CSS Physics Paper-II 2025 Solved

Question 2: Black-Body Radiation, Ultraviolet Catastrophe and Balmer Series

Q2(a): Spectral Properties of Perfect Black-Body Radiation

A perfect black body is an ideal body that absorbs all incident radiation and emits the maximum possible radiation at a given temperature. The emitted radiation depends only on temperature, not on the material or surface composition.

Main Spectral Properties

1. The spectrum is continuous and contains all wavelengths.
2. For a fixed temperature, intensity first increases with wavelength, reaches a maximum and then decreases.
3. As temperature increases, the total emitted energy increases rapidly.
4. The peak wavelength shifts toward shorter wavelengths when temperature rises.
5. The spectral distribution depends only on absolute temperature.

The shift of peak wavelength is described by Wien’s displacement law:

λmaxT = b

Total emitted power per unit area is given by Stefan-Boltzmann law:

E = σT⁴

Ultraviolet Catastrophe

Classical physics tried to explain black-body radiation using the Rayleigh-Jeans law. It worked at long wavelengths but failed badly at short wavelengths. According to the classical result, intensity should increase without limit as wavelength becomes very small.

This impossible prediction of infinite energy at ultraviolet wavelengths is called the ultraviolet catastrophe.

Planck’s Explanation

Max Planck solved the problem by proposing that energy is not emitted continuously. Instead, it is emitted and absorbed in discrete packets called quanta.

E = nhf

For one quantum:

E = hf

This quantum assumption produced the correct black-body spectrum and started quantum theory.

Q2(b): Fifth and Seventh Members of Balmer Series

The Balmer series corresponds to electronic transitions ending at n=2. The formula is:

1/λ = R(1/2² - 1/n²)

In the Balmer series:

The third member is given as:

λ3 = 1200 Å

For the third member, n=5:

1/λ3 = R(1/4 - 1/25)

1/1200 = R(25-4)/100

1/1200 = R(21/100)

So:

R = 1/[1200(21/100)]

Instead of calculating R separately, use ratio method:

λn = λ3[(1/4 - 1/25)/(1/4 - 1/n²)]

Fifth Member: n=7

λ5 = 1200[(1/4 - 1/25)/(1/4 - 1/49)]

λ5 = 1200[(0.25 - 0.04)/(0.25 - 0.02041)]

λ5 = 1200[0.21/0.22959]

λ5 ≈ 1098 Å

Seventh Member: n=9

λ7 = 1200[(1/4 - 1/25)/(1/4 - 1/81)]

λ7 = 1200[(0.21)/(0.25 - 0.01235)]

λ7 = 1200[0.21/0.23765]

λ7 ≈ 1061 Å

Question 3: Commutation, de Broglie Waves and Uncertainty Principle

Q3(a): Commutation Between Observables

In quantum mechanics, physical observables such as position, momentum, energy and angular momentum are represented by operators. Two observables commute if the order of applying their operators does not change the result.

The commutator of two operators A and B is:

[A,B] = AB - BA

If:

[A,B] = 0

then the two observables commute.

Meaning of Commuting Observables

1. They can have simultaneous eigenfunctions.
2. They can be measured simultaneously with definite values.
3. No uncertainty relation of the same type is forced between them.
4. The order of measurement does not disturb the result in the same way as non-commuting observables.

Non-Commuting Observables

If:

[A,B] ≠ 0

then the observables are generally incompatible. They cannot both have exact definite values in the same state.

A famous example is position and momentum:

[x,px] = iℏ

This non-zero commutator leads to the uncertainty relation:

ΔxΔp ≥ ℏ/2

Q3(b): de Broglie Wave-Particle Duality and Uncertainty Principle

De Broglie proposed that every moving particle has an associated matter wave. The wavelength of this wave is:

λ = h/p

where h is Planck’s constant and p is momentum.

This means that particles such as electrons are not merely localized points. They have wave character. A perfectly definite momentum corresponds to a single wavelength, but a single wavelength extends over all space and cannot represent a sharply localized particle.

Wave Packet Argument

To localize a particle, many waves of different wavelengths must be superposed to form a wave packet. The narrower the wave packet in position, the wider the spread of wavelengths and momenta.

Therefore:

• Small uncertainty in position gives large uncertainty in momentum.
• Small uncertainty in momentum gives large uncertainty in position.

This leads to Heisenberg uncertainty principle:

ΔxΔp ≥ ℏ/2

It is not a weakness of measuring instruments. It is a fundamental property of matter waves.

Question 4: Nuclear Magic Numbers and Nuclear Quadrupole Moment

Q4(a): Magic Numbers in Nuclear Physics

Magic numbers are special numbers of protons or neutrons that make a nucleus unusually stable. They are explained by the nuclear shell model, which says that nucleons occupy shells in a way similar to electrons in atomic shells.

