CSS Physics Paper-II 2023 Solved

Question 2: Gaussian Wave State and Uncertainty

Q2(a): Normalization Constant A

The given wave function is interpreted as:

ψ(x,t) = A exp[-a(mx²/ℏ + it)]

This can be written as:

ψ(x,t) = A exp[-(am/ℏ)x²] exp[-iat]

Let:

α = am/ℏ

Then:

ψ(x,t) = A e^{-αx²} e^{-iat}

The probability density is:

|ψ(x,t)|² = A² e^{-2αx²}

Normalization requires:

∫₋∞∞ |ψ(x,t)|² dx = 1

So:

A² ∫₋∞∞ e^{-2αx²} dx = 1

Using the Gaussian integral:

∫₋∞∞ e^{-βx²} dx = √(π/β)

Here β=2α, therefore:

A² √(π/(2α)) = 1

Thus:

A² = √(2α/π)

and:

A = (2α/π)¹/⁴

Substitute α=am/ℏ:

A = (2am/(πℏ))¹/⁴

Q2(b): Potential Energy Function V(x)

The time-dependent Schrödinger equation is:

iℏ ∂ψ/∂t = [-(ℏ²/2m) ∂²/∂x² + V(x)]ψ

For the time part:

ψ(x,t) = ψ(x) e^{-iat}

Therefore:

∂ψ/∂t = -ia ψ

Then:

iℏ ∂ψ/∂t = iℏ(-ia)ψ = ℏa ψ

So the energy eigenvalue is:

E = ℏa

The spatial wave function is:

ψ(x) = A e^{-αx²}

Differentiate:

dψ/dx = -2αx ψ

Differentiate again:

d²ψ/dx² = (-2α + 4α²x²)ψ

Put this in the time-independent Schrödinger equation:

[-(ℏ²/2m)d²/dx² + V(x)]ψ = Eψ

Substitute d²ψ/dx²:

-(ℏ²/2m)(-2α + 4α²x²)ψ + V(x)ψ = Eψ

So:

[(ℏ²α/m) - (2ℏ²α²/m)x² + V(x)]ψ = Eψ

Since α=am/ℏ:

ℏ²α/m = ℏa

and:

2ℏ²α²/m = 2ma²

Therefore:

[ℏa - 2ma²x² + V(x)]ψ = ℏa ψ

Hence:

V(x) = 2ma²x²

This is the potential of a harmonic oscillator because:

V(x) = (1/2)mω²x²

Comparing:

(1/2)mω² = 2ma²

So:

ω = 2a

Q2(c): Expectation Values of x, x², p and p²

The wave function is even in x. Therefore, by symmetry:

<x> = 0

For a normalized Gaussian ψ=Ae^{-αx²}:

<x²> = 1/(4α)

Since α=am/ℏ:

<x²> = ℏ/(4am)

The momentum operator is:

p = -iℏ d/dx

Again by symmetry, the average momentum is:

<p> = 0

The operator for momentum squared is:

p² = -ℏ² d²/dx²

Using:

d²ψ/dx² = (-2α + 4α²x²)ψ

we get:

<p²> = ℏ²α

Substitute α=am/ℏ:

<p²> = amℏ

Q2(d): σx, σp and Uncertainty Principle

Standard deviation is:

σx = √(<x²> - <x>²)

Since <x>=0:

σx = √(ℏ/(4am))

So:

σx = (1/2)√(ℏ/(am))

Similarly:

σp = √(<p²> - <p>²)

Since <p>=0:

σp = √(amℏ)

Now multiply:

σx σp = [(1/2)√(ℏ/(am))][√(amℏ)]

Therefore:

σx σp = ℏ/2

The uncertainty principle states:

σx σp ≥ ℏ/2

Question 3: Drift Current and Magnetic Force Between Wires

Q3(a): Current in Each Copper Wire

The drift-current formula is:

I = n e A vd

Given:

n = 8.45×10²⁸ m⁻³

e = 1.60×10⁻¹⁹ C

diameter = 1 mm = 1×10⁻³ m

radius = 0.5 mm = 0.5×10⁻³ m

vd = 0.3 cm/s = 0.003 m/s

Cross-sectional area is:

