

mpv=

h

(1:673 1027) (4:59 105) = 8:631 1013 m

(c)

=h

p=

h

mev=

h

(9:1095 1031) (10000) = 7:274 108 m (12)

(d)

60kmhour

1 m103 km

1 hour60 min

1 min60 s

=60

103 (602)= 16:67 m/s (13)

=h

p=

h

mv=

h

(800) (16:67)= 4:969 1038 m

2.7. Determine the wavelength of a 150 gram baseball traveling 90 miles/hour. Use this result to explainwhy baseballs do not seem to diract around baseball bats.

Solution:90 mileshour

1 km0:6214 miles

1 m103 km

1 hour60 min

1 min60 s

(14)

=90

0:6214 (103) (602)= 40:23 m/s (15)

=h

p=

h

mv=

h

(150 103) (40:23) = 1:098 1034 m

The de Broglie wavelength is too small to observe diraction, since one would observe diraction onsize scales of the order of . The size scale of the bat is far to large.

2

2.8. How much would the mass of a ball need to be in order for it to have a de Broglie wavelength of 1 m(at which point its wave properties would be clearly observable)? Assume that the ball is travelling 90miles/hour.

Solution: =

h

p=

h

mv=

h

m (40:23)= 1 m! m = h

40:23= 1:647 1035 kg (16)

2.9. Determine the momentum carried by a 640 nm photon. Since a photon is massless, does this momentumhave the same meaning as the momentum carried by a particle with mass?

Solution: =

h

p= 640 109 ! p = h

640 109 = 1:035 1027 Js/m=kg m/s (17)

The momentum has essentially the same meaning as for a particle having mass: the photon momentumexerts a force on objects (in general, force multiplied by time equals momentum) that can be used to,for example, move objects.

2.10. Consider a 4 eV electron, a 4 eV proton, and a 4 eV photon. For each, compute the de Brogliewavelength, the frequency, and the momentum.

Solution: For the photon,

=h

p=hc

E=

hc

4 jqej = 310:17 nm, (18)

E = hf ! f = Eh=4 jqejh

= 9:672 Hz,

p =E

c=4 jqejc

= 2:136 1027 kg m/s.

For the electron,

=hp2meE

=hp

2me4 jqej= 0:613 nm, (19)

E = hf ! f = Eh=4 jqejh

= 9:672 Hz,

p = mev = (me)1:186 106 = 1:080 1024 kg m/s, since

E =1

2mev

2 = 4 jqej ! v =s(2) (4 jqej)

me= 1:186 106 m/s

For the proton,

=hp2mpE

=hp

2mp4 jqej= 0:0143 nm, (20)

E = hf ! f = Eh=4 jqejh

= 9:672 Hz,

p = mpv = mp (27683) = 4:630 1023 kg m/s, since

E =1

2mv2 = 4 jqej ! v =

s(2) (4 jqej)

mp= 27; 683 m/s

Obviously, f is the same for all particles since E = hf . The momentum values are very small, butsmallest for the photon. The wavelength is far larger for the photon than for the electron, which itselfhas a far larger wavelength than for the proton (the proton has far greater mass than the electron).

2.11. Determine the de Broglie wavelength of an electron that has been accelerated from rest through apotential dierence of 1:5 volts.

3

Solution:E = 1:5 eV, and thus =

hp2meE

=hp

2me (1:5) jqej= 1:001 nm. (21)

2.12. Calculate the uncertainty in velocity of a 1 kg ball conned to

(a) a length of 20 m,

(b) a length of 20 cm,

(c) a length of 20 m.

(d) What can you conclude about observing quantum eectsusing 1 kg balls? What kind of objectswould you need to use to see quantum eects on these length scales?Solution: From

px }2; (22)

(a)

v }2mx

=}

2 (1) (20 106) = 2:637 1030 m/s (23)

(b)

v }2mx

=}

2 (1) (20 102) = 2:637 1034 m/s (24)

(c)

v }2mx

=}

2 (1) (20)= 2:637 1036 m/s (25)

(d) 1 kg balls are far too heavy to observe quantum eects - one would need to have masses onthe order of an atomic particle to observe quantum eects, since the value of h is so small.

2.13. If we know that the velocity of an electron is 40:23 0:01 m/s, what is the minimum uncertainty inits position? Repeat for a 150 gram baseball travelling at the same velocity.

Solution: Frommvx }

2; (26)

for the electron,

x }2me (0:01)

= 5:789 103 m. (27)

For the baseball,

x }2 (:15) (0:01)

= 3:52 1032 m. (28)

2.14. If a molecule having mass 2:31026 kg is conned to a region 200 nm in length, what is the minimumuncertainty in the molecules velocity?

Solution: Frommvx }

2; (29)

v }2mx

=}

2 (2:3 1026) (200 109) = 0:0115 m/s. (30)

2.15. Determine the minimum uncertainty in the velocity of an electron that has its position specied towithin 10 nm.

Solution: Frommvx }

2; (31)

v }2mex

=}

2me (10 109) = 5788:5 m/s. (32)

4

2.16. Explain the dierence between a fermion and a boson, and give two examples of each.

Solution: Particles with integral (in units of }) spin are bosons. Examples are photons and phonons.Particles with half-integral spin are fermions. Examples are electrons, protons, and neutrons.

3 Problems Chapter 3: Quantum Mechanics of Electrons

3.1. For the matrix operator L = 5 0

1 2

, show that eigenvalues and eigenvectors are

= 2, x =0

; (33)

= 5, x = 7

;

where ; 6= 0. That is, show that the preceding quantities satisfy the eigenvalue problem Lx = x.Solution:

Lx = x (34) 5 01 2

0

= 2

0

02

=

02

pLx = x (35) 5 0

1 2

7

= 5

7

355

=

355

p3.2. Consider the set of functions

n1p2einx; n = 0;1;2; :::

o.

(a) Show that this is an orthonormal set on the interval (; ).Solution: Z

1p2einx

1p2eimxdx =

1

2

ei(nm) ei(nm)i (nm) = 0 if n 6= m (36)Z

1p2einx

1p2einxdx = 1 if n = m

(b) On the interval (=2; =2), is the set an orthogonal set, an orthonormal set, or neither?Solution: Evaluating the integral one sees that the set is not orthogonal (and, hence, cant beorthonormal).

3.3. Consider the set of functionsnq

2 sin(nx); n = 1; 2; :::

oon the interval (0; ).

(a) Show that this is an orthonormal set.

5

Solution:Z 0

r2

sin(nx)

r2

sin(mx)dx =

1

Z 0

(cos (mx nx) cos (mx+ nx)) dx

=1

sin ((nm))

nm sin ((n+m))

n+m

= 0 if n 6= m

=2

Z 0

sin2(nx)dx =2

Z 0

1

2 12cos 2nx

dx

=2

1

2

= 1 if n = m

(b) Determine an operator (including boundary conditions) for which the preceding set are eigenfunc-tions. What are the eigenvalues?Solutions: The operator is bo = d2

dx2; (37)

the second derivative operator (d2=dx2 also works), acting on functions dened over 0 x .Every function (x) = A sin (kx) +B cos (kx) is an eigenfunction with eigenvalue = k2, since

d2

dx2(A sin (kx) +B cos (kx)) = k2 (A sin (kx) +B cos (kx)) : (38)

If k is to be an integer, k = n, and B = 0, then the boundary condition is (0) = () = 0. Theeigenvalues are simply n = 0; 1; 2; :::.

3.4. For the dierential operator L = d2=dx2, u (0) = u (a) = 0, determine eigenvalues and eigenfunc-tions u. That is, solve

Lu = u; (39)

where u (x) is a nonzero function subject to the given boundary conditions. Normalize the eigenfunc-tions, and show that the eigenfunctions are orthonormal.

Solution:

d2

dx2u u = 0 (40)

u = A sinpx+B cos

px

check: d2

dx2

A sin

px+B cos

px

A sin

px+B cos

px= 0

u (0) = B = 0

u (a) = A sinpa = 0!

pa = n, n = 0; 1; 2; :::

=na

2:

Thereforeu = A sin

n

ax (41)

To normalize the eigenfunctions, Z a0

A2 sin2(n

ax)dx = 1 (42)

A2Z a

0

1

2 12cos 2

anx

dx

= 1

A21

2a

= 1! A =

r2

a

6

Finally,

2

a

Z a0

sin(n

ax) sin(

m

ax)dx = 0 if n 6= m (43)

= 1 if n = m:

3.5. Repeat problem 3.4, but for boundary conditions u0(0) = u0(a) = 0, where u0 = du=dx.

Solution:

d2

dx2u u = 0 (44)

u = A sinpx+B cos

px

u0 (0) = A = 0

u0 (a) = Bp sin

pa = 0!

pa = n, n = 0; 1; 2; :::

=na

2:

Thereforeu = B cos

n

ax: (45)

To normalize the eigenfunctions, set Z a0

B2 cos2naxdx = 1 (46)

leading to

B =

r"na; (47)

where

"n 1; n = 02; n 6= 0 : (48)

3.6. Assume that some observable of a certain system is measured and found to be n for some integer n.By postulate 2, we know that immediately after the measurement the system is in state n, which isan eigenstate of the measurement operator bo (i.e., where bo n = n n).(a) What can we conclude about the systems state immediately before the measurement?

Solution: Nothing, other than that it was in a superposition with some content in n.

(b) Assume that the identical measurement is then performed on 100; 000 identical systems, andeach time the measurement result is the same, n. What can we infer about the systems stateimmediately before the measurement?Solution: We can reasonably assume that before the measurement the system was in state n.

3.7. Assume that an electronic state has a lifetime of 108 s. What is the minimum uncertainty in theenergy of an electron in this state?

Solution: FromEt }

2; (49)

then

E }2t

=}

2 (108)=5:273 1027

jqej J=3:291 108 eV. (50)

7

3.8. In the example of solving the one-dimensional Schrdinger equation on p. 65, we obtained the statefunctions

(x; t) = (x) eiEnt=}

where

(x) =

2

L

1=2sinnLx; n even, (51)

=

2

L

1=2cosnLx; n odd,

are eigenfunctions of the second derivative operator d2=dx2, and where energy eigenvalues were foundto be

En =}2

2m

nL

2: (52)

(a) Show that the odd eigenfunction (sine) can be written as

(x) =1p2

1

ipLei

pn} x 1

ipLei

pn} x

(53)

= + ;and determine a similar expression for the even eigenfunction. The term + ( ) represents awave propagating with positive (negative) momentum. Thus, any state described by sine andcosine can be thought of as representing a superposition of positive and negative momentumstates.Solution:

(x) =

2

L

1=2sinnLx=

2

L

1=2ei

nL x einL x

2i

; (54)

and usingk =

n

L; (55)

it was shown in the example that

p =h

=hk

2=n}L

= pn; (56)

so that

(x) =1

i

1

2L

1=2 ei

pn} x ei pn} x

: (57)

(b) Although the decomposition of a standing wave into two counterpropagating waves, as in part (a),is useful, it can be misinterpreted. Since the probability density (x; t) (x; t) is independent oftime, the expectation value of position, hxi, is independent of time, and so, really, we should notthink of the particle as bouncingback and forth in the conned space (otherwise, hxi would bea function of t). Determine the expectation value of momentum, using either (3.217) or (3.219),and discuss your answer in light of the above comment.Solution: The expectation value for momentum is,

hpxi =Z L=2L=2

i} @

@x

dx (58)

=

Z L=2L=2

1p2

1

ipLei

pn} x 1

ipLei

pn} x

i} @

@x

1p2

1

ipLei

pn} x 1

ipLei

pn} x

dx

=i

Lpn

Z L=2L=2

sin2pn}x

dx = 0:

8

Therefore, the average momentum is zero, meaning that there are an equal number of positiveand negative momentum states, and no net movement.

