Principles of Semiconductor Physics Problem Set...

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Principles of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate the Schr¨ odinger equation, - ¯ h 2 2m d 2 ψ(x) dx 2 - λδ (x)ψ(x)= (x), over x from -² to ², we get - ¯ h 2 2m dx ² -² - λ Z ² -² δ (x)ψ(x)dx = E Z ² -² ψ(x)dx. In the limit ² 0, the righthand side term vanishes, since ψ(x) is finite. Utilizing the property of δ -function, we have Z ² -² δ (x)ψ(x)dx = ψ(0). Therefore, we find lim ²0 dx ! x=+² - ˆ dx ! x=-² # = - 2¯ h 2 ψ(0). (b) The particle is free except at x = 0. The appropriate wavefunction can be chosen as ψ(x)= ( Ae kx x< 0, Ae -kx x> 0, where k = q 2m|E|/¯ h and E< 0. Therefore, lim ²0 ˆ dx ! x=+² = lim ²0 (-kAe -kx ) x=+² = -kA, and similarly, lim ²0 ˆ dx ! x=-² = lim ²0 (kAe kx ) x=-² = kA. To match the wavefunction at x = 0, lim ²0 dx ! x=+² - ˆ dx ! x=-² # = -2kA = - 2¯ h 2 A, 1

Transcript of Principles of Semiconductor Physics Problem Set...

Page 1: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Principles of Semiconductor PhysicsProblem Set #1, Solutions

1. (a) Integrate the Schrodinger equation,

− h2

2m

d2ψ(x)

dx2− λδ(x)ψ(x) = Eψ(x),

over x from −ε to ε, we get

− h2

2m

dx

∣∣∣∣∣

ε

−ε

− λ∫ ε

−εδ(x)ψ(x)dx = E

∫ ε

−εψ(x)dx.

In the limit ε → 0, the righthand side term vanishes, since ψ(x) is finite.Utilizing the property of δ-function, we have

∫ ε

−εδ(x)ψ(x)dx = ψ(0).

Therefore, we find

limε→0

[(dψ

dx

)

x=+ε

−(

dx

)

x=−ε

]= −2mλ

h2 ψ(0).

(b) The particle is free except at x = 0. The appropriate wavefunctioncan be chosen as

ψ(x) =

{Aekx x < 0,Ae−kx x > 0,

where k =√

2m|E|/h and E < 0. Therefore,

limε→0

(dψ

dx

)

x=+ε

= limε→0

(−kAe−kx)x=+ε = −kA,

and similarly,

limε→0

(dψ

dx

)

x=−ε

= limε→0

(kAekx)x=−ε = kA.

To match the wavefunction at x = 0,

limε→0

[(dψ

dx

)

x=+ε

−(

dx

)

x=−ε

]= −2kA = −2mλ

h2 A,

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Page 2: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

noticing that ψ(0) = A, based on our assumption. Therefore,

k =mλ

h2 ,

E = − h2k2

2m= −mλ2

2h2 ,

which is the only bound state of the δ-potential.

2. The particle is free of potential except at x = ±a. For a bound statewith E < 0, we can assume the wavefunction is

ψ(x) =

Aekx x < −aBekx + Ce−kx −a < x < aDe−kx x > a,

where k =√−2mE/h.

The continuity of ψ at ±a requires

Ae−ka = Be−ka + Ceka (1)

De−ka = Beka + Ce−ka (2)

Generalizing the equation of discontinuity in dψ/dx obtained in problem1 to x = ±a, we get

limε→0

[(dψ

dx

)

x=±a+ε

−(

dx

)

x=±a−ε

]= −2mλ

h2 ψ(±a),

which leads to

(kBe−ka − kCeka)− kAe−ka = −2mλ

h2 Ae−ka (3)

−kDe−ka − (kBeka − kCe−ka) = −2mλ

h2 Deka (4)

Eliminating A from Eq. (1) and (3), we get

kBe−ka − kCeka = (k − 2mλ

h2 )(Be−ka + Ceka),

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Page 3: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

which can be rearranged as

B

C=

(2k − 2mλh2 )eka

2mλh2 e−ka

. (5)

Similarly, from Eq. (2) and (4), we can get

B

C=

2mλh2 e−ka

(2k − 2mλh2 )eka

.