A = πr²

A = π(0.5×10⁻³)²

A = 7.85×10⁻⁷ m²

Now calculate current:

I = (8.45×10²⁸)(1.60×10⁻¹⁹)(7.85×10⁻⁷)(0.003)

I ≈ 31.9 A

Q3(b): Magnetic Force Between the Wires

The magnetic force per unit length between two long parallel wires is:

F/L = μ0 I1 I2/(2πd)

Since both wires carry the same current:

I1 = I2 = I = 31.9 A

Given separation:

d = 20 cm = 0.20 m

Also:

μ0 = 4π×10⁻⁷ N/A²

Substitute:

F/L = (4π×10⁻⁷)(31.9)²/(2π×0.20)

F/L ≈ 1.02×10⁻³ N/m

If the currents are in the same direction, the force is attractive. If the currents are in opposite directions, the force is repulsive.

Q3(c): Electric Current in Terms of Charges and Drift Velocity

Electric current is the rate of flow of charge through a cross-section of a conductor.

I = dQ/dt

For a wire with charge-carrier number density n, cross-sectional area A, charge of each carrier e, and drift velocity vd, the current is:

I = n e A vd

The current density is:

J = I/A = n e vd

Question 4: Fundamental Constants Expressions

Q4(a): Energy Needed to Ionize Hydrogen Atom

The ground-state energy of hydrogen-like atom is:

E1 = -2π²k²μe⁴/h²

Therefore, the ionization energy is the magnitude of the ground-state energy:

EI = 2π²k²μH e⁴/h²

For hydrogen:

μH = me mp/(me + mp)

Therefore:

EI(H) = [2π²k²e⁴/h²][me mp/(me + mp)]

Q4(b): Difference in Frequency of Lyman-Alpha Line in Hydrogen and Deuterium

The Lyman-alpha line corresponds to the transition:

n = 2 → n = 1

The frequency for a transition is:

ν = ΔE/h

For the Lyman-alpha transition:

ΔE = (3/4)(2π²k²μe⁴/h²)

Thus:

ν = (3/4)(2π²k²μe⁴/h³)

For hydrogen:

μH = me mp/(me + mp)

For deuterium, using deuteron mass approximately 2mp:

μD = me(2mp)/(me + 2mp)

Therefore, the difference in frequency is:

Δν = (3/4)(2π²k²e⁴/h³)(μD - μH)

Substitute reduced masses:

Δν = (3/4)(2π²k²e⁴/h³)[2me mp/(me+2mp) - me mp/(me+mp)]

Q4(c): Magnetic Moment of the Electron

The standard magnetic moment unit for the electron is the Bohr magneton:

μB = eℏ/(2me)

Since:

ℏ = h/(2π)

we get:

μB = eh/(4πme)

Q4(d): Spread in π0 Mass from Lifetime τ

The energy-time uncertainty relation gives natural energy width:

ΔE ≈ ℏ/τ

Since:

E = mc²

the corresponding mass spread is:

Δm ≈ ΔE/c²

Therefore:

Δm ≈ ℏ/(τc²)

Using ℏ=h/(2π):

Δm ≈ h/(2πτc²)

Question 5: Photon Absorption and Hydrogen Atom Transitions

Q5(a): Photon Frequency Required for Absorption

Initial atom is at rest with rest energy:

Ei = Mc² + hν

The photon momentum is:

pγ = hν/c

After absorption, the excited atom has rest mass M+Δ and recoils with momentum equal to photon momentum:

p = hν/c

The final relativistic energy is:

Ef = √[(M+Δ)²c⁴ + p²c²]

Substitute p=hν/c:

Ef = √[(M+Δ)²c⁴ + (hν)²]

Energy conservation gives:

Mc² + hν = √[(M+Δ)²c⁴ + (hν)²]

Let:

x = hν

Then:

Mc² + x = √[(M+Δ)²c⁴ + x²]

Square both sides:

M²c⁴ + 2Mc²x + x² = (M+Δ)²c⁴ + x²

Cancel x²:

2Mc²x = [(M+Δ)² - M²]c⁴

Simplify:

2Mc²x = (2MΔ + Δ²)c⁴

Therefore:

x = Δc²(1 + Δ/(2M))

Since x=hν:

