PHY380

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PHY380 TURN OVER

Data Provided: A formula sheet and table of physical constants is attached to this paper. DEPARTMENT OF PHYSICS AND ASTRONOMY Autumn (2016) SOLID STATE PHYSICS 2 HOURS Instructions: The paper is divided into 5 questions. Answer question ONE (Compulsory), and TWO other questions. The total number of marks available for the exam is 50. Question ONE is marked out of 20. Questions 2–5 are marked out of 15. The breakdown on the right-hand side of the paper is meant as a guide to the marks that can be obtained from each part. Please clearly indicate the question numbers on which you would like to be examined on the front cover of your answer book. Cross through any work that you do not wish to be examined.

PHY380

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PHY380 CONTINUED

Question 1 COMPULSORY

a) What are typical electron densities in metals, semiconductors and insulators? Explain

how electron densities may be controlled in one of the three cases.

[3]

b) The electron configuration in potassium is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1. Explain which

electrons contribute to conduction. What is the role of the electrons in the other shells.

[3]

c) A metal has face centred cubic structure with lattice constant equal to 3 x 10-10 m.

Each atom contributes one electron to conduction. Calculate the free electron density

in the metal.

[3]

d) Explain the concept of donors and acceptors in semiconductors. Why does doping

increase significantly the conductivity of semiconductors?

[2]

e) Sketch a graph showing the variation of the Fermi level as a function of temperature

in an n-type semiconductor doped with both donors and acceptors. Explain the

temperature behaviour of the Fermi level at low and high temperatures.

[3]

f) The magnetisation of a paramagnet as a function of temperature T is given by

M(T) = NgBJBJ(y), where g is the Landé factor, J is the quantum number for the total

angular momentum of the atoms, B is the Bohr magneton, N is the number of

magnetic atoms per unit volume and BJ(y) is the Brillouin function with

y = gBJB/(kBT), where B is the value of magnetic field. Explain why the

magnetization at temperature T can be related to the magnetisation M(0) at

temperature T = 0 by M(T)/M(0) = BJ(y).

[2]

g) Mn2+ ions have electronic configuration 3d5. Use Hund’s rule to calculate the

quantum numbers J, L and S. Determine the magnitude of the magnetic moment of

the ion in units of the Bohr magneton. In your answer you may use the expression

for the Landé g-factor

𝑔 = (3

2+

𝑆(𝑆+1)−𝐿(𝐿+1)

2𝐽(𝐽+1)).

[2]

h) Explain why most metals have silvery (metallic) colour and some metals, e.g. gold or

copper have yellowish colour.

[2]

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PHY380 TURN OVER

Question 2 OPTIONAL

a) Explain the meaning of the terms:

i) Fermi surface

ii) Bands and band gaps

iii) Brillouin zone

[1]

[1]

[1]

b) Explain why different bands in solids have different widths in energy. How do you expect

the widths of the bands to vary as a function of energy, from low energy bands to high

energy bands?

[2]

c) Using the expression 22 Gk.G where k is a wavevector and G a reciprocal lattice

vector with magnitude G, using geometrical constructions, derive the shapes of the first

and second Brillouin zones for a simple cubic lattice.

[3]

d)

A simple metal has Fermi wavevector of magnitude kF in the range 𝜋/𝑎 < 𝑘𝐹 < √2𝜋/𝑎,

where a is the crystal lattice constant. Deduce and sketch which states are occupied in the

first and second Brillouin zones. You should use the reduced zone scheme in your

answer.

[3]

e) An electric field of magnitude 10 V/m is applied to a metal with electron mobility

10-3 m2/Vs. Calculate the resulting displacement of the Fermi surface.

[2]

f) Describe a measurement technique which may be used to deduce the value of the Fermi

energy in a metal.

[2]

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PHY380 CONTINUED

Question 3 OPTIONAL

a) Indium antimonide has a dielectric constant = 17. The donor ionisation energy in this

material is ED = 6.6×10-4 eV. Calculate the electron effective mass and the radius of the

ground state orbit.

[3]

b) In a heavily doped sample of GaAs the onset of optical absorption occurs at a wavelength

of 700 nm at 300 K. Calculate the electron Fermi k-vector. The electron (me) and hole

(mh) effective masses in GaAs are 0.067 m0 and 0.45 m0, respectively, where m0 is the free

electron mass. The band gap of GaAs is 1.43 eV at 300 K.

[4]

c) Provide an explanation of how the sign and the effective mass of charge carriers in a

semiconductor can be obtained from cyclotron resonance measurements.

[3]

d) A piece of copper is placed in an external magnetic field H, whose magnitude inside the

copper is 104 A/m. The magnetic moment per unit volume in copper is 0.05 A/m.

Calculate the magnetic susceptibility of copper.

[1]

e) Provide a qualitative explanation of the mechanism responsible for spontaneous

alignment of magnetic dipole moments in a ferromagnetic material. The Curie

temperature of pure iron is Tc = 1043 K. Assuming that the magnetic dipole moment per

atom in iron is 2.2 Bohr magnetons estimate the value of the effective “internal” magnetic

field in iron.