Therefore, [2mλ

h2 e−ka

]2

=

[(2k − 2mλ

h2 )eka

]2

,

which implies, as expected,

[h2k

mλ− 1

]2

= e−4ka. (6)

(b) Eq. (6) can be rewritten as

h2k

mλ− 1 = ±e−2ka, (7)

where ± correspond to symmetric and antisymmetric state, respectively.The ground state correspond to + sign. Eq. (5) becomes

B

C= 1,

which implies A = D through Eq. (1) and (2). This state, as expected, hasan symmetric wavefunction,

ψ0(−x) = ψ0(x),

with energy E0 = − h2k20

2m, where h2k0

mλ− 1 = e−2k0a. The ground state is

guaranteed to exist, as we can see from the sketch, there is always a cross-point of the linear function h2k

mλ− 1 and the exponential function e−2ka, no

matter what a and λ are.

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Page 4: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

(c) The second state, corresponding to the − sign in Eq. (7), is anantisymmetric state,

ψ1(−x) = −ψ1(x),

since we can show B = −C and A = −D through Eq. (5), (1) and (2).However, the excited state exists only when the coupling between the two

δ-functions is relatively weak. Perturbatively, when the excited state energyis very close to the critical value E = 0, the wavefunction is, to the lowestorder,

ψ(x) =

ka x < −a−kx −a < x < a−ka x > a,

since k ∼√|E| ∼ 0. (Practically, |x| is taken within some cutoff ∼ 1

k.) The

discontinuity in dψ/dx at x = a becomes

0− (−k) = −2mλ

h2 (−ka),

which implies the critical coupling is thus

λa =h2

2m.

The same equation can be obtained from the first order Taylor expansion ofEq. (7) with the negative sign,

h2k

mλ− 1 = −(1− 2ka).

When the coupling of the two δ-functions is weaker, or a > h2

2mλ, there exists

the second antisymmetric bound state with energy E1 = − h2k21

2m, where 1 −

h2k1

mλ= e−2k1a.

Graphically, the critical point is approached as the two functions 1− h2kmλ

and e−2ka are tangent at k = 0.(d) The energy diagram is shown in Fig. 2. When the coupling is very

weak (a À h2

2mλ), the two energies are nearly degenerate, which is close to

the bound state energy of a single δ-well as we obtained in problem 1.When the coupling becomes stronger, the two states begin to repulse each

other.

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Page 5: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Figure 1: Evolution of the two energy states in delta-function molecule.Insert shows how the two states are obtained.

When the coupling is less than critical value (a = h2

2mλ), the excited state

disappears.When a = 0, the two δ-wells collapse on top of each other, becoming a

single δ-well with a doubled strength 2λ, the ground state energy is expectedto be scaled by a factor of 4.

2. (a) Assume a solution for E < 0,

ψ(x) = AeKx + Be−Kx, 0 < x < a,

where K =√−2mE/h. Using the Bloch condition

ψ(x + a) = eikaψ(x),

we write for −a < x < 0,

ψ(x) = e−ika(AeK(x+a) + Be−K(x+a)

).

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Page 6: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Matching boundary conditions at x = 0, we have

A + B = Ae(−ik+K)a + Be(−ik−K)a

KA−KB =[KAe(−ik+K)a −KBe(−ik−K)a

]− 2mλ

h2 (A + B).

Recall, from single δ-function,

limε→0

[(dψ

dx

)

x=+ε

−(

dx

)

x=−ε

]= −2mλ

h2 ψ(0).

Rearranging the boundary conditions, we get

A(1− e−ikaeKa

)= B

(−1 + e−ikae−Ka

)

A

(1− e−ikaeKa +

2mλ

h2K

)= B

(1− e−ikae−Ka − 2mλ

h2K

),

which lead to

1 +2mλh2K

1− e−ikaeKa= −1 +

2mλh2K

1− e−ikae−Ka,

that is,

2 =2mλ

h2K

(1

1− e−ikae−Ka− 1

1− e−ikaeKa

)

=2mλ

h2K

e−ika(e−Ka − eKa)

(1− e−ikaeKa)(1− e−ikae−Ka)

=2mλ

h2K

e−ika(e−Ka − eKa)

(1 + e−2ika)− e−ika(e−Ka + eKa)

=2mλ

h2K

− sinh Ka

cos ka− cosh Ka.

Therefore, energy E = −h2K2/2m satisfies

cos ka = cosh Ka− mλ

h2Ksinh Ka.