ν = [Δc²/h][1 + Δ/(2M)]

Q5(b): Bohr Derivation of Hydrogen Energy Levels

For an electron in a circular orbit around the proton, Coulomb force provides centripetal force:

ke²/r² = mv²/r

Thus:

mv² = ke²/r

Bohr’s angular momentum quantization condition is:

mvr = nℏ = nh/(2π)

From this:

v = nh/(2πmr)

Substitute into mv² = ke²/r:

m[n²h²/(4π²m²r²)] = ke²/r

Simplify:

n²h²/(4π²mr²) = ke²/r

Therefore:

rn = n²h²/(4π²mke²)

Total energy is:

E = K + U

From the force equation:

K = (1/2)mv² = ke²/(2r)

Potential energy is:

U = -ke²/r

Therefore:

E = ke²/(2r) - ke²/r

E = -ke²/(2r)

Substitute rn:

En = -2π²mk²e⁴/(n²h²)

For hydrogen, this is commonly written as:

En = -13.6 eV/n²

Q5(c): Frequency and Wavelength for n=109 to n=108

For hydrogen spectral transitions:

ν = Rc(1/nf² - 1/ni²)

Here:

ni = 109

nf = 108

R = 1.097×10⁷ m⁻¹

c = 3.00×10⁸ m/s

Therefore:

ν = (1.097×10⁷)(3.00×10⁸)(1/108² - 1/109²)

Calculate:

ν ≈ 5.15×10⁹ Hz

Wavelength is:

λ = c/ν

λ = (3.00×10⁸)/(5.15×10⁹)

λ ≈ 5.82×10⁻² m

λ ≈ 5.82 cm

Question 6: Elastic Vibrations of Crystal

Q6(a): Frequency of Elastic Wave in a Monoatomic Crystal

A crystal with one atom in the primitive cell has three acoustic vibration branches: one longitudinal and two transverse branches. In the long-wavelength limit, lattice vibrations behave like elastic waves in a continuous medium.

The equation of motion for an elastic medium is:

ρ ∂²ui/∂t² = Cijkl ∂²uk/(∂xj∂xl)

Here:

• ρ is mass density.
• ui is displacement component.
• Cijkl are elastic constants.

Assume a plane-wave solution:

ui = ui0 exp[i(k·r - ωt)]

Then:

∂²ui/∂t² = -ω²ui

and:

∂²uk/(∂xj∂xl) = -kjkl uk

Substitute into the elastic equation:

ρ ω² ui = Cijkl kj kl uk

This is the Christoffel equation:

(Cijkl kj kl - ρ ω² δik)uk = 0

For non-zero displacement, determinant must vanish:

det|Cijkl kj kl - ρ ω²δik| = 0

This equation gives the allowed frequencies ω for a given wavevector k.

Isotropic Solid Result

For an isotropic medium, there are longitudinal and transverse sound waves.

Longitudinal wave:

ωL = vL k

vL = √[(λ + 2μ)/ρ]

Transverse wave:

ωT = vT k

vT = √(μ/ρ)

Here λ and μ are Lamé elastic constants. In a cubic crystal along a principal axis, the common simplified forms are:

ωL = k√(C11/ρ)

ωT = k√(C44/ρ)

Q6(b): Vibrations of Crystal

Crystal vibrations are collective oscillations of atoms about their equilibrium positions in the lattice. Atoms in a crystal are not fixed permanently; they vibrate due to thermal energy and interatomic forces.

Acoustic and Optical Modes

If a crystal has one atom per primitive cell, it has only acoustic branches. If it has more than one atom per primitive cell, it can also have optical branches.

Phonons

Quantized crystal vibrations are called phonons. A phonon is a quantum of lattice vibrational energy.

E = ℏω

Phonons are important because they explain:

1. Specific heat of solids.
2. Thermal conductivity in insulators.
3. Sound propagation in crystals.
4. Electron-phonon interaction.
5. Thermal expansion.

Question 7: Density of States and Debye Model

Q7(a): Density of States in Three Dimensions

Density of states means the number of allowed quantum states available in a given interval of wavevector, frequency or energy.

