UPSC Civil Services (Main) Examination 2026 — Physics Optional Paper II: Questions with Model Answers | Plutus IAS

08 Sep UPSC Civil Services (Main) Examination 2026 — Physics Optional Paper II: Questions with Model Answers | Plutus IAS

The questions below are from Physics Optional Paper II of UPSC Civil Services (Main) Examination 2026 (held 2026-08-30) — the actual paper, which is public. Each carries a model answer written by Aanya in Plutus IAS teaching style, to the marks and word limit.

Official source: official (upsc.gov.in).

Q1. (i) The de Broglie wavelength of a particle in thermal equilibrium at a temperature of $27^{\circ}\text{C}$ is $\lambda$ . At what temperature will it be $\frac{\lambda}{2}$ ? (ii) The wave function of a spin $\frac{1}{2}$ particle is given by $\left(\frac{-i}{2}\right)$ . What is the probability of finding the particle in the spin-up state? 5+5 (b) The OH-radical has a moment of inertia of $1.48 \times 10^{-46} \text{ kg.m}^2$ . For $J = 6$ , calculate its angular velocity and angular momentum. Find the energy absorbed in the $J = 6 \rightarrow J = 7$ transition. (c) In a Stern-Gerlach experiment, the magnetic field varies with distance in the $z$ -direction according to $\frac{\text{dB}_z}{\text{d}z} = 1.5 \text{ T/mm}$ . The speed of silver atoms coming from the oven is $725 \text{ m/s}$ . Silver atoms travel a distance of $3 \text{ cm}$ through the magnet. Calculate the separation of the two beams of atoms when they leave the magnet. [Given : The mass of a silver atom = $1.8 \times 10^{-25} \text{ kg}$ and its magnetic moment is 1 Bohr magneton.] (d) If the position and momentum operators in 1-D are represented by $\hat{x}$ and $\hat{p}$ respectively, calculate the commutator $[\hat{x}^2, \cos(\hat{p})]$. (e) Find the eigenvalues and eigenfunctions of a system described by Hamiltonian $H = ae^{b\sigma_x}$, where a, b are constants and $\sigma_x$ is the x-component of Pauli matrix $\vec{\sigma}$. (15 marks)

How to approach this question

This is a two-part problem testing de Broglie wavelength and spin-state probability. The examiner is checking understanding of thermal de Broglie wavelength and spin superposition. A top answer must (1) derive the temperature dependence of λ from kinetic theory, (2) compute the new temperature for λ/2, and (3) normalize the spin wave function and compute the projection probability. The common mistake is forgetting the 3/2 factor in the average kinetic energy per translational degree of freedom.

Model answer

(i) The de Broglie wavelength of a particle in thermal equilibrium is

λ = h / √(2 m k_B T).

Solving for T,

T = h² / (2 m k_B λ²).

For λ → λ/2, T scales as (λ/2)⁻², so

T₂ = T₁ × (λ / (λ/2))² = 4 T₁.

Given T₁ = 27 °C = 300 K, T₂ = 1200 K = 927 °C.

(ii) The spin wave function is |ψ⟩ = (−i/2)|↑⟩ + (√3/2)|↓⟩. Normalization:

(|−i/2|² + |√3/2|²) = (1/4 + 3/4) = 1.

The probability of finding the particle in the spin-up state is |⟨↑|ψ⟩|² = |−i/2|² = 1/4.

Q2. (a) A particle is in the ground state of a 1-D simple harmonic oscillator potential, $V(x) = \frac{1}{2} m\omega^2 x^2$. Calculate the position uncertainty of the particle. (b) $$H = \frac{p^2}{2m} + V(x), \text{ जहाँ}$$ $$V(x) = \frac{1}{2} m\omega^2 x^2 \quad x \ge 0 \text{ के लिए}$$ $$= \infty \quad x < 0 \text{ के लिए}$$ (i) A particle is in the normalized state $\psi$ which is a superposition of the energy eigenstates $\psi_1$ and $\psi_2$ with energies 10 eV and 30 eV, respectively. The average value of the energy of the particle in the state $\psi$ is 24 eV. Find the state $\psi$ in terms of $\psi_1$ and $\psi_2$. (ii) Obtain the allowed eigenenergies of the half 1-D simple harmonic oscillator defined by the Hamiltonian $$\begin{array}{l} H = \frac{p^2}{2m} + V(x), \text{ where} \\ V(x) = \frac{1}{2} m\omega^2 x^2 \quad \text{for } x \ge 0 \\ = \infty \quad \text{for } x < 0 \end{array}$$ Express its eigenstates in terms of eigenstates of the full 1-D oscillator. 10+10 (c) (i) Why does the electron paramagnetic resonance (EPR) spectroscopy utilize the microwave region of the electromagnetic spectrum, while nuclear magnetic resonance (NMR) uses radio waves ? (ii) Why can the experimental observation of the Lamb shift not be explained by the Dirac theory ? 7+8 (15 marks)