Similarly, for E = h2q2/2m > 0, substituting K by iq, we get

cos ka = cos qa− mλ

h2qsin qa.

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Page 7: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Figure 2: Energy bands and gaps of the delta-function solid.

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(b) See Fig. 2. The key point is | cos ka| ≤ 1. Not all K’s or q’s cansatisfy the corresponding equation. Those that satisfy form bands, whilethose not form gaps.

(c) See Fig. 2. For larger energy, the particle is less affected by the δ-function potential. Therefore, the band diagram is expected to look morelike E = h2k2/2m folded into [−π/a, π/a] (first Brillouin zone). And the gapis expected to be narrower as energy goes up, while the band becomes wider.

4. (a) Given ml = 1.6me and mt = 0.08me, the isotropic mass in eachvalley is

m∗e = (mlm

2t )

1/3 = 0.217me.

Extending results from class notes to the model, we find, in the coductionband,

gc(E) = 4V m∗

e

π2h3

√2m∗

e(E − Ec),

where the factor of 4 comes from the four minima. In the valence band,similarly,

gv(E) = 2V m∗

h

π2h3

√2m∗

h(Ev − E),

where the factor of 2 comes from the double degeneracy. Therefore, followingthe class notes,

nc(T ) =4√2

(m∗

ekBT

πh2

)3/2

e−(Ec−µ)/kBT ,

nv(T ) =2√2

(m∗

hkBT

πh2

)3/2

e−(µ−Ev)/kBT ,

and

ni(T ) =√

nc(T ) · nv(T ) = 2(m∗em

∗h)

3/4

(kBT

πh2

)3/2

e− Eg

2kBT

Given at room temperature, Eg = 0.67eV , ni = 2.4 × 1013cm−3 = 2.4 ×1019m−3, we have

m∗h =

1

m∗e

ni(T )e

Eg2kBT

2(

kBTπh2

)3/2

4/3

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Page 9: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

=1

0.217× 9.1× 10−31

2.4× 1019exp(

0.67×1.6×10−19

2×1.38×10−23×300

)

2(

1.38×10−23×3003.14×(1.055×10−34)2

)3/2

4/3

= 3.11× 10−31Kg = 0.342me.

(b) Since the parabolic energy band for larger mass is more flat, we ex-pected that more holes come from the band with mass 0.28me. Therefore, m∗

h

defined in (a) would be in between 0.04me and 0.28me, and closer to 0.28me.

5. (a) Given ml = 0.9me and mt = 0.2me, the isotropic mass in eachenergy valley is

m∗e = (mlm

2t )

1/3 = 0.33me.

Assuming complete ionization,

nc(T ) =6√2

(m∗

ekBT

πh2

)3/2

e−(Ec−µ)/kBT = ND,

where the factor of 6 comes from six conduction band minima. Therefore,for ND = 2× 1016cm−3 = 2× 1022m−3

Ec− µ = kBT ln

6√

2ND

(m∗

ekBT

πh2

)3/2

= 1.38× 10−23 × 300×

ln

6√

2× 2× 1022

(0.33× 9.1× 10−31 × 1.38× 10−23 × 300

3.14× (1.055× 10−34)2

)3/2

= 3× 10−20J = 0.19eV.

The complete ionization is justified since ED − µ = 0.14eV À kBT =0.025eV .

(b) Similarly, for ND = 2× 1018cm−3,

Ec − µ = 0.07eV,

which implies the Fermi energy is very close to the donor level. Hence, we cannot use the complete ionization assumption. However, it is practical to use

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Page 10: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Boltzmann distribution in the conduction band, and neglect the contributionfrom valence band. The charge neutrality equation becomes

nc(T ) = Nce−(Ec−µ)/kBT = ND − ND

1 + 12e(ED−µ)/kBT

,

where the last term is the number of electrons remained on the donor leveland

Nc =6√2

(m∗

ekBT

πh2

)3/2

= 2.85× 1025m−3

is referred to as conduction band effective density of states. Introducing x =e−(Ec−µ)/kBT and t = e−(Ec−ED)/kBT = 0.145, the charge neutrality equationbecomes

x2 +t

2x− NDt

2Nc

= 0.

Therefore,

x = − t

4+

√t2/16 +

NDt

2Nc

= 0.044,

orEc − µ = 0.08eV.

Actually, about 60 percent of the donors are ionized.