[4]

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PHY380 TURN OVER

Question 4 OPTIONAL

a) Electron (ne) and hole (np) concentrations in intrinsic semiconductors are given by the

following expressions

𝑛𝑒 = 2 (2𝜋𝑚𝑒𝑘𝐵𝑇

ℎ2)

3/2

exp (𝐸𝐹 − 𝐸𝐺

𝑘𝐵𝑇),

𝑛𝑝 = 2 (2𝜋𝑚𝑝𝑘𝐵𝑇

ℎ2)

3/2

exp (−𝐸𝐹

𝑘𝐵𝑇).

where me (mp) is the electron (hole) effective mass, EG the band gap of the material, EF

the Fermi level and T the temperature. Derive an expression for EF as a function of EG,

me, mp, and T only. How does the Fermi level change with T in the cases of me < mp and me > mp?

[4]

b) A sample of silicon contains 1018 m-3 donors. Estimate the temperature above which it

begins to show intrinsic behaviour. The intrinsic carrier concentration in silicon at 300 K

is 2×1016 m-3. The value of the band gap is 1.1 eV.

[3]

c) A live frog is seen to levitate above a powerful permanent magnet. Explain the origin of

this phenomenon.

[2]

d) The most important contribution to the paramagnetism of CuSO4 comes from Cu2+ ions

for which the magnetic moment is due to single unpaired electrons with zero orbital

angular momentum. Write down the probabilities at temperature T that the moment lies

parallel and antiparallel to the field. Hence show that the magnetization for N ions per

unit volume in field B is M = NBtanh(BB/(kBT)), where B is the Bohr magneton.

[3]

e) Where is the Fermi level located for the case of a semiconductor with energy gap EG doped

with

i) donors

ii) acceptors

iii) both donors and acceptors, with the concentration of donors being larger

than that of acceptors?

You may assume that the temperature is very low such that kBT << EA, ED, EG, where EA

and ED are binding energies of the acceptors and donors, respectively.

[1]

[1]

[1]

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PHY380 CONTINUED

Question 5 OPTIONAL a) A sample of sodium with a width of w = 3 mm and thickness of t = 0.5 mm is placed in a

perpendicular magnetic field of B = 0.1 T. The Hall voltage generated across the width of

the sample is 4.9 nV when a current of I = 100 mA is passed along it. Estimate the

electron concentration in sodium.

[4]

b) A sample of germanium is doped with a single type of donor, the concentration of which

is 1020 m-3. Estimate the minimum value of the magnetic field required to observe cyclotron

resonance in this sample at 4 K. In your answer you should take into account that the

collision diameter of a donor, which determines the donor scattering cross-section, is equal

to 30 nm and the effective mass of electrons is me = 10-31 kg.

[5]

c) Describe the domain theory of ferromagnetism using references to suitable diagrams.

Include in your discussion:

(i) the reversible and irreversible motion of magnetic domains in an applied

magnetic field;

(ii) the mechanisms responsible for the establishment of a stable domain structure.

[4]

d) Estimate the electron concentration in a doped semiconductor with background dielectric

constant ∞ and effective electron mass 0.02m0 (m0 is the free electron mass), where

the plasma angular frequency is 2.54 × 1013 rad/s.

[2]

END OF EXAMINATION PAPER

PHYSICAL CONSTANTS &MATHEMATICAL FORMULAE

Physical Constants

electron charge e = 1.60×10−19 Celectron mass me = 9.11×10−31 kg = 0.511MeV c−2proton mass mp = 1.673×10−27 kg = 938.3MeV c−2neutron mass mn = 1.675×10−27 kg = 939.6MeV c−2Planck’s constant h = 6.63×10−34 J sDirac’s constant (~ = h/2π) ~ = 1.05×10−34 J sBoltzmann’s constant kB = 1.38×10−23 J K−1 = 8.62×10−5 eVK−1speed of light in free space c = 299 792 458 ms−1 ≈ 3.00×108 ms−1permittivity of free space ε0 = 8.85×10−12 Fm−1permeability of free space µ0 = 4π×10−7 Hm−1Avogadro’s constant NA = 6.02×1023 mol−1gas constant R = 8.314 Jmol−1K−1ideal gas volume (STP) V0 = 22.4 l mol−1gravitational constant G = 6.67×10−11 Nm2 kg−2Rydberg constant R∞ = 1.10×107 m−1Rydberg energy of hydrogen RH = 13.6 eVBohr radius a0 = 0.529×10−10 mBohr magneton µB = 9.27×10−24 J T−1fine structure constant α ≈ 1/137Wien displacement law constant b = 2.898×10−3 mKStefan’s constant σ = 5.67×10−8 Wm−2K−4radiation density constant a = 7.55×10−16 Jm−3 K−4mass of the Sun M� = 1.99×1030 kgradius of the Sun R� = 6.96×108 mluminosity of the Sun L� = 3.85×1026 Wmass of the Earth M⊕ = 6.0×1024 kgradius of the Earth R⊕ = 6.4×106 m

