CSS Physics Paper-II 2021 Solved

Question 2: Cylindrical Charged Shell, Capacitor Energy Density and Lorentz Force

Q2(a): Electric Field of Cylindrical Shell with Axial Line Charge

The charge distribution is cylindrically symmetric. Therefore, the electric field is radial and depends only on the distance r from the common axis. We use Gauss law with a coaxial cylindrical Gaussian surface of radius r and length L.

∮E·dA = Qenc/ε0

For a cylindrical Gaussian surface, the curved area is 2πrL, and the electric field is constant on that surface.

∮E·dA = E(2πrL)

Case 1: a < r < b

For a point inside the material of the cylindrical shell, the Gaussian surface encloses the axial line charge and the volume charge from radius a to radius r.

Charge due to line charge:

Qline = λL

Charge due to cylindrical volume from a to r:

Qvolume = ρπ(r²-a²)L

Total enclosed charge:

Qenc = λL + ρπ(r²-a²)L

Apply Gauss law:

E(2πrL) = [λL + ρπ(r²-a²)L]/ε0

Cancel L:

E(2πr) = [λ + ρπ(r²-a²)]/ε0

Therefore:

E = [λ + ρπ(r²-a²)]/(2πε0r)

Case 2: r > b

For a point outside the shell, the Gaussian surface encloses the entire shell and the axial line charge.

Volume charge of the full shell per length L is:

Qvolume = ρπ(b²-a²)L

Total enclosed charge is:

Qenc = [λ + ρπ(b²-a²)]L

Apply Gauss law:

E(2πrL) = [λ + ρπ(b²-a²)]L/ε0

Cancel L:

E = [λ + ρπ(b²-a²)]/(2πε0r)

Q2(b): Energy Density for a Capacitor

Energy stored in a capacitor is:

U = (1/2)CV²

For a parallel plate capacitor:

C = εA/d

The electric field between the plates is:

E = V/d

Therefore:

V = Ed

Substitute C=εA/d and V=Ed in the energy expression:

U = (1/2)(εA/d)(Ed)²

U = (1/2)(εA/d)(E²d²)

U = (1/2)εE²Ad

The volume between the plates is:

Volume = Ad

Energy density is energy per unit volume:

u = U/(Ad)

Therefore:

u = (1/2)εE²

Q2(c): Lorentz Force

The Lorentz force is the total force acting on a charged particle moving in electric and magnetic fields.

F = q(E + v×B)

Here q is charge, E is electric field, v is velocity and B is magnetic field.

Electric Part

The electric force is:

FE = qE

It acts along the electric field for a positive charge and opposite to the field for a negative charge. It can change both the speed and direction of a charged particle.

Magnetic Part

The magnetic force is:

FB = q(v×B)

Its magnitude is:

FB = qvB sinθ

The magnetic force is perpendicular to both v and B. Therefore, it changes the direction of velocity but not the kinetic energy if only the magnetic field acts.

Question 3: Magnetic Energy Density, Lenz Law and Magnetic Work

Q3(a): Magnetic Energy Density of an Inductor

The energy stored in an inductor carrying current I is:

U = (1/2)LI²

To find the magnetic energy density, consider a long solenoid. Its inductance is:

L = μ0n²Al

where n is number of turns per unit length, A is cross-sectional area and l is length.

The magnetic field inside a long solenoid is:

B = μ0nI

So:

I = B/(μ0n)

Substitute L and I into U=(1/2)LI²:

U = (1/2)(μ0n²Al)(B²/μ0²n²)

Simplify:

U = B²Al/(2μ0)

The volume of the solenoid is:

Volume = Al

Therefore, magnetic energy density is:

uB = U/(Al)

uB = B²/(2μ0)

In a magnetic medium of permeability μ:

uB = B²/(2μ)

Q3(b): Lenz’s Law

Lenz’s law states that the direction of induced current is always such that it opposes the change in magnetic flux that produces it.

Faraday’s law is:

ε = -N dΦB/dt

The negative sign represents Lenz’s law.