The common nuclear magic numbers are:

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

When a nucleus has a magic number of protons or neutrons, a nuclear shell is complete. Such nuclei are more stable than neighbouring nuclei.

Evidence for Magic Numbers

1. Nuclei with magic numbers have higher binding energy.
2. They are often more abundant in nature.
3. They have lower tendency to capture additional nucleons.
4. They often have spherical shapes.
5. They show extra stability against nuclear deformation.

Examples

Q4(b): Nuclear Quadrupole Moment

Nuclear quadrupole moment measures how much the charge distribution of a nucleus deviates from spherical symmetry. If a nucleus is perfectly spherical, its quadrupole moment is zero.

In simplified form, quadrupole moment is related to charge distribution by:

Q = ∫ρ(r)(3z²-r²)dτ

where ρ(r) is nuclear charge density.

Physical Meaning

Significance

1. It gives information about nuclear shape.
2. It helps test the nuclear shell model and collective model.
3. It explains hyperfine splitting in atomic spectra.
4. It shows interaction of nuclear charge distribution with electric field gradients.
5. It helps identify nuclear deformation.

Question 5: Electric Dipole Moment and Biot-Savart Law

Q5(a): Electric Dipole Moment

An electric dipole consists of two equal and opposite charges separated by a small distance. If charges +q and -q are separated by vector distance d, then electric dipole moment is defined as:

p = qd

The direction of electric dipole moment is from negative charge to positive charge.

Derivation from Torque in Uniform Electric Field

Place an electric dipole in a uniform electric field E. The positive charge experiences force:

F = qE

The negative charge experiences equal and opposite force:

F = -qE

The net force is zero, but the two forces form a couple and produce torque.

If the dipole makes angle θ with electric field, torque magnitude is:

τ = qEd sinθ

Since:

p = qd

we get:

τ = pE sinθ

Vector form:

τ = p×E

Potential energy of a dipole in electric field is:

U = -p·E

Two Uses in Modern Science

Q5(b): Biot-Savart Law

Biot-Savart law gives the magnetic field produced by a small current element. It is the magnetic equivalent of Coulomb-type field calculation for steady currents.

The magnetic field due to current element I dl is:

dB = (μ0/4π)(I dl×r̂/r²)

Magnitude form:

dB = (μ0/4π)(I dl sinθ/r²)

Direction is given by right-hand rule.

Features of Biot-Savart Law

1. Magnetic field is directly proportional to current I.
2. It is directly proportional to length element dl.
3. It is directly proportional to sinθ.
4. It is inversely proportional to r².
5. The field direction is perpendicular to the plane containing dl and r̂.

Integral Form

For a complete current distribution:

B = (μ0/4π) ∫ I dl×r̂/r²

Question 6: Hall Effect, Sign of Charge Carriers and Carrier Concentration

Q6(a): Hall Effect and Sign of Charge Carriers

Hall effect occurs when a current-carrying conductor or semiconductor is placed in a magnetic field perpendicular to the current. Moving charge carriers experience magnetic force and accumulate on one side of the sample, creating a transverse voltage called Hall voltage.

The magnetic force on charge carriers is:

F = q(vd×B)

This force pushes positive and negative carriers in opposite directions. Therefore, the sign of Hall voltage tells whether the majority charge carriers are positive or negative.

Carrier Sign Identification

Q6(b): Hall Effect and Carrier Concentration

Consider a rectangular sample carrying current I along the x-direction. Magnetic field B is applied along the z-direction. Charge carriers are deflected sideways and a Hall electric field EH develops along the y-direction.

At equilibrium, electric and magnetic forces balance:

qEH = qvdB

Cancel q:

EH = vdB

Current density is:

J = nqvd

So:

vd = J/(nq)

Substitute into Hall field:

EH = (J/(nq))B

Hall coefficient is defined as:

RH = EH/(JB)

Therefore:

RH = 1/(nq)

Thus carrier concentration is:

n = 1/(qRH)

Using Hall Voltage

If sample thickness is t, Hall voltage is:

VH = EH w

Using standard Hall geometry, carrier concentration can be written as:

n = IB/(qVHt)

where I is current, B is magnetic field, VH is Hall voltage and t is sample thickness.

Question 7: Photomultiplier Tube and Proton-Electron Energies

Q7(a): Photomultiplier Tube Construction and Working

A photomultiplier tube is a very sensitive light detector. It converts weak light into an electrical signal and amplifies that signal through secondary electron emission.