6. (a) The Schrodinger equation for a 1-D infinite square well leads toeigenenergies

En =π2h2n2

2mL2, n = 1, 2, 3, ...,

where L is the length of the system. Therefore, number of states belowenergy E is

N (E) =L

πh(2mE)1/2,

hence, density of states

g(E)dE = dN (E) =L

2πh(2m)1/2E−1/2dE.

There will be an additional factor of 2 if spin is considered,

g(E)dE = dN (E) =L

πh(2m)1/2E−1/2dE.

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Page 11: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

(b) At T = 0K, electrons occupy the lowest N states.

N =∫ EF (0)

0g(E)dE = 2N (EF (0)),

or

EF (0) =N2π2h2

8mL2.

Figure 3: Occupied states at T=0.

(c) Average energy

E =1

N

∫ EF (0)

0Eg(E)dE

=1

N

L

πh(2m)1/2

∫ EF (0)

0E1/2dE

=1

N

L

πh(2m)1/2 2

3[EF (0)]3/2

=1

3EF (0).

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Page 12: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Note that from (b),

N =L

πh[8mEF (0)]1/2.

(d) For N/L = 0.5electrons/A,

EF (0) =(

N

L

)2 π2h2

8m= 3.8× 10−19J = 2.36eV.

Figure 4: Occupied states at finite temperature.

(e) First, Fermi function has the following property,

f(EF + ∆E) = 1− f(EF + ∆E),

where 1−f(E) can be interpret as probability of a state being NOT occupied.Second, density of states g(E) decreases as E increases. At finite (but low)temperature, if the Fermi energy had not move, the number of states occupiedabove EF ∫ ∞

EF

f(E)g(E)dE.

would not be enough to compensate the number of states unoccupied belowEF ∫ EF

0[1− f(E)]g(E)dE.

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Page 13: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Therefore, EF would move up to maintain the conservation of particle.7. (a) The solutions of Schrodinger equation for 2D particle-in-a-box

problem areψ(x, y) = B sin kxx sin kyy,

with

Ekx,ky =h2

2m(k2

x + k2y) =

h2k2

2m,

wherekx =

nxπ

x0

, ky =nyπ

y0

.

Therefore, each state (kx, ky) occupies area of

π

x0

π

y0

=π2

A

in k-space, where A is the area of the 2D square well. The total number ofstates below k =

√2mE/h is

N (k) =14πk2

π2

A

=Ak2

4π,

or

N (E) =mAE

2πh2 .

The 14

factor comes from the fact kx, ky > 0. Notice the spin degeneracy isneglected, since it is not specified. Therefore, the density of states,

g(E) =dN (E)

dE=

mA

2πh2 .

Note g(E) is a constant in 2-D.(b) At T = 0K,

N =∫ EF (0)

0g(E)dE = N (EF (0)),

or

EF (0) =2πh2

m

N

A.

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Page 14: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

(c) At T > 0,

N =∫ ∞

0f(E)g(E)dE

=∫ ∞

0

1

e(E−EF )/kBT + 1

mA

2πh2dE.

Making substitutionsx = e−(E−EF )/kBT + 1,

and

dx = −e−(E−EF )/kBT

kBTdE,

we get

N =mA

2πh2

∫ 1

eEF /kBT +1−kBT

dx

x

=mAkBT

2πh2 ln(eEF /kBT + 1).

Therefore, using from (b),

N =mA

2πh2EF (0),

we getEF (T ) = kBT ln(eEF (0)/kBT − 1).

(d) Using Maxwell-Boltzmann distribution

fMB(E) = e−(E−µ)/kBT ,

N =∫ ∞

0fMB(E)g(E)dE

=mA

2πh2 eµ/kBT∫ ∞

0e−E/kBT dE

=mAkBT

2πh2 eµ/kBT

Therefore

µ(T ) = kBT ln

(2πh2

mkBT

N

A

).

14

Page 15: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

8. (a) The group velocity

~v(~k) =1

h~∇kE(~k)

= 2a

2hE0

(sin

kxa

2cos

kxa

2x + sin

kya

2cos

kya

2y + sin

kza

2cos

kza

2z

)

=a

2hE0(sin kxax + sin kyay + sin kzaz)

Figure 5: Contour plot of the equal energy surface when kza = 0

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Page 16: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Figure 6: When Ef is very small (kfa << 1), the Fermi surface is almostsperical.