Explanation

If magnetic flux through a coil increases, the induced current flows in such a direction that its own magnetic field opposes the increase. If magnetic flux decreases, the induced current flows in such a direction that its own magnetic field tries to maintain the original flux.

Physical Basis

Lenz’s law is a consequence of conservation of energy. If induced current helped the change that produced it, energy would be created from nothing. Since this is impossible, the induced effect must oppose the cause.

Example

When the north pole of a magnet moves toward a coil, the coil produces a north pole facing the magnet to oppose the approach. When the magnet moves away, the coil produces a south pole to oppose the separation.

Q3(c): Why Magnetic Field Does No Work

The magnetic force on a moving charge is:

FB = q(v×B)

This force is always perpendicular to velocity v. Work done is:

dW = F·dr

Since displacement dr is along velocity and magnetic force is perpendicular to velocity:

FB·dr = 0

or:

W = Fs cos90° = 0

Therefore, a magnetic field cannot change the kinetic energy of a charged particle. It can only change the direction of motion.

Question 4: Dual Nature, de Broglie Hypothesis and Photoelectric Effect

Q4(a): Dual Nature of Electron and Light

Dual nature means that microscopic entities can show both particle-like and wave-like behaviour. Electrons and light both show this dual character.

Dual Nature of Electron

Dual Nature of Light

Q4(b): de Broglie Hypothesis

De Broglie proposed that every moving particle has an associated matter wave. The wavelength associated with a particle is called de Broglie wavelength.

λ = h/p

For a non-relativistic particle:

p = mv

Therefore:

λ = h/(mv)

Here, h is Planck’s constant and p is momentum.

Significance

1. It connected matter and waves.
2. It explained why electrons produce diffraction patterns.
3. It became a foundation of wave mechanics.
4. It was experimentally confirmed by electron diffraction experiments.

Q4(c): Photoelectric Effect Numerical

Given:

φA = 2.2 eV

φB = 3.6 eV

φC = 4.8 eV

λ = 320 nm

Photon energy in electron-volts is:

E = hc/λ

Using:

hc = 1240 eV nm

So:

E = 1240/320

E = 3.875 eV

Photoelectric emission occurs if:

E ≥ φ

Metal A

Kmax = E - φA

Kmax = 3.875 - 2.2

Kmax = 1.675 eV

Metal B

Kmax = E - φB

Kmax = 3.875 - 3.6

Kmax = 0.275 eV

Metal C

E = 3.875 eV < 4.8 eV

Therefore, metal C does not emit photoelectrons.

Question 5: Rectangular Barrier and Commuting Operators

Q5(a): Transmission Coefficient for Rectangular Barrier

The barrier has height V0 and extends from x=-a to x=+a. Therefore, its width is:

L = 2a

The particle energy is less than the barrier height:

E < V0

Classically, the particle cannot cross the barrier. Quantum mechanically, the wave function penetrates the barrier and there is a finite probability of transmission.

Define:

k = √(2mE)/ℏ

Inside the barrier:

κ = √[2m(V0-E)]/ℏ

The exact transmission coefficient for a rectangular barrier of width L is:

T = [1 + {V0²sinh²(κL)}/{4E(V0-E)}]⁻¹

Since L=2a:

T = [1 + {V0²sinh²(2κa)}/{4E(V0-E)}]⁻¹

Thick Barrier Approximation

For a thick barrier, the approximate expression is:

T ≈ [16E(V0-E)/V0²] e^(-2κL)

Since L=2a:

T ≈ [16E(V0-E)/V0²] e^(-4κa)

Q5(b): Meaning of [Ĥ,Â]=0

The commutator of two operators is:

[Ĥ,Â] = ĤÂ - ÂĤ

If:

[Ĥ,Â] = 0

then the Hamiltonian and operator Â commute.

Main Conclusions

1. Simultaneous eigenstates: The Hamiltonian Ĥ and operator Â can have common eigenstates.
2. Conserved quantity: If Â has no explicit time dependence, then the expectation value <Â> is constant in time.
3. Compatible observables: The physical quantities represented by Ĥ and Â can be measured simultaneously with definite values in common eigenstates.