Main Parts of PMT

Working

1. A photon enters the PMT and strikes the photocathode.
2. The photocathode emits a photoelectron due to photoelectric effect.
3. The electron accelerates toward the first dynode.
4. On striking the dynode, it releases several secondary electrons.
5. These electrons move to the next dynode and multiply again.
6. After several stages, a large number of electrons reach the anode.
7. The anode current pulse is measured electronically.

Applications

• Scintillation counters.
• Nuclear radiation detection.
• Low-light measurement.
• Spectroscopy.
• Medical imaging and photon counting.

Q7(b): Energies of Proton and Electron with Same de Broglie Wavelength

Given de Broglie wavelength:

λ = 1 Å = 1×10⁻¹⁰ m

de Broglie relation is:

λ = h/p

So momentum is:

p = h/λ

Using:

h = 6.626×10⁻³⁴ J s

we get:

p = (6.626×10⁻³⁴)/(1×10⁻¹⁰)

p = 6.626×10⁻²⁴ kg m/s

For non-relativistic kinetic energy:

K = p²/(2m)

Electron Energy

Electron mass:

me = 9.11×10⁻³¹ kg

Electron kinetic energy:

Ke = p²/(2me)

Ke = (6.626×10⁻²⁴)²/[2(9.11×10⁻³¹)]

Ke ≈ 2.41×10⁻¹⁷ J

Convert into electron-volts:

Ke = (2.41×10⁻¹⁷)/(1.602×10⁻¹⁹) eV

Ke ≈ 150 eV

Proton Energy

Proton mass:

mp = 1.67×10⁻²⁷ kg

Proton kinetic energy:

Kp = p²/(2mp)

Kp = (6.626×10⁻²⁴)²/[2(1.67×10⁻²⁷)]

Kp ≈ 1.31×10⁻²⁰ J

Convert into electron-volts:

Kp = (1.31×10⁻²⁰)/(1.602×10⁻¹⁹) eV

Kp ≈ 0.082 eV

Comparison

For the same wavelength, electron and proton have the same momentum, but kinetic energy depends inversely on mass:

K = p²/(2m)

Since proton is much heavier than electron, its kinetic energy is much smaller.

The ratio is:

Ke/Kp = mp/me ≈ 1836

Question 8: Reduced Mass, Electron Microscope and Heisenberg Principle

Q8(a): Reduced Mass Correction in Spectroscopy

In simple Bohr theory, the nucleus is often assumed to be fixed and the electron is assumed to revolve around it. In reality, both electron and nucleus move around their common center of mass. Therefore, electron mass must be replaced by reduced mass.

Reduced mass is:

μ = mM/(m+M)

where m is electron mass and M is nuclear mass.

For hydrogen-like atoms, energy levels become:

En = -[μk²e⁴/(2ℏ²)](1/n²)

The Rydberg constant is also corrected:

RM = R∞(μ/me)

Significance

1. It improves accuracy of spectral line calculations.
2. It explains isotope shifts in hydrogen and deuterium spectra.
3. It shows that the nucleus is not perfectly fixed.
4. It is important in high-resolution spectroscopy.

Q8(b): Electron Microscope

An electron microscope uses accelerated electrons instead of visible light to form highly magnified images. It works because electrons have de Broglie wavelength:

λ = h/p

When electrons are accelerated to high speed, their wavelength becomes much smaller than visible light wavelength. Since resolving power improves when wavelength decreases, electron microscopes can resolve much smaller details than optical microscopes.

Main Parts

• Electron gun
• Accelerating voltage
• Magnetic or electrostatic lenses
• Specimen chamber
• Detector or fluorescent screen
• Vacuum system

Types

Advantages

1. Very high resolving power.
2. Useful in materials science, biology and nanotechnology.
3. Can show structures far smaller than optical microscopes can resolve.

Q8(c): Heisenberg Uncertainty Principle

Heisenberg uncertainty principle states that certain pairs of physical quantities cannot both be measured with unlimited precision at the same time.

For position and momentum:

ΔxΔp ≥ ℏ/2

For energy and time:

ΔEΔt ≥ ℏ/2

Physical Meaning

The principle is not merely due to poor instruments. It is a basic property of quantum systems. A particle with a very well-defined position must be represented by a narrow wave packet, and a narrow wave packet requires many wavelengths. Many wavelengths mean a spread in momentum.

Connection with de Broglie Waves

Since:

λ = h/p

uncertainty in wavelength corresponds to uncertainty in momentum. A sharply localized particle therefore cannot have one exact wavelength or one exact momentum.

Importance

1. It explains why electrons cannot be treated as classical particles in atoms.
2. It supports the wave-packet view of matter.
3. It is central to quantum mechanics.
4. It helps explain zero-point energy and atomic stability.
5. It can be used to show why electrons do not pre-exist inside the nucleus as ordinary confined particles.