For a particle or wave in a cube of volume V=L³, allowed states in k-space are separated by:

2π/L

The volume occupied by one state in k-space is:

(2π/L)³ = (2π)³/V

The number of states in a spherical shell between k and k+dk is:

g(k)dk = [V/(2π)³] 4πk² dk

Therefore:

g(k)dk = V k² dk/(2π²)

This is the density of states in wavevector space per polarization or per independent branch.

For Free Particles in Energy Space

For free particles:

E = ℏ²k²/(2m)

This means:

k = √(2mE)/ℏ

Therefore, the three-dimensional density of states in energy is proportional to:

g(E) ∝ √E

Q7(b): Debye Model for Density of States

The Debye model describes lattice vibrations in a solid by treating the solid as a continuous elastic medium at long wavelengths. It replaces the real phonon spectrum with acoustic waves up to a maximum cutoff frequency called the Debye frequency.

For phonons with linear dispersion:

ω = vk

Using k=ω/v and dk=dω/v, the density of states becomes:

g(ω)dω = 3Vω² dω/(2π²v³)

The factor 3 accounts for three acoustic branches.

The total number of vibrational modes in a solid containing N atoms must be:

3N

Therefore, Debye chooses a cutoff frequency ωD such that:

∫₀ωD g(ω)dω = 3N

Using the Debye density of states:

∫₀ωD [3Vω²/(2π²v³)] dω = 3N

This gives:

ωD³ = 6π²Nv³/V

The Debye density of states can also be written as:

g(ω) = 9Nω²/ωD³

for:

0 ≤ ω ≤ ωD

Q7(c): Phonon Heat Capacity

Phonon heat capacity is the contribution of lattice vibrations to the heat capacity of a solid.

The phonon internal energy is:

U = ∫₀ωD [ℏω/(e^(ℏω/kBT)-1)] g(ω)dω

The phonon heat capacity at constant volume is:

CV = (∂U/∂T)V

Important Results

At high temperature, Debye model gives the Dulong-Petit law:

CV → 3NkB

At low temperature, it gives Debye’s T³ law:

CV = (12π⁴/5)NkB(T/θD)³

Question 8: Maxwell Equations, Magnetic Materials and Black-Body Radiation

Q8(a): Maxwell’s Equations

Maxwell’s equations are the four fundamental equations of classical electromagnetism. They combine electricity, magnetism and optics into one unified theory.

Prediction of Electromagnetic Waves

In free space, Maxwell’s equations lead to electromagnetic waves travelling with speed:

c = 1/√(μ0ε0)

This result showed that light is an electromagnetic wave.

Q8(b): Magnetic Materials — Diamagnetic, Paramagnetic and Ferromagnetic

Magnetic materials are classified according to their response to an external magnetic field.

Diamagnetic Materials

Diamagnetic materials develop weak magnetization opposite to the applied magnetic field. Their susceptibility is negative.

Paramagnetic Materials

Paramagnetic materials have atoms or ions with unpaired magnetic moments. These moments align weakly with an applied magnetic field.

Ferromagnetic Materials

Ferromagnetic materials contain magnetic domains. When domains align, strong magnetization appears. They show hysteresis and can form permanent magnets.

Q8(c): Black-Body Radiation

A black body is an ideal body that absorbs all radiation incident on it and emits radiation depending only on its temperature. The radiation emitted by a black body is called black-body radiation.

Experimental Features

1. The spectrum is continuous.
2. As temperature increases, total emitted energy increases.
3. The peak wavelength shifts toward shorter wavelengths at higher temperature.
4. The spectrum depends only on temperature, not on material.

Stefan-Boltzmann Law

Total energy radiated per unit area per unit time is:

E = σT⁴

Wien’s Displacement Law

The wavelength of maximum intensity is:

λmax T = b

Rayleigh-Jeans Law and Ultraviolet Catastrophe

Classical physics predicted that energy density should increase without limit at short wavelengths. This failure was called the ultraviolet catastrophe.

Planck’s Quantum Hypothesis

Planck solved the problem by assuming that radiation energy is emitted and absorbed in discrete packets:

E = nhf

where n=1,2,3,.... The energy of one quantum is:

E = hf

Planck Radiation Law

Planck’s formula correctly describes black-body radiation:

u(ν,T)dν = [8πhν³/c³][1/(e^(hν/kBT)-1)]dν