(c) Assume that the particle is in a state composed of the rst two eigenfunctions,

(x; t) =1p2

2

L

1=2cosLxei

E1t} +

2

L

1=2sin

2

Lx

ei

E2t}

!: (59)

Show that the expectation value of position as a function of time is

hxi = 169

L

2cos

3

2

}m

2

L2t

: (60)

Interpret this solution, compared with the expectation value of position for a single stationarystate n, which is time-independent.Solution:

hxi =Z L=2L=2

x dx (61)

=1

2

2

L

Z L=2L=2

x

cosLxei

E1t} + sin

2

Lx

ei

E2t}

cosLxe

iE1t} + sin

2

Lx

ei

E2t}

dx

=1

L

Z L=2L=2

x

cos2

Lx+ cos

Lxsin

2

Lx

ei

(E1E2)t}

+ cosLxsin

2

Lx

ei

(E2E1)t} + sin2

2

Lx

dx

=2

Lcos

(E1 E2) t

}

Z L=2L=2

x cosLxsin

2

Lx

dx

=16L

92cos

(E1 E2) t

}

;

where we used the fact that the integral of an odd function over symmetric limits is zero. Since

(E1 E2)}

=1

}

}2

2m

L

2 }

2

2m

2

L

2!(62)

=}2m

L

22

L

2!= 3}

2m

2

L2;

then

hxi = 16L92

cos

3}2m

2

L2t

: (63)

3.9. Since Schrdingers equation is a homogeneous equation, the most general solution for the state functionis a sum of homogeneous solutions (3.142),

(r; t) =Xn

an n (r) eiEnt=}: (64)

Show that if (r; 0) is known then an expression for the weighting amplitudes an can be determined.Assume that the eigenfunction n form an orthonormal set. Hint: multiply

(r; 0) =Xn

an n (r) (65)

9

by m (r) and integrate. What is the interpretation of janj2?Solution: Z

m (r) (r; 0) dr3 =

Xn

an

Z m (r)

n (r) dr

3 (66)

=Xn

annm = am: (67)

So, since

janj2 =Z n (r) (r; 0) dr32 (68)

and

P (n) =

Z (r; t) n (x) dx2 ; (69)then janj2 is the probability that a measurement will nd that the initial state of the particle is n.

3.10. Consider a particle with time-independent potential energy, and assume that the initial state of theparticle is

(r; t) = a1 1 (r; t) + a2 2 (r; t) ; (70)

such that P (1) = ja1j2 = P1, P (2) = ja2j2 = P2, and ja1j2 + ja2j2 = 1. Show thathEi = P1 hE1i+ P2 hE2i . (71)

Solution:

hEi =Z (r; t) (i})

@

@t(r; t) d3r (72)

=

Z(a1

1 (r; t) + a

2

2 (r; t)) (i})

@

@t(a1 1 (r; t) + a2 2 (r; t)) d

3r

= ja1j2Z 1 (r; t) (i})

@

@t 1 (r; t) d

3r + ja2j2Z 2 (r; t) (i})

@

@t 2 (r; t) d

3r

= P1 hE1i+ P2 hE2i :3.11. For the example of solving the one-dimensional Schrdingers equation on p. 65, determine the proba-

bility of observing the particle near the boundary wall, x = L=2. If the particle is in the n = 2 state,where is the particle most likely to be found?

Solution: At x = L=2, = 0, so probability is zero. It would be better to say that the wavefunctionis very small near the boundary, and thus, as one considers a region near the boundary, it is veryunlikely to nd the particle there.

If the particle is in the n = 2 state, where is the particle most likely to be found? Even though theexpectation value of position is zero,

hxi =Z L=2L=2

(x; t)x (x; t) dx (73)

=2

L

Z L=2L=2

x sin2nLxdx = 0;

the n = 2 state peaks away from the origin. The particle is most likely to be found where thewavefunction is largest,

d

dxsinnLx=

Ln cos

Lnx = 0 (74)

! cos L2x = 0

! x = L4

10

3.12. For the example of solving the one-dimensional Schrdingers equation on p. 65, assume that theparticle is in the n = 2 state. What is the probability that a measurement of energy will yield

E2 =}2

2m

2

L

2? (75)

Solution: Since the particle is already in the n = 2 state, by postulate 2 the probability that anenergy measurement will yield E2 is 100 %.

What is the probability that a measurement of energy will yield

E3 =}2

2m

3

L

2? (76)

Solution: Since the particle is already in the n = 2 state, by postulate 2 the probability that anenergy measurement will yield E3 is 0 %. This can easily be seen from orthogonality,

P (E3) =

2

L

2 Z L=2L=2

sin

2

Lx

cos

3

Lx

dx

2

= 0: (77)

3.13. Consider a quantum encryption scheme using photons. Assume that a photon can only exist in eitherstate 1, 1, having energy E1, or state 2, 2, having energy E2, or in a superposition of the two states, = a 1 + b 2. Assume that the states are orthonormal.

(a) If a photon exists in the superposition state = a 1 + b 2, what is the relationship between aand b?

(b) If a photon exists in the superposition state = a 1+b 2, determine the probability of measuringenergy E2. Show all work, and/or explain your answer.

(c) If the photon in a superposition state is sent over a network, explain how undetected eavesdroppingwould be impossible.Solution: (a) Since the sum of probabilities must be unity, then jaj2 + jbj2 = 1.(b)

P (E2) =

Z a0

(x) 2 (x) dx2 = Z a

0

fa 1 + b 2g 2 (x) dx2 = jbj2 : (78)

(c) Undetected eavesdropping is impossible, since any measurement would collapse the statefunction.

3.14. In Chapter 6, the reection and transmission of a particle across a potential barrier will be considered.For now, assume that a potential energy discontinuity is present at x = a, and that to the left of thediscontinuity the wavefunction is given by

(x; t) =eikx +R eikx

eiEt=}; (79)

and to the right of the discontinuity,

(x; t) = TeiqxeiEt=}; (80)

where R and T are reection and transmission coe cients, respectively, which will depend on theproperties of the dierent regions and on the discontinuity in potential at x = a. Determine theprobability current density on either side of the discontinuity.

11

Solution:

J (r; t) =i}2m

( (r; t)r(r; t)(r; t)r (r; t)) (81)

=i}2m

bx (r; t) @@x(r; t)(r; t) @

@x (r; t)

=i}2m

bxeikx +R eikx @@x

eikx +R eikx

eikx +R eikx @

@x

eikx +R eikx

= bx}k

m

1 jRj2

;

on the left, and, on the right,

J (r; t) =i}2m

( (r; t)r(r; t)(r; t)r (r; t)) (82)

=i}2m

bx (r; t) @@x(r; t)(r; t) @

@x (r; t)

=i}2m

bxT eiqx @@x

T eiqx

T eiqx @@x

T eiqx

= bx}q

mjT j2 :

3.15. In the example of solving the one-dimensional Schrdingers equation on p. 65, we obtained the statefunctions

(x; t) = (x) eiEnt=} (83)

where

(x) =

2

L

1=2sinnLx; n even, (84)

=

2

L

1=2cosnLx; n odd,

and where

En =}2

2m

nL

2: (85)

Determine the probability current density. Discuss your result.

Solution: Since is real-valued, the probability current density is zero (since in this case = ).This means that there is no net current; these states are called stationary states.

3.16. Assume that the wave function (z; t) = 200ei(kz!t) (86)

describes a beam of 2 eV electrons having only kinetic energy. Determine numerical values for k and!, and nd the associated current density in A/m.

12

Solution:

E = }! = 2 jqej (87)

v =

r2E

me=

s2 (2 jqej)me

= 8:387 105 m/s

=h

p=

h

mev=

h

me (8:387 105) = 0:8673 nm

k =2

=

2

0:8673 109 = 7:244 109 m1

so, (z; t) = 200eikzei!t = 200ei(7:244109)zei(

2jqej} )t;

and

J (z; t) =i}2me

bz (z; t) @@z (z; t) (z; t) @

@z (z; t)

(88)

=i}2me

bz200ei(7:244109)zei( 2jqej} )t @@z

200ei(7:24410

9)zei(2jqej} )t

200ei(7:24410

9)zei(2jqej} )t

@@z

200ei(7:24410

9)zei(2jqej} )t

(89)

= bz 3:355 1010 A/m.4 Problems Chapter 4: Free and Conned Electrons

4.1. Write down the wavefunction (z; t) for a 3 eV electron in an innite space, travelling along thepositive z axis. Assume that the electron has only kinetic energy. Plug your answer into Schrdingerstime-dependent equation to verify that it is a solution.

Solution:

E = }! = 3 jqej (90)

v =

r2E

me=

s2 (3 jqej)me

= 1:027 106 m/s

=h

p=

h

mev=

h

me (1:027 106) = 0:7082 nm

k =2

=

2

0:7082 109 = 8:872 109 m1

or, use (4.5), k =

r2me}2

3 jqej = 8:87 109 (91)

so, (z; t) = Aeikzei!t = Aei(8:872109)zei(

3jqej} )t

i}@ (z; t)

@t=

}

2

2m

d2

dz2+ V

(z; t) : (92)

4.2. Determine the wavefunction (z; t) for a 3 eV electron in an innite space, travelling along the zaxis at a velocity of 105 m/s. Determine the particles potential energy, and plug your answer intoSchrdingers time-dependent equation to verify that it is a solution.