How to approach this question

The directive word is calculate, find, and obtain, so the examiner tests (i) analytical derivation of quantum uncertainties and eigenstates, (ii) superposition coefficients from energy averages, and (iii) boundary-condition modification of the harmonic-oscillator spectrum. A top answer must (1) state the governing equations, (2) impose the physical constraints (normalization, boundary), and (3) present explicit algebraic results with units. The common mistake is to overlook the half-line boundary condition in part (b)(ii) and to omit the explicit form of the superposition state in (b)(i).

Model answer

Part (a)

For a one-dimensional simple harmonic oscillator in its ground state, the normalized wave-function is

ψ₀(x) = (mω/πħ)^¼ exp(–mωx²/2ħ).

The position uncertainty Δx = √⟨x²⟩ is obtained by evaluating the expectation value ⟨x²⟩ = ∫₋∞^∞ x²|ψ₀|² dx = ħ/(2mω). Hence

Δx = √(ħ/2mω).

Part (b)(i)

Let ψ = c₁ψ₁ + c₂ψ₂ with energies E₁ = 10 eV and E₂ = 30 eV. Normalization |c₁|² + |c₂|² = 1 and the average energy ⟨E⟩ = |c₁|²E₁ + |c₂|²E₂ = 24 eV give the linear system

• 10|c₁|² + 30|c₂|² = 24
• |c₁|² + |c₂|² = 1

Solving yields |c₂|² = 0.7 and |c₁|² = 0.3. Choosing real coefficients for simplicity, ψ = √0.3 ψ₁ + √0.7 ψ₂.

Part (b)(ii)

The infinite wall at x = 0 imposes ψ(0) = 0. The full-oscillator eigenstates φₙ(x) = Nₙ Hₙ(ξ) e^–ξ²/2 with ξ = x√(mω/ħ) satisfy φₙ(0) = 0 only when n is odd. Thus the allowed eigenstates are φ₁, φ₃, φ₅, … and the corresponding eigenenergies remain Eₙ = (n + ½)ħω with n = 1, 3, 5, … . In other words, the half-line oscillator inherits every other level of the full oscillator.

Part (c)(i)

EPR uses microwave photons (≈ 0.1–100 GHz) because the electron magnetic moment μ_e ≈ 9.27×10^–24 J T^–1 couples to fields of order tesla, giving resonance frequencies in the GHz range. Nuclear moments μ_N ≈ 5.05×10^–27 J T^–1 are ∼10^–3 smaller, so NMR resonances fall in the radio-frequency band (≈ 1–100 MHz).

Part (c)(ii)

The Lamb shift, a small energy difference between 2S_½ and 2P_½ in hydrogen, arises from vacuum fluctuations and virtual electron-positron pairs. Dirac theory, which treats the electron as a point Dirac particle in a Coulomb potential, cannot account for these radiative corrections because it lacks second-quantized electromagnetic interactions.

Q3. (i) Calculate the number of permitted electrons in a sub-shell and a shell in an atom. (ii) Explain how Pauli's exclusion principle helps in determining the electronic configuration in a many-electron system. 8+12 (b) (i) Why is the relaxation process so important in nuclear magnetic resonance (NMR) ? (ii) Why are $^{12}\text{C}$ and $^{16}\text{C}$ nuclei not suitable for the study of NMR ? 10+5 (c) (i) The wave function of a hydrogen atom is written as $\psi_{n, l, m} (r, \theta, \phi)$. Explain the quantum numbers $n, l, m$ and mention their ranges. Show that $\psi_{n, l, m} (r, \theta, \phi)$ is $n^2$ degenerate. (ii) The electron in the hydrogen atom is found in the state $\psi (r, \theta, \phi) = A R(r) \sin \theta \cos \theta e^{-i\phi}$. Find the z-component of angular momentum of the electron. 10+5 (15 marks)

How to approach this question

The directive word “calculate” in (i) demands precise numerical derivation using quantum rules; “explain” in (ii) requires a logical argument linking Pauli’s principle to real electronic configurations. Examinees must (1) state the formula for sub-shell and shell capacities, (2) derive degeneracy via quantum-number ranges, and (3) illustrate with concrete examples such as carbon or oxygen. The common mistake is to quote degeneracy without proving it from the quantum-number ranges, or to confuse the roles of n, l, and m.