Conversion Factors1 u (atomic mass unit) = 1.66×10−27 kg = 931.5MeV c−2 1 Å (angstrom) = 10−10 m1 astronomical unit = 1.50×1011 m 1 g (gravity) = 9.81 ms−21 eV = 1.60×10−19 J 1 parsec = 3.08×1016 m1 atmosphere = 1.01×105 Pa 1 year = 3.16×107 s

Polar Coordinates

x = r cos θ y = r sin θ dA = r dr dθ

∇2 =1

r

∂

∂r

(r∂

∂r

)+

1

r2∂2

∂θ2

Spherical Coordinates

x = r sin θ cosφ y = r sin θ sinφ z = r cos θ dV = r2 sin θ dr dθ dφ

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

Calculusf(x) f ′(x) f(x) f ′(x)

xn nxn−1 tanx sec2 x

ex ex sin−1(xa

)1√

a2−x2

lnx = loge x1x

cos−1(xa

)− 1√

a2−x2

sinx cosx tan−1(xa

)a

a2+x2

cosx − sinx sinh−1(xa

)1√

x2+a2

coshx sinhx cosh−1(xa

)1√

x2−a2

sinhx coshx tanh−1(xa

)a

a2−x2

cosecx −cosecx cotx uv u′v + uv′

secx secx tanx u/v u′v−uv′v2

Definite Integrals∫ ∞0

xne−ax dx =n!

an+1(n ≥ 0 and a > 0)

∫ +∞

−∞e−ax

2 dx =

√π

a∫ +∞

−∞x2e−ax

2 dx =1

2

√π

a3

Integration by Parts:∫ b

a

u(x)dv(x)dx

dx = u(x)v(x)∣∣∣ba−∫ b

a

du(x)dx

v(x) dx

Series Expansions

Taylor series: f(x) = f(a) +(x− a)

1!f ′(a) +

(x− a)2

2!f ′′(a) +

(x− a)3

3!f ′′′(a) + · · ·

Binomial expansion: (x+ y)n =n∑k=0

(n

k

)xn−kyk and

(n

k

)=

n!

(n− k)!k!

(1 + x)n = 1 + nx+n(n− 1)

2!x2 + · · · (|x| < 1)

ex = 1+x+x2

2!+x3

3!+ · · · , sinx = x− x

3

3!+x5

5!−· · · and cosx = 1− x

2

2!+x4

4!−· · ·

ln(1 + x) = loge(1 + x) = x− x2

2+x3

3− · · · (|x| < 1)

Geometric series:n∑k=0

rk =1− rn+1

1− r

Stirling’s formula: logeN ! = N logeN −N or lnN ! = N lnN −N

Trigonometry

sin(a± b) = sin a cos b± cos a sin b

cos(a± b) = cos a cos b∓ sin a sin b

tan(a± b) = tan a± tan b

1∓ tan a tan b

sin 2a = 2 sin a cos a

cos 2a = cos2 a− sin2 a = 2 cos2 a− 1 = 1− 2 sin2 a

sin a+ sin b = 2 sin 12(a+ b) cos 1

2(a− b)

sin a− sin b = 2 cos 12(a+ b) sin 1

2(a− b)

cos a+ cos b = 2 cos 12(a+ b) cos 1

2(a− b)

cos a− cos b = −2 sin 12(a+ b) sin 1

2(a− b)

eiθ = cos θ + i sin θ

cos θ =1

2

(eiθ + e−iθ

)and sin θ =

1

2i(eiθ − e−iθ

)cosh θ =

1

2

(eθ + e−θ

)and sinh θ =

1

2

(eθ − e−θ

)Spherical geometry:

sin a

sinA=

sin b

sinB=

sin c

sinCand cos a = cos b cos c+sin b sin c cosA

Vector Calculus

A ·B = AxBx + AyBy + AzBz = AjBj

A×B = (AyBz − AzBy) i+ (AzBx − AxBz) j+ (AxBy − AyBx) k = εijkAjBk

A×(B×C) = (A ·C)B− (A ·B)C

A · (B×C) = B · (C×A) = C · (A×B)

gradφ = ∇φ = ∂jφ =∂φ

∂xi+

∂φ

∂yj+

∂φ

∂zk

divA = ∇ ·A = ∂jAj =∂Ax∂x

+∂Ay∂y

+∂Az∂z

curlA = ∇×A = εijk∂jAk =

(∂Az∂y− ∂Ay

∂z

)i+

(∂Ax∂z− ∂Az

∂x

)j+

(∂Ay∂x− ∂Ax

∂y

)k

∇ · ∇φ = ∇2φ =∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2

∇×(∇φ) = 0 and ∇ · (∇×A) = 0

∇×(∇×A) = ∇(∇ ·A)−∇2A