13

Solution:

E = }! = 3 jqej = V + EKE = V + 12mev

2 = V +1

2me1052

(93)

! V = 3 jqej 12me1052= 2:972 eV

=h

p=

h

mev=

h

me105= 7:274 nm

k =2

=

2

7:274 109 = 8:638 108 m1

so, (z; t) = Aeikzei!t = Aei(8:638108)zei(

3jqej} )t

Plugging into Schrdingers time-dependent equation we obtain

i}@ (z; t)

@t=

}

2

2m

d2

dz2+ V

(z; t) (94)

i}d

dt

ei(8:63810

8)zei(3jqej} )t

= }

2

2me

@2

@z2

ei(8:63810

8)zei(3jqej} )t

+ V

V = 2:971 eV (95)

4.3. For classical light the expressionc = f (96)

holds, where c is the speed of light, f is the frequency, and is the wavelength. This expression alsoholds for quantized light (i.e., photons), where the classical wavelength is the same as the de Brogliewavelength. Does the expression

v = f; (97)

where v is velocity, hold for electrons if is the de Broglie wavelength and f = !=2 is the frequency,such that E = }!? Assume that the electrons have only kinetic energy.Solution: No.

v = f =h

p

E

}2=E

p=

12mv

2

mv=v

2(98)

The velocity obtained from v = f is the phase velocity, which is not the meaningful velocity. For anelectron with kinetic energy E one needs to use the classical expression for kinetic energy, E = (1=2)mv2

to obtain v. More generally, one needs to us the concept of group velocity,

vg =@!

@k=@ (E=})@k

: (99)

4.4. Consider an electron in a room of size 10 10 10 m3. Assume that within the room potential energyis zero, and that the walls and ceilings of the room are perfect (so that the electron can not escape fromthe room). If the electrons energy is approximately 5 eV, what is the state index n2 = n2x + n

2y + n

2z?

What is the approximate energy dierence E2;1;1 E1;1;1?Solution:

En =}22

2meL2n2x + n

2y + n

2z

= 5 jqej (100)

! n =r5 jqej 2me102

}22= 3:65 1010:

What is the approximate energy dierence E2;1;1 E1;1;1?

E2;1;1 E1;1;1 = }22

2me10222 + 12 + 12

12 + 12 + 12 (101)= 1:8 1039 J=1:124 1020 eV

Therefore, the states essentially form a continuum.

14

4.5. Repeat problem 4.4 for an electron conned to a nanoscale space, 109 109 109 m3.Solution:

En =}22

2meL2n2x + n

2y + n

2z

= 5 jqej (102)

! n =s5 jqej 2me (109)2

}22= 3:65:

What is the approximate energy dierence E2;1;1 E1;1;1?

E2;1;1 E1;1;1 = }22

2me (109)2

22 + 12 + 12

12 + 12 + 12 (103)= 1:8 1019 J = 1:12 eV

These states will be easily observed to be discrete.

4.6. Consider an electron having kinetic energy 2:5 eV. What size space does the electron need to be connedto in order to observe clear energy discretization? Repeat for a proton having the same kinetic energy.

Solution:

E =1

2mev

2 = 2:5 jqej ! v =s2 (2:5 jqej)

me= 9:378 105 m/s (104)

p = mev = me9:378 105 = 8:543 1025 kg m/s

=h

p=

h

8:543 1025 = 0:776 nm

In order to observe clear energy discretization, the space must be on the order of = 0:776 nm, orsmaller.

For a proton having the same kinetic energy,

E =1

2mpv

2 = 2:5 jqej ! v =s2 (2:5 jqej)

mp= 21885 m/s (105)

p = mpv = mp (21885) = 3:661 1023 kg m/s =

h

p=

h

3:661 1023 = 0:0181 nm,

and thus the space must be on the order of 0:0181 nm.

4.7. Consider a 44 kg object (perhaps your desk) in a typical room with dimensions 10 10 m2. Assumethat within the room, potential energy is zero, and that the walls of the room are perfect (so that theroom can be modeled as an innitely deep potential well).

(a) If the desk is in the ground state, what is the velocity of the desk?Solution:

E =1

2mv2 = En =

}22

2mL2(1 + 1 + 1) =

}22

2 (44) 102(1 + 1 + 1) = 3:7 1071 (106)

v =

r2 (3:7 1071)

44= 1:3 1036 m/s

(b) If the desk is moving at 0:01 m/s, what is the desks quantum state (i.e., what is the state indexn)?

15

Solution:

E =1

2mv2 =

1

2(44) (:01)

2= 0:0022 =

}22

2mL2n2 (107)

! n =s(0:0022) 2 (44) (10)

2

}22= 1:3 1034:

4.8. Consider the one-dimensional, innite square well existing from 0 x L. The wavefunction for aparticle conned to the well is (4.34),

(x; t) =

2

L

1=2sinnLxeiEnt=}: (108)

If the particle is in the ground state, determine the probability, as a function of time, that the particlewill be in the right half of the well (i.e., from L=2 x L).Solution: From (3.6),

P =

Z LL=2

(x; t) (x; t) dx (109)

=

Z LL=2

2

L

1=2sinnLxeiEnt=}

2

L

1=2sinnLxeiEnt=}dx

=2

L

Z LL=2

sin2Lxdx =

1

2:

4.9. Consider an electron in the rst excited state (n = 2) of an innitely-high square well of length 2:3nm. Assuming zero potential energy in the well, determine the electrons velocity.

Solution: We have

En =}2

2me

nL

2(110)

E2 =}2

jqej 2me

2

2:3 1092= 0:284 eV. (111)

Then,

k =

r2meE2}2

; (112)

and

ve =p

me=}kme

=}me

r2me (0:284) jqej

}2= 3:161 105 m/s. (113)

4.10. Consider an electron conned to an innite potential well having length 2 nm. What wavelengthphotons will be emitted from transitions between the lowest three energy levels

Solution:

En =}2

2me

nL

2; (114)

so that

E3 E1 = }2

jqej 2me

2 1092

32 12 = 0:752 eV, (115)E3 E2 = }

2

jqej 2me

2 1092

32 22 = 0:470 eV,E2 E1 = }

2

jqej 2me

2 1092

22 12 = 0:282 eV,16

4.11. Consider the one-dimensional, innite square well existing from 0 x L. Recall that the energyeigenfunctions for a particle conned to the well are (4.31),

n (x) =

2

L

1=2sinnLx; (116)

n = 1; 2; 3; :::. Now, assume that the particle state is

(x) = A (x (L x)) ; (117)

such that (0) = (L) = 0.

(a) Determine A such that the particles state function is suitably normalized.Solution: Z L

0

A2 (x (L x))2 dx = 1 (118)

A21

30L5 = 1! A =

r30

L5:

(b) Expand (x) in the energy eigenfunctions, i.e., use orthogonality to nd cn such that

(x) = A (x (L x)) =1Xn=1

cn n (x) (119)

Solution:Z L0

m (x) (x) dx =

Z L0

m (x)

r30

L5(x (L x)) dx (120)

=

Z L0

m (x)1Xn=1

cn n (x) dx =1Xn=1

cn

Z L0

m (x) n (x) dx = cm

cn =

Z L0

n (x) (x) dx =

Z L0

2

L

1=2sinnLxr 30

L5(x (L x)) dx

=

r2

L

r30

L5

Z L0

sinnLx(x (L x)) dx

= p2

s1

L

p30

s1

L5

L3n sinn + 2 cosn 2

n33

= 4p151 cosnn33

:

(c) Determine the probability of measuring energy En, and use this result to determine the probabilityof measuring E1 through E6.

17

Solution:

P (En) =

Z L0

(x) n (x) dx

2

(121)

=

Z L0

1Xm=1

cm m (x) n (x) dx

2

= jcnj2 =4p151 cosnn33

2=

16(60)n66 ; n odd0; n even

P (E1) = jc1j2 = 16 (60)166

= 0:99856 = 99:8%

P (E3) = jc3j2 = 16 (60)366

= 0:00137 = 0:137%

P (E5) = jc5j2 = 16 (60)566

= 6:391 105 = 0:0064%

4.12. Repeat problem 4.11 if the particle state is

(x) = A (x (L x))2 : (122)

(a) Determine A such that the particles state function is suitably normalized.Solution: Z L

0

A2 (x (L x))4 dx = 1

A21

630L9 = 1! A =

r630

L9:

(b) Expand (x) in the energy eigenfunctions, i.e., use orthogonality to nd cn such that

(x) = A (x (L x))2 =1Xn=1

cn n (x) (123)

Solution: Z L0

m(x) dx =

Z L0

2

L

1=2sinnLxr630

L9(x (L x))2 dx (124)

=1Xn=1

cn

Z L0

m (x) n (x) dx

cn =

Z L0

2

L

1=2sinnLxr630

L9(x (L x))2 dx (125)

=

r1260

L10

Z L0

sinnLx(x (L x))2 dx

=

r1260

L10

1

5n524L5 24L5 cosn 22L5n2 + 22L5n2 cosn

=

(0; n = 0; 2; 4; :::q

1260L10

1

5n5

48L5 42L5n2 ; n = 1; 3; 5; :::

18

(c) Determine the probability of measuring energy En, and use this result to determine the probabilityof measuring E1 through E6.Solution:

P (En) =

Z L0

(x) n (x) dx

2

(126)

=

Z L0

1Xm=1

cm m (x) n (x) dx

2

= jcnj2

=

1260

1

5n5

48 42n22 ; n odd

0; n even

P (E1) = jc1j2 = 1260

1

51548 42122 = 0:9770 = 97:7%

P (E3) = jc3j2 = 1260

1

53548 42322 = 0:0215 = 2:15%

P (E5) = jc5j2 = 1260

1

55548 42522 = 0:00122 = 0:122%

4.13. Consider the one-dimensional, innite square well existing from 0 x L. If nine electrons are in thewell, what is the ground state energy of the system?

Solution: Using

En =}2

2me

nL

2; (127)

then for nine electrons, since two electrons can be in the same state (one of each spin), the n = 1; 2; 3; 4states are lled, and the 9th electron is in the n = 5 state. Therefore, the ground state energy of thesystem is

2E1 + 2E2 + 2E3 + 2E4 + E5 (128)

=}2

2me

L

2 212 + 22 + 32 + 42

+ 52

= 85

}2

2me

L 1092

1

jqej =31:963

L2eV

where L is in nm.

4.14. Consider a one-dimensional quantum well of length L = 10 nm containing 11 non-interacting electrons.Determine the chemical potential of the system.

Solution: With two electrons per state, N = 10 electrons and N = 12 electrons will have energy

Et (10) = 2E1 + 2E2 + 2E3 + 2E4 + 2E5; (129)

Et (12) = 2E1 + 2E2 + 2E3 + 2E4 + 2E5 + 2E6 (130)

Then,

=Et (12) Et (10)

2= E6 =

}2

2me

6

L

2=

}2

2me jqej

6

10 1092= 0:135 eV:

4.15. Apply boundary conditions (4.81) to the symmetric form of (4.78) (i.e., set C = 0 in (4.78)) for thenite rectangular potential energy well considered in Section 4.5.1 to show that the wavefunction (4.82)results, where (4.83) must be satised.