Model answer

Part (i): Permitted electrons in a sub-shell and a shell

For a given orbital angular momentum quantum number l, the magnetic quantum number m can take (2l+1) integer values from –l to +l. Each orbital can accommodate 2 electrons (spin up and spin down). Hence the number of electrons in a sub-shell is

N_sub-shell = 2(2l + 1).

For a principal shell with quantum number n, l ranges from 0 to (n–1), giving

N_shell = Σ_l=0^n–1 2(2l + 1) = 2n².

Part (ii): Pauli’s exclusion principle and electronic configuration

Pauli’s exclusion principle states that no two electrons in a system can share the same set of four quantum numbers (n, l, m, m_s). This directly limits each orbital to exactly two electrons with opposite spins. In many-electron atoms, electrons therefore fill the lowest-energy orbitals first (Aufbau principle), then pair only when necessary (Hund’s rule). For example, carbon (Z=6) places two 1s electrons (n=1, l=0, m=0, m_s=±½), two 2s electrons (n=2, l=0, m=0, m_s=±½), and two of the six available 2p orbitals (n=2, l=1, m=–1,0,+1; each with paired spins), yielding the configuration 1s² 2s² 2p². Without Pauli’s principle, all six electrons could occupy the 1s orbital, contradicting observed chemical periodicity and spectroscopic data.

Part (b)(i): Importance of relaxation in NMR

Relaxation restores the spin system to thermal equilibrium after radio-frequency excitation. Longitudinal (T₁) relaxation realigns spins with the static magnetic field B₀, while transverse (T₂) relaxation refocuses coherence loss, enabling signal detection and contrast in MRI. Efficient relaxation shortens repetition times, improves signal-to-noise ratio, and prevents saturation, making it indispensable for high-resolution spectra and functional imaging.

Part (b)(ii): Why ¹²C and ¹⁶O are NMR-inactive

¹²C has zero nuclear spin (I=0), so it lacks a magnetic dipole moment and cannot couple to B₀. ¹⁶O also has I=0, despite even mass number, because its protons and neutrons pair to zero net spin. Both nuclei therefore produce no NMR signal under standard conditions.

Part (c)(i): Quantum numbers n, l, m and degeneracy

The principal quantum number n (n=1,2,…) sets the energy and the radial extent; l (0≤l≤n–1) defines the orbital shape; m (–l≤m≤+l) fixes the z-component of angular momentum. For each n, there are n possible l-values, each with (2l+1) m-values, giving a total degeneracy

D = Σ_l=0^n–1(2l+1) = n².

Thus ψ_n,l,m is n²-fold degenerate.

Part (c)(ii): z-component of angular momentum

The factor e^–iφ corresponds to m=–1, so the z-component of angular momentum is

L_z = ħ m = –ħ.

Q4. (i) What are hot bands in vibrational spectroscopy ? Why are they called so ? (ii) The fundamental and 2$^{nd}$ harmonic transitions of $^{14}\text{N}^{16}\text{O}$ are recorded at $1876.06 \text{ cm}^{-1}$ and $3724.2 \text{ cm}^{-1}$, respectively. Calculate the equilibrium vibrational frequency, the anharmonicity constant and the zero-point energy of the molecule. [Given : Mass of $^{14}\text{N}$ = $23.25 \times 10^{-27} \text{ kg}$, Mass of $^{16}\text{O}$ = $26.56 \times 10^{-27} \text{ kg}$] 8+7 (b) A particle of mass m is in the ground state of a 1-D box of infinite potential of length a. If the length of the box is changed to 2a symmetrically without disturbing the wave function of the particle, find the probability that the particle will be in the ground state of the new box. (c) (i) Explain why the rotational transition J = 0 → J = 1 is often not the most intense. (ii) Find the positions of the first four rotational Raman lines in the spectrum of H₂, if its bond length is 0·742 Å. What will be the effect of nuclear spin on the spectrum ? [Given : Mass of ¹H = 1·673 × 10⁻²⁷ kg] 8+7 (15 marks)

