PHY250B

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Data Provided: A formula sheet and table of physical constants are attached to this paper. DEPARTMENT OF PHYSICS AND ASTRONOMY Autumn Semester (2016-2017) DEPARTMENT OF PHYSICS AND ASTRONOMY PHY250 - From electromagnetism to quantum mechanics. Paper B – Solids, electromagnetism and relativity. 3 HOURS

There are three sections to this paper.

Each section also has a different weighting.

Section 1 – Solids (weighting 34/100) Section 2 – Electromagnetism (weighting 50/100) Section 3 – Relativity (weighting 16/100)

Sections 1 and 2 have one compulsory and two optional questions. You

should answer the compulsory question and one optional question.

Sections 3 has one compulsory question, which you should answer.

Answer each section in a separate book.

Clearly indicate the question numbers on which you would like to be examined on the front cover of your answer books.

Cross through any work that you do not wish to be examined.

PHY250B

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SECTION 1 - Solids

COMPULSORY QUESTION 1. (a) Calculate the atomic packing fraction for a body-centred cubic lattice. [4]

(b) Draw a hexagonal close-packed (hcp) structure. Why can this structure not be described by a Bravais lattice? What are the primitive vectors for an hcp lattice? [4]

(c) The Young modulus of copper is 0.12 TPa. What is the extension of a

2 m long wire of diameter 4 mm, if a (stretching) force of 6 N is applied to it? [3]

(d) What is the utility of Miller indices in crystallography? Draw a cubic

unit cell and indicate the location of the (101) plane. [2]

(e) An X-ray diffraction experiment is used to determine the (simple cubic) crystal lattice of CsI (parameter a = 457 pm). The second and third scattering peaks are observed at angles θ = 10.8° and 13.3° respectively. At what angle is the first peak located? What is the wavelength of the X-rays? [4]

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OPTIONAL QUESTIONS – ANSWER EITHER QUESTION 2 OR QUESTION 3 2. (a) Give one advantage and one disadvantage of using (i) neutrons,

(ii) X-rays, and (iii) electrons to determine crystal structure. [6]

(b) X-rays of wavelength 121 pm have been used to determine the simple cubic caesium chloride lattice, and second order diffraction peaks were observed at 17.07° and 30.55°. What are the Miller indices of the corresponding planes? What is the lattice parameter of CsCl? [5]

(c) What is the de Broglie wavelength of electrons with kinetic energy of

300 keV? [4] (d) What is the scattering angle of these 300 keV electrons when they are

passed through the (100) plane of a crystal with lattice parameter 285.6 pm? Comment on the difference between your answer for this and the information given in part (b). [2]

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3. (a) Describe, with the aid of a sketch of interatomic energies, what we mean by the harmonic approximation in the context of crystal dynamics. [4]

(b) Why is an understanding of crystal dynamics important? [3]

(c) The dispersion relation in a monatomic lattice can be shown to be

€

ω2m = 4Ksin2 ka/2( ) , where ω and k are the angular frequency and wave number of the

travelling waves, m is the mass of the atoms in the lattice, a is the interatomic spacing, and K is the lattice “spring constant”. Find the speed of sound associated with this lattice, stating clearly your approximations. [4]

(d) What is the speed of sound in diamond if the maximum angular

frequency and wave number from the dispersion relation are 4.9 × 1013 rad/s and 4.1 × 109 m–1 respectively? Comment on your result. [3]

(e) Sketch the dispersion relations for a diatomic lattice. Why does the

diatomic nature of this lattice make a difference to the dispersion relation? [3]

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SECTION 2 - Electromagnetism

COMPULSORY QUESTION 4. (a) Two concentric spherical shells of radii a = 10.00 cm and b = 10.10 cm

and negligible thickness carry charges of +15 nC and −15 nC respectively. Calculate, showing all your working:

(i) The magnitude of the electric field at r = 5 cm, r =

12 cm, and r = 10.07 cm. [3] (ii) The magnitude of the potential difference between the

spheres. [2] (iii) The capacitance of the system. [1] (b) A layer of dielectric material having a relative permittivity of 12 and

thickness 0.5 mm is deposited on the surface of the inner sphere in part (a). This is followed by a further layer of identical thickness, but a higher relative permittivity of 20, so that the entire region between the spheres is filled with dielectric. Calculate, showing all your working:

(i) The values of both the electric field and the displacement

field at r = 10.02 cm and r = 10.07 cm. [4]