19

4.16. Repeat problem 4.15 for the antisymmetric form of (4.78) (i.e., with D = 0), showing that (4.84)results, where the wavenumbers must obey (4.85).

4.17. Comparing energy levels in one-dimensional quantum wells, for an innite-height well of width 2Lenergy states are given by (4.35) with L replaced by 2L,

En =}2

2me

n2L

2; (131)

and for a nite-height well of width 2L the symmetric states are given by a numerical solution of (4.83),

k2 tan (k2L) = k1: (132)

Assume that L = 2 nm, and compute E1 and E3 (the second symmetric state) for the innite-heightwell. Then compare it with the corresponding values obtained from the numerical solution of (132) forthe nite-height well. (You will need to use a numerical root-solver.) For the nite-height well, assumebarrier heights V0 = 1000; 100; 10; 1; 0:5, and 0:2 eV. Make a table comparing the innite-height andnite-height results (with percent errors), and comment on the appropriateness, at least for low energystates, of the much simpler innite well model for reasonable barrier heights, such as V0 = 0:5 eV.

Solution. For the innite-height well,

E1 =}2

jqej 2me

2 (2 109)2= 0:0235 eV (133)

E3 =}2

jqej 2me

3

2 (2 109)2= 0:21152 eV. (134)

For the nite-height well,

k2 tan (k2L) = k1; k22 =

2meE

}2; k21 =

2me (V0 E)}2

(135)

r2meE

}2tan

r2meE

}2L

!=

r2me (V0 E)

}2(136)r

2meE jqej}2

tan

r2meE jqej

}22 109!r2me ((V0) jqej E jqej)

}2= 0: (137)

The table below shows that the simple innite-height well is a reasonable model as a rst approximation,and gives an order-of-magnitude estimate, although the percent error is probably too large to providea quantitative model, especially for low barrier heights.

state E1V0 E

nite-heightnumerical %Error

1000 0:02336 0:62100 0:02305 1:9610 0:02212 6:271 0:01950 20:540:5 0:01812 29:700:2 0:01576 49:13

state E3V0 E

nite-heightnumerical %Error

1000 0:21022 0:62100 0:20745 1:9610 0:19900 6:231 0:17459 21:150:5 0:16084 31:510:2 0:13414 57:68

4.18. For a 1s electron in the ground state of hydrogen, determine the expectation value of energy.

Solution: Since 100 =

1p4

2

a3=20

era0 eiE1t=}; (138)

20

then, from (3.116)

hEi =Z 20

Z 0

Z 10

(r; t)i}@

@t

(r; t) r2 sin drdd (139)

=1

4

2

a3=20

!2 Z 20

Z 0

Z 10

era0 eiE1t=}

i}@

@t

e

ra0 eiE1t=}r2 sin drdd

=1

4

2

a3=20

!22 (E1)

Z 0

Z 10

e2ra0 r2 sin drd = E1:

4.19. For an electron in the (n; l;m) = (2; 0; 0) state of hydrogen, determine the expectation value of position.

Solution: Since

200 =1p4

1

(2a0)3=2

2 r

a0

e

r2a0 ; (140)

then, from (3.112)

hri =Z 20

Z 0

Z 10

(r; t) r(r; t) r2 sin drdd (141)

=

Z 10

1p4

1

(2a0)3=2

2 r

a0

e

r2a0

!(r)

1p4

1

(2a0)3=2

2 r

a0

e

r2a0

!r2dr

=1

(2a0)3

Z 10

2 r

a0

e

r2a0

2 r

a0

e

r2a0

r3dr

=1

(2a0)3

1

a048a50

= 6a0:

4.20. Repeat problem 4.19 for an electron in the (n; l;m) = (2; 1; 0) state of hydrogen.

Solution:

2;1;0 =1p

3 (2a0)3=2

r

a0e

r2a0

r3

4cos ; (142)

hri =Z 20

Z 0

Z 10

(r; t) r(r; t) r2 sin drdd

=

Z 20

Z 0

Z 10

1p

3 (2a0)3=2

r

a0e

r2a0

r3

4cos

!2r3 sin drdd

=

Z 20

d

Z 0

sin cos2

d

Z 1p

3 (2a0)3=2

r

a0e

r2a0

r3

4

!2r3dr

= 43

1

32a40

120a50 = 5a0:4.21. For an electron in the (nx; ny) = (1; 1) subband of a metallic quantum wire having Lx = Ly = 1 nm,

if the total energy is 1 eV, what is the electrons longitudinal (i.e., zdirected) group velocity?Solution: From (4.133),

E =}22

2me

nxLx

2+

nyLy

2!+

}2

2mek2z (143)

=}22

jqej 2me

1

1 1092+

1

1 1092!

+}2

jqej 2me k2z

= En + Econt = 1

21

En = 0:7521, so Econt = E En = 1 0:7521 = 0:2479. Using E = }!,

0:2479 =}2

jqej 2me k2z ! kz =

r0:2479 jqej 2me

}2= 2:551 109 (144)

}2

2mek2z = }!

(vg)z =@!

@kz=

}2me

2kz =}2me

22:551 109 = 2:953 105 m/s.

4.22. A 3 eV electron is to be conned in a square quantum dot of side L. What should L be in order forthe electrons energy levels to be well-quantized?

Solution: From (2.14),

e =h

p=

h

mev=

h

me

q2Eme

=h

me

q23jqejme

= 0:708 nm; (145)

and we need L e.

5 Problems Chapter 5: Electrons Subject to a Periodic PotentialBand Theory of Solids

5.1. To gain an appreciation of the important role of surface eects at the nanoscale, consider building upa material out of bcc unit cells. (See Section 5.1). For one bcc cube, there would be 9 atoms, 8 onthe outside and one interior, as depicted on p. 134. If we constrain ourselves to only consider cubesof material, the next largest cube would consist of 8 bcc unit cells, and so on. If one unit cell is 0:5nm, how long should the materials side be in order for there to be more interior atoms then surfaceatoms?

Solution:# unit cells in the cube Surface Atoms Interior Atoms Ratio (S/I)1 8 1 823 = 8 26 9 2:933 = 27 56 35 1:643 = 64 98 91 1:153 = 125 152 189 0:8

so that we need 53 unit cells, leading to a material cube having side length 5 (0:5) = 2:5 nm.

5.2. Consider the Kronig-Penney model of a material with a1 = a2 = 5 and V0 = 0:5 eV. Determinenumerically the starting and ending energies of the rst allowed band.

Solution: From

cos kT = cos (a) cosh (b) 2 22

sin (a) sinh (b) ; (146)

if 0 < E < V0, and

cos kT = cos (a) cos (b) 2 + 2

2sin (a) sin (b) ; (147)

if E > V0. Then, for 0 < E < V0 the plot is Band edge energy occurs when cos kT is 1. It is foundnumerically that cos kT = 1 when E = 0:459, cos kT = 0 when E = 0:300, and cos kT = 1 whenE = 0:217. Therefore, the band edges are at E = 0:217 eV and E = 0:459 eV.

5.3. Use the equation of motion (5.34) to show that the period of Bloch oscillation for a one-dimensionalcrystal having lattice period a is

=h

eEa: (148)

22

Solution: Use}dk

dt= eE ; (149)

and assume that is the time required for the electron to accelerate across the full Brillouin zone.Then,

}Z =20

dk

dtdt =

Z =20

eEdt (150)

}k2

k (0)

= eE

2

}a 0= eE

2

=h

eEa:

5.4. Determine the probability current density (A/m) from (3.187) for the Bloch wavefunction

(x) = u (x) eikxei!t; (151)

where u is a time-independent periodic function having the period of the lattice,

u (x) = u (x+ a) : (152)

Solution:

J (r; t) =i}2m

( (r; t)r(r; t)(r; t)r (r; t)) (153)

=i}2m

u (x) eikxei!tr u (x) eikxei!t u (x) eikxei!tr u (x) eikxei!t

= bxi}2m

u (x) eikxei!t

d

dx

u (x) eikxei!t

u (x) eikxei!t ddx

u (x) eikxei!t

= bxi}

2m

u2ik + u0u

u2ik + u0u= bxi}

2m

2u2ik

= bx}k

mu2 = bx p

mu2:

5.5. If an energy-wavevector relationship for a particle of mass m has the form

E =}2

3mk2; (154)

determine the eective mass. (Use (5.29)).

Solution:

m = }2@2E

@k2

1= }2

0@@2}23mk

2

@k2

1A1 = 32m: (155)

5.6. If the energy-wavenumber relationship for an electron in some material is

E =}2

2mcos (k) ;

determine the eective mass and the group velocity. (Use (5.29).) Describe the motion (velocity,direction, etc.) of an electron when a d.c. (constant) electric eld is applied to the material, such thatthe electric eld vector points right to left (e.g., an electron in free space would then accelerate towardsthe right). In particular, describe the motion as k varies from 0 to 2. Assume that the electron doesnot scatter from anything.

23

Solution:

m = }2@2E

@k2

1= }2

0@@2}22m cos (k)

@k2

1A1 = 2mcos k

; (156)

vg =1

}@E

@k=1

}

@}22m cos (k)

@k

= 12

}msin k; (157)

For small positive k values the electron moves in the direction of the eld (to the left), and as k increasesthrough positive value from k = 0 to k = =2, the electron increases its velocity and mass. At k = =2,velocity is maximum and the eective mass is innite. As k changes from =2 to , the velocitydecreases, as does the eective mass. At k = , the velocity is zero. Then, as k increases further, theelectron reverses direction, and its velocity increases again, reaching a maximum at k = 3=2, thendecreasing till k = 2.

5.7. If the energy-wavenumber relationship for an electron in some material is

E = E0 + 2A cos (ka) ; (158)

determine the electrons position as a function of time. Ignore scattering.

Solution: The solution of the equation of motion (ignoring scattering) is (5.36),

k (t) = k (0) +qeE}t: (159)

Velocity is given as

vg (k (t)) =1

}@E (k (t))

@k= 2Aa

}sin (k (t) a) (160)

= 2Aa}sin

qeEa}

t

(161)

and position can be determined from the relationship v = dx (t) =dt as

x (t) =

Z t0

vg (t) dt = Z t0

2Aa

}sin

qeEa}

t

dt (162)

=2A

Eqe

cos

qeEa}

t 1; (163)

and therefore the electron Bloch oscillates in time.

5.8. Consider an electron in a perfectly periodic lattice, wherein the energy-wavenumber relationship in therst Brillouin zone is

E =}2k2

5me;

where me is the mass of an electron in free space. Write down the time-independent eective massSchrdingers equation for one electron in the rst Brillouin zone, ignoring all interactions exceptbetween the electron and the lattice. Dene all terms in Schrdingers equation.