How to approach this question

The directive word “calculate” in part (ii) signals that the examiner is testing your ability to translate spectroscopic data into molecular constants using the anharmonic oscillator model. A top answer must (1) define hot bands and explain their origin, (2) set up and solve the anharmonic oscillator equations for the given NO transitions to extract ω_e, x_eω_e and ZPE, and (3) show the quantum‐mechanical probability calculation for the sudden box expansion. The common mistake is to confuse the fundamental with the equilibrium frequency or to misapply the harmonic‐oscillator approximation when anharmonicity is explicitly required.

Model answer

Hot bands are transitions in vibrational spectroscopy that originate from excited vibrational levels (v > 0) rather than the ground state (v = 0). They appear at slightly lower wavenumbers than the fundamental band because the upper state has a higher average bond length and hence a smaller force constant. The name “hot” reflects the requirement that the initial vibrational level be thermally populated, so their intensity increases with temperature.

Part (ii)
For the anharmonic oscillator, the vibrational term values are

• G(v) = ω_e(v + ½) – x_eω_e(v + ½)².

Given the fundamental (0→1) and second harmonic (0→2) transitions:

• ν₀₁ = G(1) – G(0) = ω_e – 2x_eω_e = 1876.06 cm^–1
• ν₀₂ = G(2) – G(0) = 2ω_e – 6x_eω_e = 3724.2 cm^–1

Solving simultaneously gives ω_e = 1904.26 cm^–1 and x_eω_e = 14.10 cm^–1. The zero-point energy is G(0) = ½ ω_e – ¼ x_eω_e = 947.60 cm^–1.

Part (b)
The original ground-state wavefunction ψ₁(x) = √(2/a) sin(πx/a) is expanded symmetrically to length 2a. In the new basis {φ_n(x)}, the probability of finding the particle in the new ground state φ₁ is |⟨φ₁₁

Part (c)(i)
The J = 0→1 transition is weak because the initial state has no permanent dipole moment, so the transition moment vanishes in the rigid-rotor approximation; intensity arises only through centrifugal distortion or centrifugal distortion–induced dipole moments.

Part (c)(ii)
For H₂ the rotational constant B = ħ/(4πcI) with I = μr² gives B = 60.8 cm^–1. Raman selection rules ΔJ = 0, ±2 give Stokes lines at 2B, 4B, 6B, 8B, i.e. 121.6, 243.2, 364.8 and 486.4 cm^–1. Because H₂ nuclei are identical fermions (I = ½), the spectrum alternates in intensity (ortho–para alternation) with even-J lines enhanced.

Q5. Are the nucleons inside the nucleus governed by the laws of quantum physics ? Justify your answer. (b) Binding energy and rest mass energy of a two-nucleon bound state are denoted by B and $Mc^2$, respectively, where c is the speed of light. Calculate the minimum energy of a photon required to dissociate this bound state in terms of B and $Mc^2$. (c) A nucleus of rest mass M is initially in an excited state whose energy is $\Delta E$ above its ground state. The nucleus emits a $\gamma$-ray of energy $h\nu$ and makes a transition to its ground state. Calculate the fractional change in energy for the nucleus. (d) (i) What is the effective mass of an electron ? (ii) What are the physical reasons that the effective mass of an electron can be infinite and negative ? (iii) The energy near the valence band edge of a crystal is given by $E = -10^{-39} \text{ k}^2 \text{ Jm}^2$. An electron with wave vector $10^{10} \hat{k}_x \text{ m}^{-1}$ is removed from an orbital in the completely filled valence band. Determine its effective mass and momentum. 2+3+5 (e) Explain briefly the two-fluid model proposed by the London brothers and obtain the expressions for the penetration depth and the number of superelectrons in a superconducting specimen. (15 marks)

How to approach this question

The directive word “Justify” signals that the examiner wants a reasoned argument, not just a yes/no answer. The question has three parts: (a) conceptual justification of quantum laws for nucleons, (b) a simple energy-balance calculation, and (c) a kinematic correction for recoil. The common mistake is to ignore the finite nuclear mass in part (c) and treat the nucleus as infinitely heavy.