The time variation of expectation value is:

d<A>/dt = (i/ℏ)<[H,A]> + <∂A/∂t>

If:

[H,A] = 0

and:

∂A/∂t = 0

then:

d<A>/dt = 0

Q5(c): Examples of Operators That Commute with Hamiltonian

Question 6: Band Theory, Diamond Structure and Silicon Carrier Concentration

Q6(a): Electrical Conduction in Solids by Band Theory

Band theory explains electrical conduction in solids using allowed and forbidden energy bands. When atoms combine to form a crystal, individual atomic energy levels split into closely spaced bands.

Important Bands

• Valence band: Highest energy band normally filled with electrons.
• Conduction band: Higher band where electrons are free to move and conduct current.
• Forbidden energy gap: Energy gap between valence and conduction bands.

Conductors

In conductors, the valence band and conduction band overlap, or the conduction band is partially filled. Therefore, many electrons are available for conduction even at ordinary temperatures.

Semiconductors

In semiconductors, the forbidden gap is small. At room temperature, some electrons can jump from the valence band to the conduction band, leaving holes behind. Both electrons and holes contribute to conduction.

Insulators

In insulators, the forbidden gap is large. Electrons cannot easily move from the valence band to the conduction band at ordinary temperatures, so conductivity is very low.

Q6(b): Crystal Structure of Diamond

Diamond has a diamond cubic crystal structure. It can be described as a face-centered cubic lattice with a two-atom basis.

Important Features

1. Each carbon atom is covalently bonded to four nearest carbon atoms.
2. The bonding is tetrahedral.
3. Carbon atoms are sp³ hybridized.
4. The coordination number is 4.
5. The structure is very rigid, making diamond extremely hard.
6. Diamond has a large band gap, so it behaves as an insulator at ordinary conditions.

Diamond Cubic Description

The diamond structure can be represented as an FCC lattice with two atoms in the basis:

(0,0,0) and (1/4,1/4,1/4)

This means the second atom is displaced by one quarter of the body diagonal from the first.

Q6(c): Carrier Concentration of Electrons in Silicon at 25°C

For intrinsic silicon at room temperature, electron concentration equals hole concentration:

n = p = ni

At about 25°C or 300 K, the commonly accepted intrinsic carrier concentration of silicon is approximately:

ni ≈ 1.5×10¹⁰ cm⁻³

Converting into m⁻³:

1 cm⁻³ = 10⁶ m⁻³

ni ≈ 1.5×10¹⁶ m⁻³

Using the Formula

The intrinsic carrier concentration is:

ni = √(NcNv)e^[-Eg/(2kT)]

For silicon near room temperature:

Eg ≈ 1.12 eV

Using standard values gives a result of the order:

ni ≈ 10¹⁰ cm⁻³

Question 7: Nuclear Stability, Chain Reaction and Potassium Radioactivity

Q7(a): Factors Contributing to Nuclear Stability

Nuclear stability depends on the balance between attractive nuclear force and repulsive electrostatic force between protons. The following factors are important.

1. Neutron-Proton Ratio

For light nuclei, stability usually occurs when:

N ≈ Z

For heavy nuclei, more neutrons are needed to reduce proton-proton repulsion, so stable nuclei have:

N > Z

2. Binding Energy per Nucleon

A nucleus is more stable if its binding energy per nucleon is high. Iron and nickel region nuclei have very high binding energy per nucleon.

3. Pairing Effect

Nuclei with even numbers of protons and even numbers of neutrons are generally more stable than odd-odd nuclei.

4. Magic Numbers

Nuclei with magic numbers of protons or neutrons show extra stability. Common magic numbers are:

2, 8, 20, 28, 50, 82, 126

5. Coulomb Repulsion

In heavy nuclei, repulsion between many protons reduces stability. Extra neutrons help provide nuclear attraction without adding electric repulsion.

