∑ ∑ SOLUTION to Exam Dec. 16, 2008

SOLUTION to Exam Dec. 16, 2008 Problem 1

a) The primitive vectors are: a=½a(-1,1,1), b=½a(1,-1,1), c=½a(1,1,-1) The angle between vectors: a· b= -½√3a½√3a cosθ = ¼a².

Therefore: cosθ =-1/3 og θ = 109.5°.

The volume of the primitive cell is a³/2 (2 atoms in the FCC cell).

The primitive reciprocal lattice vectors for the FCC lattice:

This shows that the reciprocal lattice of the BCC structure is the FCC lattice.

b) By studying the FCC structure we find: 4 3-fold and 3 4-fold rotation axes.

Also, 9 mirror planes may be identified.

c) The Laue condition for x-ray diffraction where K is the scattering vector and R_uvw is a real space lattice vector. This means that the scattering vector K must be a reciprocal lattice vector.

The extinction rules (norsk: utslokkingregler) for the BCC structure may be found by noting that this structure may be viewed as a simple cubic structure with a basis (0,0,0) and (½,½,½).

The structure factor (or scattering amplitude) is then:

which becomes zero for h+k+l = 2n+1 = an odd number.

d) = the Madelung constant,

β = constant describing the strength of the repulsive part of the potential, n = integer describing the range of the repulsive part of the potential

= is a sum over atoms (repulsive part) q = the charge on the ions (+ or -)

R = nearest neighbor separation

R₀ defines the equilibrium distance, and is given by:

The volume of the solid is given by: V = 2NR³ , where 2N = number of ions in the NaCl structure.

a∗ 2πb×c a⋅(b×c) ---

= 4π

a³

---a²(xˆ –yˆ +zˆ)×(xˆ+yˆ–zˆ) 4π

---a (zˆ+yˆ + +zˆ xˆ +yˆ–xˆ) 2π

--- 0 1 1a ( , , )

= = =

b∗ 2πc×a a⋅(b×c) ---

= 4π

a³

---a²(xˆ +yˆ –zˆ)×(–xˆ +yˆ +zˆ) 4π

---a (zˆ–yˆ+ +zˆ xˆ+yˆ +xˆ) 2π

--- 1 0 1a ( , , )

= = =

c∗ 2πa×b a⋅(b×c) ---

= 4π

a³ ---a² –ˆx

yˆ zˆ

+ +

( )×(xˆ –yˆ +zˆ) 4π

---a (zˆ+yˆ–zˆ+xˆ +yˆ +xˆ) 2π

--- 1 1 0a ( , , )

= = =

e^{iK R⋅ uvw} = 1

F_hkl f 1 e

2πi 1 2---h 1

2---k 1 2---l

+ +

⎝ ⎠

⎛ ⎞

+

⎝ ⎠

⎜ ⎟

⎜ ⎟

⎛ ⎞

f(1+e^{πi h( +k+l)})

= =

A ±1

a_ijⁿ ---

i

∑

=

b 1

a_ijⁿ ---

i

∑

=

∂R

∂U = 0⇒R₀^n–1 4πε₀nβb Aq² ---

=

To find the Bulk modulus we have to differentiate U with respect to the volume V twice:

The first term on the right hand side disappears at the equilibrium position R₀. Furthermore:

Where:

The Bulk modulus is then:

Contribution to the Madelung constant from one unit cell.

∂V

∂U

∂R

⎝∂U⎠

⎛ ⎞

∂V

⎝ ⎠∂R

= ⎛ ⎞

V²

2

∂

∂ U

⇒ ∂∂UR

⎝ ⎠

⎛ ⎞ V²

2

∂

∂ R

⎝ ⎠

⎜ ⎟

⎛ ⎞

R²

2

∂

∂ U

⎝ ⎠

⎜ ⎟

⎛ ⎞

∂V

⎝ ⎠∂R

⎛ ⎞² +

=

∂V

⎝ ⎠∂R

⎛ ⎞

0

2 1

36N²R₀⁴ ---

=

V²

2

∂

∂ U

⎝ ⎠

⎜ ⎟

⎛ ⎞

0

⇒ 1

36N²R₀⁴ ---

R²

2

∂

∂ U

⎝ ⎠

⎜ ⎟

⎛ ⎞

0

=

∂R

∂U N –βbn Rⁿ⁺¹

--- Aq² 4πε₀R² ---

⎝ + ⎠

⎜ ⎟

⎛ ⎞

=

R²

2

∂

∂ U

⇒ N βbn n( +1) Rⁿ⁺²

--- 2Aq² R³ ---

⎝ – ⎠

⎜ ⎟

⎛ ⎞

=

B V

V²

2

∂

∂ U

⎝ ⎠

⎜ ⎟

⎛ ⎞

0

1 18NR₀ ---

R²

2

∂

∂ U

⎝ ⎠

⎜ ⎟

⎛ ⎞

0

1 18R₀

--- βbn n( +1) R₀ⁿ⁺²

--- 2Aq² R₀³ ---

⎝ – ⎠

⎜ ⎟

⎛ ⎞

= = =

B 1

18R₀

--- Aq²(n+1) 4πε₀R₀³

--- 2Aq² R₀³ ---

⎝ – ⎠

⎜ ⎟

⎛ ⎞ Aq²

72πε₀R₀⁴

---(n+1–2)

= =

B Aq²(n–1) 72πε₀R₀⁴ ---

=

Consider the central atom (e.g. Na).