Solution:

m = }2@2E

@k2

1= }2

0@@2}2k25me

@k2

1A1 = 52me; (164)

and so Schrdingers equation is }

2

2md2

dx2

(x) = E (x) (165)

where m is the eective mass, } is the reduced Plancks constant, and E is the energy (V = 0 sincethere is no potential energy term; potential energy is accounted for by the eective mass).

24

5.9. Assume that a constant electric eld of strength E = 1 kV/m is applied to a material at t = 0, andthat no scattering occurs.

(a) Solve the equation of motion (5.34) to determine the wavevector value at t = 1; 3; 7; and 10 ns.

(b) Assuming that the period of the lattice is a = 0:5 nm, determine in which Brillouin zone thewavevector is in at each time. If the wavevector lies outside the rst Brillouin zone, map it intoan equivalent place in the rst zone.Solution: (a) The solution of the equation of motion is

k (t) =qeE}t; (166)

assuming k (0) = 0. Then k (1 ns) = 1:52 109 m1, k (3 ns) = 4:56 109 m1, k (7 ns) =1:06 1010 m1, and k (10 ns) = 1:52 1010 m1.(b) Brillouin zone boundaries occur at kn = n=a, where a is the period and n = 1; 2; 3; :::.Thus, k1 = 6:28 109 m1, k2 = 1:26 1010 m1, k3 = 1:89 1010 m1, etc.. Thus, k is inthe rst zone for t = 1 and 3 ns, the second zone for t = 7 ns, and the third zone for t = 10 nm.For higher zones, subtracting 2=a leads to the equivalent point in the rst zone.t (ns) k (t) (m1) zone equiv. point

in 1st zone

1 1:52 109 1 st 3 4:56 109 1 st 7 1:06 1010 2 nd 1:966 4 10910 1:52 1010 3 rd 2:63 109

5.10. Using the hydrogen model for ionization energy, determine the donor ionization energy for GaAs(me = 0:067me, "r = 13:1).

Solution:

Ed =0:067meq

4e

jqej 8 (13:1)2 "20h2= 5:32 meV. (167)

This compares well with measured values.

5.11. Determine the maximum kinetic energy that can be observed for emitted electrons when photonshaving = 232 nm are incident on a metal surface with work function 5 eV.

Solution:

E = }! = }2c

= }

2c

232 109= 8:568 1019 J (168)

= 5:347 eV

So, the maximum kinetic energy isE e = 0:347 eV. (169)

5.12. Photons are incident on silver, which has a work function e = 4:8 eV. The emitted electrons have amaximum velocity of 9 105 m/s. What is the wavelength of the incident light?Solution:

EKE =1

2mev

2 =1

2 jqejme9 1052 = 2:303 eV (170)

E = }! = e+ EK = 4:8 + 2:303 = 7:103 eV, (171)

=hc

E=

hc

7:103 jqej = 174:67 nm. (172)

25

5.13. In the band theory of solids, there are an innite number of bands. If, at T = 0 K, the uppermostband to contain electrons is partially lled, and the gap between that band and the next lowest bandis 0:8 eV, is the material a metal, an insulator, or a semiconductor?

Solution: Metal

5.14. In the band theory of solids, if, at T = 0 K, the uppermost band to have electrons is completely lled,and the gap between that band and the next lowest band is 8 eV, is the material a metal, an insulator,or a semiconductor? What if the gap is 0:8 eV.

Solution: Insulator. What if the gap is 0:8 eV. Solution: Semiconductor

5.15. Describe in what sense an insulator with a nite band gap cannot be a perfect insulator.

Solution: As long as the band gap is nite, an electron can be elevated to the conduction band,resulting in conduction.

5.16. Draw relatively complete energy band diagrams (in both real-space and momentum space) for a p-typeindirect bandgap semiconductor.

5.17. For an intrinsic direct bandgap semiconductor having Eg = 1:72 eV, determine the required wavelengthof a photon that could elevate an electron from the top of the valance band to the bottom of theconduction band. Draw the resulting transition on both types of energy band diagrams (i.e., energy-position and energy-wavenumber diagrams).

Solution:

}! = Eg = 1:72 eV, (173)

k =2

=!

c! = 2c

!=

2c

Eg=}= 721:3 nm.

5.18. Determine the required phonon energy and wavenumber to elevate an electron from the top of thevalance band to the bottom of the conduction band in an indirect bandgap semiconductor. Assumethat Eg = 1:12 eV, the photons energy is Ept = 0:92 eV, and that the top of the valance band occursat k = 0, whereas the bottom of the conduction band occurs at k = ka.

Eg Ept = Epn = 0:20 eV (174)kpn = ka:

5.19. Calculate the wavelength and energy of the following transitions of an electron in a hydrogen atom.Assuming that energy is released as a photon, using Table 3, on p. 4 classify the emitted light (e.g.,X-ray, IR, etc.).

(a) n = 2! n = 1Solution: From

En = 13:6 1n2; (175)

and so

E = 13:61

22 112

= 10:2 eV, (176)

=hc

E=

hc

10:2 jqej = 121:6 nm, between visible and UV

(b) n = 5! n = 4

E = 13:61

52 142

= 0:306 eV, (177)

=hc

E=

hc

0:306 jqej = 4054:5 nm, far-infrared

26

(c) n = 10! n = 9

E = 13:61

102 192

= 0:0319 eV, (178)

=hc

E=

hc

0:0319 jqej = 3:89 105 m, microwave

(d) n = 8! n = 2

E = 13:61

82 122

= 3:188 eV, (179)

=hc

E=

hc

3:188 jqej = 389:2 nm, visible, violet

(e) n = 12! n = 1

E = 13:61

122 112

= 13:51 eV, (180)

=hc

E=

hc

13:51 jqej = 91:83 nm, between visible and UV

(f) n =1! n = 1

E = 13:61

12 1

12

= 13:6 eV, (181)

=hc

E=

hc

13:6 jqej = 91:23 nm, between visible and UV

5.20. Excitons were introduced in Section 5.4.5 to account for the fact that sometimes when an electron iselevated from the valance band to the conduction band, the resulting electron and hole can be boundtogether by their mutual Coulomb attraction. Excitonic energy levels are located just below the bandgap, since the usual energy to create a free electron and hole, Eg, is lessened by the binding energy ofthe exciton. Thus, transitions can occur at

E = Eg mr

me"2r13:6 eV (182)

where Eg is in electron volts1 .

(a) For GaAs, determine the required photon energy to create an exciton. For mr use the average ofthe heavy and light hole masses.Solution: Using mr = 0:0502me, "r = 13:3, and Eg = 1:43 eV, we nd that

E = Eg mr

me"2r13:6 eV (183)

= 1:43 0:0502(13:3)

2 13:6 eV (184)

= 1:426 eV. (185)

(b) The application of a d.c. electric eld tends to separate the electron and the hole. Using Coulombslaw, show that the magnitude of the electric eld between the electron and the hole is

jEj =mrme

22

"3r jqejRYa0

: (186)

1Really, the quantity 13:6 should be replaced by 13:6=n2, where n is the energy level of the exciton. Here we consider thelowest level exciton (n = 1), which is dominant.

27

Solution: The electric eld due to a charge q in a medium characterized by "r is

E = br q4"r"0r2

= br jEj ; (187)where br is a unit vector that points radially outward from the charge, and r is the radial distanceaway from the charge. Making the substitutions q = qe and r = aex leads to

jEj = jqej4"r"0a2x

=

mrme

2 jqej4"3r"0a

20

(188)

=

mrme

2me jqej4"20h

2

1

"3ra0=

mrme

22RYjqej

1

"3ra0(189)

=

mrme

22

"3r jqejRYa0

: (190)

(c) For GaAs, determine jEj from (5.79). Determine the magnitude of an electric eld that wouldbreak apart the exciton.Solution:

jEj =mrme

22

"3r jqejRYa0

(191)

= (0:0502)2 2

(13:3)3 jqej

13:6 jqej0:053 109 = 5:5 10

5 V/m. (192)

An applied electric eld with a magnitude greater than jEj can break apart the exciton.5.21. The E k relationship for graphene is given by (5.62). The Fermi energy for graphene is EF = 0, and

the rst Brillouin zone forms a hexagon (as shown in Fig. 5.35), the six corners of which correspondto E = EF = 0. The six corners of the rst Brillouin zone at located at

kx = 2p3a; ky = 2

3a; (193)

andkx = 0; ky = 4

3a: (194)

(a) Verify that at these points, E = EF = 0.Solution:

E (kx; ky) = 0

vuut1 + 4 cos p3kxa2

!cos

kya

2

+ 4 cos2

kya

2

;

E

0;4

3a

= 0

s1 + 4 cos

43a

a

2

+ 4 cos2

43a

a

2

= 0s1 + 4 cos

2

3

+ 4 cos2

2

3

= 0

E (kx; ky) = 0

vuut1 + 4 cos p3kxa2

!cos

kya

2

+ 4 cos2

kya

2

;

E

2p

3a;23a

= 0

vuut1 + 4 cos 2p3a

p3a

2

!cos

23a

a

2

+ 4 cos2

23a

a

2

= 0r1 + 4 cos () cos

3

+ 4 cos2

3

= 0

28

(b) At the six corners of the rst Brillouin zone, jkj = 4=3a. Make a two-dimensional plot of theE k relationship for kx; ky extending a bit past jkj. Verify that the bonding and antibondingbands touch at the six points of the rst Brillouin zone hexagon, showing that graphene is asemi-metal (sometimes called a zero bandgap semiconductor). Also make a one-dimensional plotof E (0; ky) for jkj ky jkj, showing that the bands touch at E = 0 at ky = 4=3a.

Solution:

Using (5.62), since a =p3 (0:142 nm) = 0:246 nm, jkj = 4=3a = 17 nm1. Thus,

E mE,

in two dimensions (the bands actually touch at the corners, although in the plot a small gap is showndue to using a coarse wavenumber grid). In one-dimension, for

E (0; ky) = 0s1 + 4 cos

kya

2

+ 4 cos2

kya

2

= 2:5s1 + 4 cos

ky0:246

2

+ 4 cos2

ky0:246

2

we have

15 10 5 0 5 10 153

2

1

0

1

2

33

3-

E 0 ky,( )

E 0 ky,( )-

1717- ky

where the vertical scale is in eV and the horizontal scale is in nm1.

29

5.22. What is the radius of a (19; 0) carbon nanotube? Repeat for a (10; 10) nanotube. Consider a (n; 0)zigzag carbon nanotube that has radius 0:3523 nm. What is the value of the index n?