(b) When ~v(~k) is along the direction of ~k,

0 = ~v × ~k

=a

2hE0[(kz sin ky − ky sin kz)x

+(kx sin kz − kz sin kx)y + (ky sin kx − kx sin ky)z

It is easy to check kz sin ky = ky sin kz only if |ky| = |kz| or ky = 0 or kz = 0.

Therefore, ~v(~k) is along the direction of ~k, only if

• |kx| = |ky| = |kz|, or [111] direction

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Page 17: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

• |kx| = |ky| and kz = 0, or [110] direction

• |ky| = |kz| and kx = 0, or [011] direction

• |kz| = |kx| and ky = 0, or [101] direction

• kx = ky = 0, or [001] direction

• ky = kz = 0, or [010] direction

• kz = kx = 0, or [100] direction.

where [hkl] is a standard notation for the direction along

hx + ky + lz.

(c)1

m∗ij

=1

h2

∂2E

∂ki∂kj

.

Obviously,

i 6= j,1

m∗ij

= 0

i = j,1

m∗ii

=1

h

∂vi

∂ki

=a2E0

2h2 cos kia,

or

1

m∗ij

=

a2E0

2h2 cos kxa 0 0

0 a2E0

2h2 cos kya 0

0 0 a2E0

2h2 cos kza

.

Keep in mind that 1m∗

ijitself is a matrix, while m∗

ij is its matrix inverse.

(d) Fermi surface is not spherical, because E(~k) is not a constant over

the sphere |~k| = k0. Fig. 5 shows the contour plot of the equal energy surfacewhen kza = 0.

Figs. 6-9 show the evolution of the Fermi surface as we increase the elec-tron concentration, or equivalently, as we increase the Fermi energy. Only

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Page 18: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Figure 7: For larger Ef (kfa ∼ 1), the Fermi surface develops bulges alongthe ±x, ±y and ±z directions and troughs along x+ y + z and its equivalentdirections.

if Ef is very small, the Fermi surface would be sperical. (See part (e) fordetails.) For larger Ef , the Fermi surface develops bulges along the ±x, ±yand ±z directions and troughs along x + y + z and its equivalent directions.When Ef > 3E0/2, neighboring bulges connect into hollow necks. (NoteE(k) is actually periodic in k space, and we usually draw within only oneof the cells.) The neck structure is very similar to that occurs in the noblemetals, such as silver, copper and gold. See class notes for Fermi surface ofnoble metals, and be aware of the different locations of the necks.

To find out the Fermi energy for 0.5 electron per atom, we notice, first,

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Page 19: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Figure 8: Ef = E0, the Fermi surface looks like a cube.

the energy band is symmetric around E = 3E0/2. To see this, we can rewrite

E(k) = E0

[3

2− 1

2(cos kxa + cos kya + cos kza)

].

Second, a band would be fully filled if we had 2 electrons per atom (one spinup, another spin down). So, energy levels are filled up to the band centerEf = 3E0/2 for 1 electron per atom. Therefore, Fermi energy for 0.5 electronper atom is expected to be less than 3E0/2. It turns out, by numericalintegration, Ef ∼ E0. The Fermi surface looks like what we observe in Fig. 5but with small open tubes at the corners.

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Page 20: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

Figure 9: When Ef > E0, neighboring bulges connect into hollow necks.

(e) If Ef is small,

E(~k) ' E0

(kxa

2

)2

+

(kya

2

)2

+

(kza

2

)2

=E0a

2

4k2.

The volume of occupied states in k-space is

4

3k3

F ,

where kf = 2a

√Ef/E0 is the largest k of occupied states. Each state occupies

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Page 21: Principles of Semiconductor Physics Problem Set …zimp.zju.edu.cn/~xinwan/courses/semicond10/sol1.pdfPrinciples of Semiconductor Physics Problem Set #1, Solutions 1. (a) Integrate

volume of(2π)3

V,

where V = a3 is the volume of the system in real space. Therefore, totalnumber of states, counting spin, is

N = 243k3

F

(2π)3

V

=V

3π2k3

F .

Fermi energy

EF =E0a

2

4k2

F =E0a

2

4

[3π2N

V

]2/3

.

Take E0 = 2.0eV , a = 2.5A, and one electron per atom (therefore, N/V =1/a3),

EF =E0a

2

4

[3π2

a3

]2/3

=(3π2)2/3

4E0 = 4.79eV

The true EF is lower, since sin2 x < x2.

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