N-Z Stability Curve

Nuclei above the stability band are neutron-rich and tend to undergo beta-minus decay. Nuclei below the stability band are proton-rich and tend to undergo beta-plus decay or electron capture.

Q7(b): Chain Reaction in a Nuclear Reactor

A nuclear chain reaction occurs when neutrons produced by one fission event cause further fission events. In a reactor, this chain reaction is controlled so that energy is released steadily.

Fission Chain Reaction

For uranium-235:

²³⁵U + ¹n → fission fragments + 2 or 3 neutrons + energy

The released neutrons can strike other uranium nuclei and continue the reaction.

Multiplication Factor

The multiplication factor k describes the chain reaction:

k = neutrons in one generation / neutrons in previous generation

Role of Reactor Parts

Q7(c): Expected Radioactivity from Potassium Isotope

Stable potassium is mainly ³⁹K. If the intended comparison is between stable potassium and neutron-deficient potassium such as ³⁸K, then:

³⁹K: Z=19, N=20

³⁸K: Z=19, N=19

Compared with stable potassium, ³⁸K is neutron-deficient or proton-rich. Proton-rich nuclei move toward stability by converting a proton into a neutron.

This can occur by positron emission:

p → n + e⁺ + ν

or by electron capture:

p + e⁻ → n + ν

Therefore, the expected radioactivity is:

β⁺ decay or electron capture

Question 8: Poynting Vector, Fusion in Stars and MOSFET

Q8(a): Poynting Vector

The Poynting vector represents the flow of electromagnetic energy per unit area per unit time. It gives both magnitude and direction of energy propagation in an electromagnetic field.

S = E×H

In vacuum:

S = (1/μ0)E×B

The direction of S is perpendicular to both electric field and magnetic field. In an electromagnetic wave, E, B and the direction of propagation are mutually perpendicular.

Unit

Unit of S = W/m²

Energy Density

The electromagnetic energy density is:

u = (1/2)ε0E² + B²/(2μ0)

Poynting Theorem

Poynting theorem is:

∂u/∂t + ∇·S + J·E = 0

It expresses conservation of electromagnetic energy.

Q8(b): Fusion in Stars

Fusion in stars is the process by which light nuclei combine to form heavier nuclei, releasing enormous energy. In main-sequence stars like the Sun, hydrogen nuclei fuse to form helium.

Main Fusion Reaction

The overall proton-proton chain can be summarized as:

4¹H → ⁴He + 2e⁺ + 2ν + energy

The energy released is about:

26.7 MeV per helium nucleus formed

Why Fusion Requires High Temperature

Protons repel each other due to Coulomb repulsion. Very high temperature and pressure in stellar cores allow protons to come close enough for the nuclear force to act. Quantum tunneling also helps fusion occur at stellar temperatures.

Energy Source of Stars

The mass of the helium nucleus is slightly less than the total mass of four hydrogen nuclei. This mass defect is converted into energy according to Einstein’s relation:

E = Δmc²

Importance of Fusion

1. It powers the Sun and stars.
2. It produces heavier elements in the universe.
3. It maintains hydrostatic equilibrium in stars.
4. It explains stellar luminosity and lifetime.

Q8(c): MOSFET

MOSFET stands for Metal-Oxide-Semiconductor Field-Effect Transistor. It is a voltage-controlled device widely used in digital electronics, switching circuits and integrated circuits.

Main Terminals

• Gate: Controls the channel.
• Source: Supplies charge carriers.
• Drain: Collects charge carriers.
• Body/Substrate: Semiconductor base.

Structure

The gate is separated from the semiconductor by a thin oxide layer, usually silicon dioxide. This insulation makes gate current extremely small and input resistance very high.

Working

When gate voltage is applied, it controls the conductivity of the channel between source and drain. In an n-channel enhancement MOSFET, positive gate voltage attracts electrons and forms a conducting channel.

Types of MOSFET

Applications

1. Microprocessors and memory chips.
2. Digital logic gates.
3. Power switching circuits.
4. Amplifiers.
5. Voltage regulators and motor control.