Solution: The CNs radius is

r =

p3

2bpn2 + nm+m2; (195)

where b = 0:142 nm. Therefore,

r(19;0) =

p3

2

0:142 109p192 = 0:7437 nm

r(10;10) =

p3

2

0:142 109p102 + (10) (10) + 102 = 0:678 nm.

For (n; 0),

n =2ap3b=20:3523 109p3 (0:142 109) = 9:

5.23. Since carbon nanotubes are only periodic along their axis, the transverse wavenumber becomes quan-tized by the nite circumference of the tube. Derive (5.66) and (5.67) by enforcing the condition thatan integer number q of transverse wavelengths must t around the tube (k? = 2=?).

Solution: For the armchair tube (m = n), tube radius is r = 3nb=2. Thus,

q? = 2r = 23nb

2= 3nb

k? = kx;q =2

?=2q

n3b; q = 1; 2; :::; 2n:

For zigzag tubes, (n = 0), r =p3nb=2, and

q? = 2r = 2p3nb

2=p3nb

k? = ky;q =2

?=

2q

np3b; q = 1; 2; :::; 2n:

The limit 2n on q comes from the fact that kx;2n = 4=3b, and ky;2n = 4=p3b, and beyond these

values one is outside of the rst Brillouin zone of graphene.

5.24. Using (5.68) and (5.69), plot the dispersion curves for the rst eight bonding and antibonding bands ina (5; 5), (9; 0), and (10; 0) carbon nanotube. Let the axial wavenumber vary from k = 0 to k = =aacfor the armchair tube, and from k = 0 to k = =azz for the zigzag tube. Comment on whether eachtube is metallic of semiconducting, and identify the band (i.e., the q value) that is most important. Ifthe tube is semiconducting, determine the approximate band gap.

Solution: For the armchair tube (5; 5),

Eac (ky) = 0s1 + 4 cos

qn

cos

kya

2

+ 4 cos2

kya

2

= 0s1 + 4 cos

q5

cos

kya

2

+ 4 cos2

kya

2

< kyaac < , q = 1; 2; :::; 2n, and so

30

0 2 4 6 8 10 123

2

1

0

1

2

32.87

2.87-

E 1 k,( )

E 1 k,( )-

E 2 k,( )

E 2 k,( )-

E 3 k,( )

E 3 k,( )-

E 4 k,( )

E 4 k,( )-

E 5 k,( )

E 5 k,( )-

p.246

0 k

(vertical scale is E=0). The q = 5 band (note that n = 5!) is the most important, since these bandscross in the rst Brillouin zone (and hence, there is no band gap). The crossing point is 2=3 of the wayto the zone boundary, and so kF = 2=3a, such that the Fermi wavelength is F = 3a = 0:74 nm.

For zigzag tubes,

Ezz (kx) = 0

vuut1 + 4 cos p3kxa2

!cosqn

+ 4 cos2

qn

;

< kyazz < , q = 1; 2; :::; 2n. For the (9; 0) tube,

0 2 4 63

2

1

0

1

2

32.879

2.879-

E 1 k,( )

E 1 k,( )-

E 2 k,( )

E 2 k,( )-

E 3 k,( )

E 3 k,( )-

E 4 k,( )

E 4 k,( )-

E 5 k,( )

E 5 k,( )-

E 6 k,( )

E 6 k,( )-

E 7 k,( )

E 7 k,( )-

E 8 k,( )

E 8 k,( )-

p

3 .246

0 k

where the q = 6 bands cross at k = 0, and, hence, this tube is metallic. For the (10; 0) tube,

31

0 2 4 63

2

1

0

1

2

32.902

2.902-

E 1 k,( )

E 1 k,( )-

E 2 k,( )

E 2 k,( )-

E 3 k,( )

E 3 k,( )-

E 4 k,( )

E 4 k,( )-

E 5 k,( )

E 5 k,( )-

E 6 k,( )

E 6 k,( )-

E 7 k,( )

E 7 k,( )-

E 8 k,( )

E 8 k,( )-

p

3 .246

0 k

no bands cross, hence, the (10; 0) tube is a semiconductor. The q = 7 bands come the closest to eachother (at k = 0), and so the band gaps is 2E (k = 0) for q = 7, which is approximately 0:88 eV using

0 = 2:5 eV. It can be shown (See the book by Saito, Dresselhaus, and Dresselhaus, Reference [9] inChapter 5) that

Eg =20ap32r

;

which for the (10; 0) tube (r = 0:391 nm) leads to 0:90 eV.

6 Problems Chapter 6: Tunnel Junctions and Applications ofTunneling

6.1. Plot the tunneling probability versus electron energy for an electron impinging on a rectangular poten-tial barrier (Fig. 6.2, p. 185) of height 3 eV and width 2 nm. Assume that the energy of the incidentelectron ranges from 1 eV to 10 eV.

Solution:

T =4E (E V0)

V 20 sin2 (k2a) + 4E (E V0)

; k22 =2me (E V0)

}2(196)

T =4E jqej (E jqej 3 jqej)

32 jqej2 sin2q

2me(Ejqej3jqej)}2 (2 109)

+ 4E jqej (E jqej 3 jqej)

6.2. Plot the tunneling probability verses barrier width for a 1 eV electron impinging on a rectangularpotential barrier (Fig. 6.2, p. 185) of height 3 eV. Assume that the barrier width varies from 0 nm to3 nm.

Solution:

T =4E (E V0)

V 20 sin2 (k2a) + 4E (E V0)

; k22 =2me (E V0)

}2(197)

T =4 (1) jqej ((1) jqej 3 jqej)

32 jqej2 sin2q

2me((1)jqej3jqej)}2 (a 109)

+ 4 (1) jqej ((1) jqej 3 jqej)

32

6.3. A 6 eV electron tunnels through a 2 nm wide rectangular potential barrier with a transmission coef-cient of 108 . The potential energy is zero outside of the barrier, and has height V0 in the barrier.What is the height V0 of the barrier?

Solution:

T =4E (E V0)

V 20 sin2 (k2a) + 4E (E V0)

; k22 =2me (E V0)

}2(198)

T =4 (6 jqej) (6 jqej V0 jqej)

V 20 jqej2 sin2q

2me(6jqejV0jqej)}2 (2 109)

+ 4 (6 jqej) (6 jqej V0 jqej)

= 108; (199)

V0 = 6:858 eV.

6.4. Can humans tunnel? Consider running 1 m/s (assume that you have no potential energy) at an energybarrier of 50 joules that is 1 m thick. (a) If you weigh (i.e., your mass is) 50 kg, determine theprobability that you will tunnel through the barrier. (b) Consider that in order for tunneling to have areasonably large probability of occurring, k2a cant be too large in magnitude. Discuss the conditionsthat would result in this happening.

Solution:. (a) Your kinetic energy is

E =1

2mv2 =

1

2(50) (1)

2= 25 J. (200)

Then,

k2 =

r2m (E V0)

}2=

r2 (50) (25 50)

}2= i4:74 1035; (201)

T =4E (E V0)

V 20 sin2 (k2a) + 4E (E V0)

=4 (25) (25 50)

502 sin2 ((i4:74 1035) 1) + 4 (25) (25 50) (202)

' 2e9:481035

which is extremely small.

(b) Since

k2a =

r2m (E V0)

}2a; (203)

given the extremely small value of }2, obviously one would need an extremely small mass, a su cientlysmall value of E V0, and probably also a very small value of a.

6.5. Referring to the development of the tunneling probability through a potential barrier, as shown inSection 6.1, apply the boundary conditions (3.143) to (6.13) to obtain (6.14).

6.6. Consider the metalinsulator junction shown in Fig. 6.4 on p. 190. Solve Schrdingers equationin each region (metal and insulator), and derive tunneling and reection probabilities analogous to(6.15)(6.16) for this structure.

Solution: In region I (x < 0), where V = 0, Schrdingers equation is

}2

2m

d2

dx2 1 (x) = E 1 (x) ; (204)

which has solutions

1 (x) = Aeik1x +Beik1x; k21 =

2mE

}2: (205)

In region II (0 x), where V = V0 = Evac, Schrdingers equation is }

2

2m

d2

dx2+ V0

2 (x) = E 2 (x) ; (206)

33

which has solutions

2 (x) = Ceik2x +Deik2x; k22 =

2m (E V0)}2

(207)

Since there is no potential disturbance to reect the wave after it reaches region II, D = 0. Therefore,we have

1 (x) = Aeik1x +Beik1x; (208)

2 (x) = Ceik2x:

The boundary conditions continuity of and 0 at x = 0, lead to

A+B = C; (209)

k1A k1B = k2C;

so that

B

A=

k1 k2k1 + k2

; (210)

C

A= 2

k1k1 + k2

:

Therefore,

T =

CA2 = 2k1k1 + k2

2 ; (211)R =

BA2 = k1 k2k1 + k2

:6.7. Consider a metalinsulatormetal junction, as shown in Fig. 6.7 on p. 193, except assume two dierent

metals Fermi levels. (a) Draw the expected band diagram upon rst bringing the metals into closeproximity. (b) Because of the dierence in Fermi levels, tunneling will occur (assuming that the barrierbetween the metals is thin), and will continue until a su cient voltage is built up across the junction,equalizing the Fermi levels. This internal voltage is called the built-in voltage. Draw the energy banddiagram showing the built-in voltage in this case.

6.8. Draw the potential energy prole for a metalvacuummetal structure when a voltage V0 is appliedacross the vacuum region.

6.9. Determine the tunnelling probability for an Al-SiO2-Al system, if the SiO2 width is 1 nm and theelectron energy is 3:5 eV. Repeat for an SiO2 width of 2 nm, 5 nm, and 10 nm.

Solution: The modied work function for an Al-SiO2 junction is 3:2 eV, as given in Table 6.2 (p. 191).This plays the role of the barrier height V0 in the generic tunneling problem. Therefore,

T =4E (E V0)

V 20 sin2 (k2a) + 4E (E V0)

; k22 =2me (E V0)

}2(212)

=4 (3:5) jqej ((3:5) jqej 3:2 jqej)

3:22 jqej2 sin2q

2me((3:5)jqej3:2jqej)}2 (1 109)

+ 4 (3:5) jqej ((3:5) jqej 3:2 jqej)

= 0:791

For a = 2 nm, T = 0:515, for a = 5 nm, T = 0:293, and for a = 10 nm, T = 0:901.

6.10. Use the WKB tunneling approximation (6.30) to determine the tunneling probability for the rectangularbarrier depicted in Fig. 6.2 on p. 185.

34

Solution: From x = 0 to x = a, V (x) = V0. Thus

T ' e2R x2x1

(x)dx= e

2 R a0

q2m}2

(V0E)dx

= e2a

q2m}2

(V0E)

which has the form of (6.27).