Nearest neighbors (Cl) on face centers (0.5 of each atom inside cell) : 6 at distance R Next nearest neighbors (Na) on edges (0.25 of each atom inside cell): 12 at distance Next nearest neighbors (Cl) on corners (0.125 of each atom inside cell): 8 at distance

Therefore A = = 1.46

(A = 1.75 if the sum is over all ions in the crystal)

2R 3R 1

2--- 6 1 1--- 1

4--- 12 1 2 --- 1

8--- 8 1 3 ---

⋅ ⋅

⋅ ⋅ +

⋅ ⋅ –

The Madelung constant

where + means ions of opposite polarity and - means ions of same polarity

a_ij is the distance between ions i and j in units of R (nearest neighbor distance)

A ±1

a_ij ---

i

∑

=

Problem 2 (a)

(i) 3D free electron density of states:

(ii) Fermi-energy:

(iii) Average electron energy at T = 0 K:

(b) Fermi wave-vector in 2D (N_c is the number of electrons in unit cell):

Fermi surface:

e^ikx = e^{ik x( +L)}⇒kL = n2π⇒k n2π ---L

= periodic boundary conditions

L 2π---

⎝ ⎠⎛ ⎞³ is the number of states per unit volume in k–space g k( )dk 2 L

2π---

⎝ ⎠⎛ ⎞³ 4πk² dk

⋅ ⋅ ⋅ g E( )dE

= =

E h²k² ---2m

= ⇒g E( ) g k( ) dE ---dk

--- mL³ h²π²

---k V 2π² --- 2m

h² ---

⎝ ⎠

⎛ ⎞

3 2---

E

1 2---

= = =

N g E( )dE

0 E_F

∫

⎝ ⎠

⎜ ⎟

⎛ ⎞

E_F

⇒ h²

2m--- 3π²N ---V

⎝ ⎠

⎛ ⎞

2 3---

= = =

〈 〉E Eg E( )dE

0 E_F

∫

⎝ ⎠

⎜ ⎟

⎛ ⎞

3 2---

3 2---

⋅ 2 5---E_F

⋅ 3

5---NE_F

= = =

g'( )E ma² h²π ---

= 2D d– ensity of states per unit cell( )

N_c g'( )E dE

0 E_F

∫

h²π ---E_F

= = ⇒k_F 2mE_F

---h π ---a 2N_c

---π π --- 1 13a⋅ ,

= = = (N_c = 2)

Fermi wave vector k_F 1st Brillouin zone 2nd brillouin zone

Fermi surface (hole-pockets) Fermi surface (electron-pockets) (Repeated zone scheme)

The Fermi surface does not extend into the 3rd Brillouin zone (2 electrons in unit cell).

The 3rd BZ is the hatched areas in the figure in problem 3.

k_F

(c) Empty-lattice approximation in 1D.

The “empty lattice” approximation describes “free” electrons that are confined to a periodic lattice.

The wavevector of the electron is determined moduli a reciprocal lattice vector.

The electronic band structure E(k) for a one-dimensional system of lattice spacing a is given by:

The lowest energybands:

(d) Phonon annihilation process (two TA phonons recombine to one LA phonon):

Both crystal momentum and energy conservation laws are satisfied in this process.

E k( ) h²

2m---(k+G)²

=

where G n2π ---a

±

=

NFE model (qualitative changes)

k₃ k₁+k₂+G 2π 3a--- 2π

3a--- G

+ + 4π

---3a 2π ---a

– 2π

3a--- –

= = = = inside 1st Brillouin zone

ω₃ ω₁+ω₂ ω₀ π 3---

sin ω₀ π

3--- sin

+ 2 3 2ω₀ π

3---

⎝ ⎠– sin⎛ ⎞

= = = =

Problem 3

a) Effective electron mass is inversely proportional to the curvature of the electron band.

b) The concentration n of electrons in the conduction band of an intrinsic semiconductor at T = 300 K, a value of the energy gap of 1.2 eV (i.e. E_c-μ=0.6 eV), and an effective electron mass of 50% of the mass of a free electron:

= 7.5·10¹⁴ m^-3 = p

c) The position of the donor level relative to the bottom of the conduction band for the semiconductor as shown in the figure, when the effective mass of the electron

m_e* = 0.1 m_e and the dielectric constant of the semiconductor ε = 10 ε₀ may be found by using the expression for the Rydberg constant:

d) The potential across the junction is caused by diffusion of electrons from the n-side to the p- side and diffusion of holes from the p-side to the n-side. Equilibrium is established between the recombination currents (electrons recombinates with holes) and generating currents (thermal excitation of electrons to the conduction band) to prevent build up of charges.

The electric field across the depletion layer removed electrons and holes, and therefore the number of charge carriers is low and the resistance is high.

π/a k -π/a

Ε(k)

π/a k -π/a

m(k)

n 2 2πm_e∗k_BT h² ---

⎝ ⎠

⎜ ⎟

⎛ ⎞

3 2---

e^{–(Ec–μ)⁄kBT}

=

E_n = –R₀⁄n² where R₀ e⁴m_e 32π²ε₀²h²

--- 13 6, eV

= =

E_d R₀ m_e∗ m_e --- ε₀

---ε

⎝ ⎠⎛ ⎞²

⋅ ⋅ R₀

1000--- 13 6meV,

= = =

p-type n-type E_F

E_F before contact

after contact

E_c

E_v

E_v E_c E_F

depletion region -

- - -

+ + + +