(ii) The new capacitance of the system. [3] (c) State Ampère’s circuital law for magnetism and use it to obtain an

expression for the magnetic field produced along the axis of a long solenoid with n turns per unit length, and carrying a current I. Fully explain your reasoning at every stage of the derivation. [4]

(d) An electron with velocity 2.5 ×105 i m s-1 enters a region of space where

there is a uniform magnetic field. If the instantaneous magnetic force which acts on the electron is 6.4 ×10-14 j N find the magnitude and direction of the magnetic field. [3]

(e) The speed of light in a non-magnetic material is measured to be

1.7 ×108 m s-1. What is the relative permittivity of the material? [2]

(f) A shiny coin is placed at a distance of 10 m from a 150 W light bulb. Obtain a quantitative estimate of the force on the coin due to the electromagnetic radiation emitted by the bulb. State any assumptions that you make. [3]

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OPTIONAL QUESTIONS – ANSWER EITHER QUESTION 5 OR QUESTION 6

5. (a) Point charges of +7 pC, −8 pC and +12 pC are placed at the Cartesian coordinates (2,0), (-1,0) and (0,1) respectively, where all distances

are measured in centimetres. Find the magnitude and direction of the electric field produced at the origin. [6]

(b) What electrostatic force will act on an electron placed at the origin in the

charge system in part (a)? [2]

(c) Calculate the energy required to move the electron in part (b) from the origin to the point (0,−2). [6]

(d) A charge system consists of a line charge of total charge +2Q formed into

a ring of radius a and a point charge –Q placed at the centre of the ring. Show that the electric field which acts along the axis of the ring a distance x from the centre of the ring is given by [7]

𝑄4𝜋𝜀!

2𝑥

𝑥! + 𝑎!!!−1𝑥! .

(e) A very long cylinder of length l and radius a (l >> a) carries a uniform charge density ρ. Show that the electric field at a distance r from the axis

of the cylinder for r < a is given by [4]

𝐸 =𝜌𝑟2𝜀!

.

.

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6 (a) Write down the four Maxwell equations that apply for E and B fields in a vacuum. [4]

(b) Show how these equations lead to the conclusion that the fields can

propagate as electromagnetic waves, obeying the equations

∇!𝐸 = 𝜇!𝜀!𝜕!𝐸𝜕𝑡!

∇!𝐵 = 𝜇!𝜀!𝜕!𝐵𝜕𝑡!

[NB A full derivation is required for only one of these equations.] [6] (c) If it is assumed that the electromagnetic waves in part (b) are unbounded plane waves, propagating in the x-direction and linearly polarized in the y-direction the following equations are obeyed:

𝜕!𝐸!𝜕𝑥! =

1𝑐!𝜕!𝐸!𝜕𝑡! (A)

𝜕𝐸!𝜕𝑥 = −

𝜕𝐵!𝜕𝑡 (B)

(i) Explain the meanings of the terms in underlined italics above. [3] (ii) Show that µ0ε0 has the same dimensions as c-2, where c is the

speed of light. [4] (iii) Write down the equivalent wave equation to equation (A) above,

but for the magnetic field. [1] (iv) Show that the waveforms for the electric and magnetic fields are in phase, and that their amplitudes are related by E0 = cB0. [4]

(d) A 10 W laser beam with a diameter of 3 mm propagates in a vacuum. Calculate the average amplitudes of the E and B fields associated with the laser light. [3]

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SECTION 2 – Relativity

COMPULSORY QUESTION

7. An electron and a positron can annihilate and produce a pair of photons.

Consider the case shown in the figure below, where an accelerated electron (e-) collides with a positron (e+) at rest in the laboratory and produces two photons (γ1 and γ2)

(a) Using four-vectors to represent the energy and momentum of e-, e+,

γ1 and γ2, write down a simple relation showing conservation of energy-momentum for the above scenario. [2]

(b) Write down the relativistic energy-momentum relation for photon γ2. How

does this relation compare to the square of the photon’s four-vector? Also consider what you know regarding the mass of the photon. [3]

(c) Since the positron e+ is at rest in the laboratory frame, what mathematical

statement can be made regarding its momentum? What statement can be made regarding its energy? [2]

(d) Derive a relation for the energy of photon γ1 as a function of its angle θ

with respect to the incoming electron e-. Express your answer in terms of me, Ee-, pe-, c and θ. You might find some of the relations and statements you derive in parts a) to c) above to be helpful. [9]

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