6.11. Plot the tunneling probability versus electron energy for an electron impinging on a triangular potentialbarrier (Fig. 6.9, p. 196), where e = 3 eV and the electric eld is 109 V/m. Assume that the energydierence (E EF ) ranges from 0 to 3 eV.Solution: Since

T = exp

4p2me3 jqeEj } (e (E EF ))

3=2

(213)

= exp

4p2me3 jqe109j } (3 jqej Ed jqej)

3=2

6.12. Consider the double barrier structure depicted in Fig. 6.22 on p. 206.

(a) Plot the tunneling probability versus electron energy for an electron impinging on the doublebarrier structure. The height of each barrier is 0:5 eV, each barrier has width a = 2 nm, and thewell has width 4 nm. Assume that the energy of the incident electron ranges from 0:1 eV to 3 eV,and that the eective mass of the electron is 0:067me in all regions.

(b) Verify that the rst peak of the plot corresponds to an energy approximately given by the rstdiscrete bound state energy of the nite-height, innitely-thick-walled well formed by the twobarriers. Use (4.83) adopted to this geometry, i.e.,

k2 tan (k2L=2) = k1; (214)

where L = 4 nm and

k2 =

r2meE}2

; k1 =

r2me (V0 E)

}2: (215)

Solution:

T =

1 +

4R1T 21

sin2 (k1L )1

; (216)

where

T1 =4E (E V0)

V 20 sin2 (k2a) + 4E (E V0)

; (217)

R1 =V 20 sin

2 (k2a)

V 20 sin2 (k2a) + 4E (E V0)

; (218)

are the transmission and reection coe cients for a single barrier of width a, L is the length of thewell between the barriers, and

tan =2k1k2 cos (k2a)

(k21 + k22) sin (k2a)

; (219)

where

k1 =

r2 (0:067)meE jqej

}2; k2 =

r2 (0:067)me (E jqej 0:5 jqej)

}2: (220)

A plot of T vs. E is shown above.

35

10.750.50.25

1

0.75

0.5

0.25

0

x

y

x

y

(b) r2 (0:067)meE jqej

}2tan

r2 (0:067)meE jqej

}2

4

2 109

!(221)

r2 (0:067)me (j0:5 jqej E jqejj)

}2= 0

Root is: E = 0:143 63 eV.

6.13. Derive the tunneling probability (6.38) for the double barrier junction depicted in Fig. 6.22 on p. 206.

6.14. Research how tunneling is utilized in ash memories, and describe one such commercial ash memoryproduct.

6.15. Research how eld emission is used in displays, and summarize the state of display technology basedon eld emission.

6.16. Explain how a negative resistance device can be used to make an oscillator.

Solution: An LC circuit can exhibit a pure resonance, although the presence of a ordinary resistor willdissipate energy and damp the oscillation. A negative resistance can cancel the ordinary resistance,and lead to a sustained oscillation. Alternatively, a circuit consisting of an inductor, a capacitor, anda negative resistance can, in theory, exhibit oscillations that grow in time.

7 Problems Chapter 7: Coulomb Blockade and the Single Elec-tron Transistor

7.1. For a tunnel junction with C = 0:5 aF and Rt = 100 k, what is the RC time constant? What doesthis value mean for the tunnel junction circuit?

Solution: = RtC =

0:5 1018 100 103 = 5:0 1014 s (222)

is the characteristic time between tunneling events.

7.2. For a tunnel junction having C = 0:5 aF and Rt = 100 k, what is the maximum temperature atwhich you would expect to nd Coulomb blockade? Repeat if C = 1:2 pF.

Solution: We must haveq2e2C

kBT; (223)

36

so that

T q2e

2CkB=

1:6 10192

2 (0:5 1018) (1:38 1023) = 1855K. (224)

For C = 1:2 pF,

T q2e

2CkB=

1:6 10192

2 (1:2 1012) (1:38 1023) = 7:73 104 K. (225)

7.3. For the capacitor depicted in Fig. 7.2 on p. 215, for an electron to tunnel from the negative terminalto the positive terminal we determined the condition

V >qe2C

(226)

(see (7.8)). Show, including all details, that for an electron to tunnel from the positive terminal to thenegative terminal we need

V 0, thenwe nd that

qe (Q qe=2)C

> 0; (230)

! Q < qe2

(231)

for tunneling to occur. Thus,V 0; (235)

37

then

Q > N2qe (236)

or

V > N2C

qe: (237)

7.5. There is always a capacitance between conductors separated by an insulating region. For the case ofconductors associated with dierent circuits (i.e., circuits that should operate independently from oneanother), this is called parasitic capacitance, and, generally, C / 1=d, where d is some measure of thedistance between the conductors. For two independent circuits d is often fairly large, and thus verysmall parasitic capacitance are generally present.

(a) Using the formula for the impedance of a capacitor, show that even for very small values of C,at su ciently high frequencies the impedance of the capacitor can be small, leading to signicantunintentional coupling between the circuits.

(b) Describe why the small values of parasitic capacitance, which can easily be aF or less, do not leadto Coulomb blockade phenomena.Solution: (a).

Zc =1

j!C; (238)

and, therefore, even if C is very small, if ! is su ciently large jZcj can be small.(b). If C is small due to large separation of the conductors, Rt will be too large (i.e., tunnelingwill not occur), and no Coulomb blockade phenomena will occur.

7.6. For the SET oscillator shown in Fig. 7.10 on p. 224, derive the oscillation period given by (7.31).

Solution: Using (7.25),

V (t) =1

C

Z t0

Is dt =IsCt; (239)

then

V

T

2

=IsC

T

2=

e

2C; (240)

so that

T =e

Is=jqejIs: (241)

7.7. As another way of seeing that Coulomb blockade is di cult to observe in the current-biased junctionshown in Fig. 7.9 on p. 223, consider the equivalent circuit shown in Fig. 7.11 on p. 225. If Rb issu ciently large so that it can be ignored, and if jZLj is su ciently small, construct an argument, froma total capacitance standpoint, for why the amplitude of the SET oscillations will tend towards zero.

Solution: Referring to Fig. 7.11, if we ignore Rb and ZL, then we have simply a current source inparallel with a capacitance and a tunnel junction. The tunnel junction itself consists of a capacitorin parallel with a tunnel resistance, and therefore the two capacitors combine in parallel (i.e., alge-braically), resulting in, typically, a large capacitance. The amplitude of the oscillation is e2=2C, andfor large C this amplitude is small.

7.8. For the quantum dot circuit depicted in Fig. 7.13 on p. 226, assume that initially there are n = 100electrons on the dot. If Ca = Cb = 1:2 aF, what is the condition on Vs for an electron to tunnel ontothe dot through junction b?

Solution: From

Vs >qeCa

n+

1

2

; (242)

so that

Vs >qeCa

100 +

1

2

: (243)

38

For Ca = 1:2 aF, then

Vs >qe

1:2 1018100 +

1

2

= 13:42 V (244)

7.9. For the quantum dot circuit depicted in Fig. 7.13 on p. 226, we found that for an electron to tunnelonto the dot through junction b, and then o of the dot through junction a, we need

Vs >qe2C

(245)

if initially there were no electrons on the dot (n = 0), where Ca = Cb = C. It was then stated that forthe opposite situation, where an electron tunnels onto the dot through junction a, and then o of thedot through junction b, we would need

Vs 0; (252)

such that

Vs qe (n+ 1=2) CgVg

Ca: (256)

we have

Vs >qe (175 + 1=2)

1:4 1016 (0:1)

10 1018 = 1:41 V. (257)

7.12. For the SET we considered electrons tunneling onto the island through junction b, then o of the islandthrough junction a, resulting in positive current ow (top-to-bottom). Explicit details were providedfor the generation of Coulomb diamonds for I > 0, and the results merely stated for I < 0. Fill in thedetails of the derivation predicting Coulomb diamonds for I < 0.

7.13. Draw the energy band diagram for a SET (a) under zero bias (Vs = Vg = 0), (b) when Vs > 0 andVg = 0, and (c), when Vs > 0 and Vg = jqej = (2Cg).

7.14. Consider an electron having kinetic energy 5 eV.

(a) Calculate the de Broglie wavelength of the electron.

(b) If the electron is conned to a quantum dot of size L L L, discuss how big the dot should befor the electrons energy levels to be well-quantized.

(c) To observe Coulomb blockade in a quantum dot circuit, is it necessary to have energy levels onthe dot quantized? Why or why not?Solution: (a)

EKE =1

2mev

2 ! v =r2EKEme

(258)

=h

p=

h

mev=

h

me

q2EKEme

=hp

2meEKE jqej(259)

= 0:548 nm.

The electron will act like a classical particle when L .(b) L .(c) It is not necessary to have energy levels on the dot quantized. The developed formulas didnot assume energy quantization.

7.15. Assume that a charge impurity q = 2qe resides in an insulating region. Determine the force on anelectron 10 nm away from the impurity. Repeat for the case when the electron is 10 m away from theimpurity.

Solution: The force on an electron by a charge 2qe is

F = qeE = brqe 2qe4"0r2

! jFj = 2q2e

4"0 (10 109)2= 4:617 1012 N; (260)

40

and at 10 m,

jFj = 2q2e

4"0 (10 106)2= 4:617 1018 N. (261)

7.16. Universal conductance uctuations (UCF) occur in Coulomb blockade devices due to the interferenceof electrons transversing a material by a number of paths. Using other references, write one-half toone page on UCFs, describing the role of magnetic elds and applied biases.

7.17. Summarize some of the technological hurdles that must be overcome for molecular electronic devicesto be commercially viable.

7.18. There is currently a lot of interest in spintronics, which rely on the spin of an electron to carryinformation (see Section 10.4). In one-half to one page, summarize how spin can be used to providetransistor action.

8 Problems Chapter 8: Particle Statistics and Density of States

8.1. Energy levels for a particle in a three-dimensional cubic space of side L with hard walls (boundaryconditions (4.50)) were found to be (4.54),

En =}22

2meL2n2x + n

2y + n

2z

; (262)

nx;y;z = 1; 2; 3; :::, which leads to the density of states (8.6). Using periodic boundary conditions (4.55),energy levels were found to be (4.59)

En =2}22

meL2n2x + n

2y + n

2z

; (263)

nx;y;z = 0;1;2; :::. Following a derivation similar to the one shown for (8.6), show that the samedensity of states arises from (8.65).

Solution: In the hard wall case we only count the rst octant of the aforementioned sphere, since signchanges dont lead to additional states. For the case of periodic BCs, sign is important, and so we areinterested in the total number of states NT having energy less than some value (but with E E1).This is approximately the volume of the sphere

NT =4

3n3 =

4

3

E

E1

3=2(264)

(a factor of 1=8 was present for the hard-walled case). The total number of states having energy in therange (E;E E) is

NT =4

3

E3=21

(E)

3=2 (E E)3=2

(265)

=4

3

E3=21

(E)

3=2 E3=21 E

E

3=2!

' 43

E3=2

E3=21

1

1 3

2

E

E

=4

3

E3=2

E3=21

3

2

E

E

=2E1=2

E3=21

(E) ;

where we used (1 x)p ' 1 px for x 1. The density of states (DOS), N (E), is dened as thenumber of states per unit volume per unit energy around an energy E. The total number of states in

41

a unit volume in an energy interval dE around an energy E is (replacing NT , E with dNT , dE,respectively)

dNT = N (E) dE =2E1=2

E3=21

dE; (266)

N (E) =2E1=2

E3=21

=m3=2e E1=2

21=2}32

(since the density of states is per-unit-volume, we set L3 = 1 in the expression for E1). Accounting forspin, we multiply by 2, such that

N (E) =21=2m

3=2e E1=2

}32: (267)

Finally, if the electron has potential energy V0, we have

N (E) =21=2m

3=2e (E V0)1=2}32

: (268)

8.2. Derive (8.16), the density of states in two-dimensions.

Solution: From the equation for energy the number of states below a certain energy En is equal tothe number of states inside a circle of radius

n =qn2x + n

2y =

rEnE1

; (269)

Considering the hard-wall case with non-periodic boundary conditions and counting the rst octant ofthe circle (since sign changes dont lead to additional states), the total number of states NT havingenergy less than some value (but with E E1) is approximately the area of the octant,

NT =1

4n2 =

4

E

E1

: (270)

The total number of states having energy in the range (E;E E) is

NT =

4E1(E (E E)) (271)

=

4E1E:

The density of states, N (E), is dened as the number of states per unit area per unit energy aroundan energy E. The total number of states in a unit area in an energy interval dE around an energy Eis (replacing NT , E with dNT , dE, respectively)

dNT = N (E) dE =

4E1dE; (272)

N (E) =

4E1=

me2}2

(273)

(setting L = 1 in the expression for E1). Accounting for spin, we multiply by 2, such that

N (E) =me}2

; (274)

which is the desired result.

8.3. The density of states in a one-dimensional system is given by (8.15),

N (E) =

p2me}

E1=2; (275)

assuming zero potential energy.

42

RaymondHighlight

RaymondSticky Notemistake?

(a) Use this formula to show that the Fermi energy in terms of the total number of lled states atT = 0 K, Nf , is (8.40),

EF =}2

2me

Nf2

2: (276)

Solution:

Nf =

Z EF0

N (E) dE =

p2me}

Z EF0

E1=2dE =p2me}

2pEF ;

EF =

}Nf2p2me

2=

}2

2me

Nf2

2(b) If there are N electrons in a one dimensional box of length L, show that the energy level of

the highest energy electron is EF , given by (276). Use the fact that the energy levels in a one-dimensional box are (4.35),

En =}2

2me

nL

2; n = 1; 2; 3; ::: (277)

Solution:

En =}2

2meINT

N

2

2= EF (278)

where INT is the integer part, rounded up. N=2 is used, rather than N , to account for spin, andL = 1 since the Fermi energy is per unit length.

8.4. Assume that the density of states in a one-dimensional system is given by

N (E) =

p2m

}E1=3 (279)

at zero potential energy.

(a) Use this formula to obtain the Fermi energy.

(b) What is the relationship between the de Broglie wavelength and the Fermi wavelength?Solution: (a)

Nf =

Z EF0

N (E) dE =

p2m

}

Z EF0

E1=3dE =p2m

}3

2E2=3F ; (280)

EF =

2}Nf3p2m

3=2=

hNf

3p2m

3=2:

(b) The Fermi wavelength is the de Broglie wavelength at the Fermi energy.

8.5. The electron carrier concentration in the conduction band can be determined by multiplying the elec-tron density of states in the conduction band, Ne (E), and the probability that a state is occupied,f (E;EF ; T ), and then summing over all energies to yield (8.54) on p. 275. Perform the analogouscalculations for determining the hole carrier concentration in the valance band, leading to (8.56).

Solution: Start with

p =

Z Nv1

Np (E) (1 f (E;EF ; T )) dE; (281)

where f (E;EF ; T ) is the Fermi-Dirac function. This leads to

p =

Z Ev1

p2m

3=2p (Ev E)1=2

2~3

!1

eEFEkBT + 1

dE (282)

=1

22

2mp~2

3=2 Z Ev1

(Ev E)1=2

eEFEkBT + 1

!dE:

43

If (EF E) = (kBT ) 1 (Boltzmann approximation), then

p =1

22

2mp~2

3=2eEFkBT

Z Ev1

(Ev E)1=2 eE

kBT dE: (283)

With Z Ev1

(Ev E)1=2 e

EkBT

dE = e

EvkBT

Z Ev1

(Ev E)1=2 eEvEkBT dE (284)

= eEvkBT (kBT )

1=2(kBT )

Z 10

u1=2eudu

= eEvkBT (kBT )

3=2 1

2

p;

using the change of variables u = (Ev E) = (kBT ), du = dE= (kBT ), andZ 10

u1=2eudu =1

2

p; (285)

then

p =1

22

2mp~2

3=2eEFkBT e

EvkBT (kBT )

3=2 1

2

p (286)

= 2

mpkBT2~2

3=2eEvEFkBT (287)

= NveEvEFkBT :

8.6. The Fermi wavelength in three-dimensional copper is F = 0:46 nm. Determine the Fermi wavelengthin two-dimensional copper.

Solution:(2d)F =

2

k(2d)F

=2

(2N)1=2

=2

1092 (8:45 1028)2=3

1=2 = 0:571 nm. (288)8.7. Determine the Fermi wavelength of electrons in three-dimensional aluminum, and zinc.

Solution: Fromk(3d)F =

3N2

1=3; F =

2

kF: (289)

From the appendix, for aluminum, N = 18:06 1028 m3, and for zinc, N = 13:10 1028 m3.Therefore,

k(3d)F =

318:06 1028 21=3 = 1:749 1010 m1, aluminum (290)

=313:10 1028 21=3 = 1:571 1010 m1, zinc, (291)

and so

F = 0:359 nm for aluminum (292)

= 0:399 nm for zinc (293)

8.8. What is the electron concentration in an n-type semiconductor at room temperature if the material isdoped with 1014 cm3 donor atoms? How would one determine the hole concentration?

Solution: Doping is su ciently heavy such that

n ' Nd = 1014 cm3 (294)The hole concentration is given by (8.58), utilizing the equation after (8.62). One would need to knowEv.

44

8.9. The concept of the Fermi energy can be used to give some condence of electron-electron screening inconductors (i.e., the ability to ignore interactions among electrons), previously described in Sections3.5 and 4.2. To see this, approximate the electrons kinetic energy by the Fermi energy. Then, for anelectron density N m3, assuming that the average distance between electrons is N1=3, show that theratio of Coulomb potential energy to kinetic energy goes to zero as N goes to innity, showing thatthe electrons essentially screen themselves.

Solution: For a three-dimensional material EF is

EF =

}2

2me

3N2

2=3 ' EKE : (295)For an electron density N m3, assuming that the electrons uniformly ll the space, the averagedistance between electrons is N1=3. The potential energy between electrons separated by a distanced is

V (d) =e2

4"0d; (296)

and, with d = N1=3 we have

V (N) =e2

4"0N1=3: (297)

Therefore, the ratio of an electrons potential to kinetic energy is

V

EKE/ N

1=3

N2=3= N1=3; (298)

and so V

EKE

! 0 as N !1: (299)

Therefore, as the electron density becomes large, the electrons essentially screen themselves, such thatwe can often ignore electron-electron interactions.

8.10. Constructive and destructive interference of electromagnetic waves (light, radio-frequency signals, etc.)is one of the most commonly exploited phenomena in classical high-frequency devices. For example,resonance eects result from wave interference, and are used to form lters, impedance transformers,absorbers, and antennas, etc. In contrast to these electromagnetic (i.e., photon) devices, which obeyBose-Einstein statistics, electron waves are fermions, and obey Fermi-Dirac statistics and the exclusionprinciple. Describe how it is possible for a large collection of bosons to form a sharp interferencepattern, yet this will not be observed by a large collection of electrons in a solid. Does this make senseconsidering Fig. 2.6 on p. 29, where sharp electron interference was, in fact, observed?

Solution: Since photons/electromagnetic waves dont obey the Pauli principle, they can all have thesame energy (i.e., the same frequency), and thus, via the de Broglie wavelength, they can all havethe same wavelength. Since interference is based on wavelength, a large group of photons, all havingthe same wavelength, will all contribute to forming an interference pattern. However, electrons in agroup must all have dierent quantum numbers by the Pauli exclusion principle. Excepting for spin,free electrons will all have dierent energies, and hence dierent wavelengths. Thus, although thedierence in wavelengths may be small, a sharp (due to electrons with precisely the same energy) andstrong (contributed by many electrons) interference pattern will not result. This is not at odds withFig. 2.6 on p. 29, since in electron interference experiments electrons are emitted one-at-a-time. Sincethey are not present at the same time, they can have the same energy and wavelength.

45

9 Problems Chapter 9: Generic Models of Semiconductor Quan-tum Wells, Quantum Wires, and Quantum Dots

9.1. A nite region of space is said to be eectively two-dimensional if Lx F Ly; Lz. However, Fwas computed from the three-dimensional result (9.3)

F =2

(3Ne2)13

; (300)

where Ne is the electron density m3. Is it then permissible to use this formula to conclude that astructure is two- or one-dimensional? Shouldnt the two- or one-dimensional F be used? Why or whynot.

Solution: From (8.42) we have

F =2

(3N3d2)1=3

in 3d, (301)

=2

(2N2d)1=2

in 2d, (302)

=4

N1din 1d. (303)

Assuming N2d =N3d

2=3, and N1d =

N3d

1=3, then

F =2

(3N3d2)1=3

=2

(32)1=3(N3d)

1=3= 2:031

1

(N3d)1=3

in 3d, (304)

=2

2 (N3d)2=31=2 = 2p

2 (N3d)1=3

= 2:5071

(N3d)1=3

in 2d, (305)

=4

(N3d)1=3

= 41

(N3d)1=3

in 1d, (306)

and the various values for F are in close agreement.

9.2. At the beginning of this chapter, it was stated that the Fermi wavelength is the important parameter indeciding if quantum connement eects are important, and for determining the eective dimensionalityof a system. Another method to characterize when quantum connement eects are important is to saythat if the dierence in adjacent energy levels for a nite space is large compared with other energiesin the system (thermal, etc.), then connement eects are important. Considering a one-dimensionalinnite-height potential well of width L, as considered in Section 4.3.1, show that connement eectswill be important compared with thermal energy when

L 1

2kBT; (309)

L