Model answer

Nucleons inside the nucleus are governed by quantum physics. Experimental evidence shows discrete nuclear energy levels (e.g., the 14.4 keV transition in ⁵⁷Fe observed via Mössbauer spectroscopy), quantized angular momentum (nuclear shell model), and Pauli exclusion that stabilizes neutron stars. Moreover, the de Broglie wavelength of a nucleon with momentum ~100 MeV/c is ~1 fm, comparable to nuclear dimensions, so wave-like behaviour is essential. Thus, nucleons obey the Schrödinger (or Dirac) equation inside the potential well of the strong force.

(b) Photon energy to dissociate a two-nucleon bound state. Energy conservation requires the photon to supply both the binding energy B and the rest-mass energy of the separated nucleons, Mc². Hence, the minimum photon energy is

E_γ,min = B + Mc².

(c) Fractional change in nuclear energy. After γ-emission the nucleus recoils with momentum p = hν/c. Its kinetic energy is p²/2M = (hν)²/2Mc². Energy conservation gives

hν + (hν)²/2Mc² = ΔE.

Solving to first order in the small recoil term yields

hν ≈ ΔE – (ΔE)²/2Mc².

The fractional change is therefore

ΔE – hν / ΔE ≈ (ΔE)/(2Mc²).

Q6. By considering the three-dimensional harmonic oscillator potential, explain the energy levels of nucleons inside the nucleus. Also, derive the zero-point energy of nucleons. Why did this assumption fail to explain the existence of nuclei with higher magic number ? (b) What do you understand by the packing fraction 'f' and the binding energy ' $E_b$ ' of a nucleus ? Draw the graphs for packing fraction f versus mass number (A) and binding energy fraction $f_b \left( = \frac{E_b}{A} \right)$ versus mass number (A). Further, explain how the graphs of the variation of f and $f_b$ with A have complementary approaches. (c) (i) What are the different methods used to increase the probability of exposure for the atomic planes with right orientation to X-rays in the X-rays diffraction studies ? (ii) Explain the principle and working of the Laue's diffraction method. Comment on the origin of Laue spots and the utility of Laue's diffraction pattern. 5+15 (15 marks)

How to approach this question

This is a 15-mark, three-part question testing conceptual clarity in nuclear physics. The examiner wants: (a) an explanation of the 3D harmonic oscillator potential for nucleons, derivation of zero-point energy, and identification of its failure at higher magic numbers; (b) definitions and graphical trends of packing fraction and binding energy fraction, with an explanation of their complementary roles; (c) a two-subpart answer on X-ray diffraction—methods to increase exposure probability and a detailed explanation of Laue’s method, including spot origin and utility. The common mistake is to treat each part in isolation without linking the physics narrative across subparts.

Model answer

(a) Energy levels of nucleons in a 3D harmonic oscillator potential

In the independent-particle shell model, nucleons move in an effective potential approximated by a three-dimensional isotropic harmonic oscillator: V(r) = ½ mω²r². The Schrödinger equation yields quantized energy levels E = ħω(n + 3/2), where n = nₓ + nᵧ + n_z = 0, 1, 2, … is the principal quantum number. Each level is (n+1)(n+2)/2-fold degenerate, leading to closed shells at n = 0, 2, 5, 7, … corresponding to magic numbers 2, 8, 20, 28, …. The zero-point energy, obtained at n = 0, is E₀ = (3/2)ħω. However, this model fails beyond A ≈ 40 because spin–orbit coupling, omitted here, splits levels and generates the higher magic numbers 50, 82, 126 observed experimentally.

(b) Packing fraction f and binding energy E_b

Packing fraction is defined as f = (M − A)/A, where M is the atomic mass in u and A is the mass number. It measures mass defect per nucleon and is negative for stable nuclei. The binding energy fraction is f_b = E_b/A, where E_b = [Zm_p + (A−Z)m_n − M]c². Graphically, f starts near zero for light nuclei, dips to a minimum around A ≈ 56 (Fe), then rises slowly; f_b peaks near A ≈ 56 and gently declines thereafter. These trends are complementary: f tracks the fractional mass defect, while f_b reflects the energy needed to remove a nucleon; together they encapsulate nuclear stability and the liquid-drop model’s saturation property.

(c)(i) Methods to increase exposure probability in X-ray diffraction

• Rotation method: Continuously rotate the crystal to bring different planes into reflecting orientation.
• Powder method: Use a polycrystalline sample so that randomly oriented crystallites ensure some planes satisfy Bragg’s law.
• Fiber diffraction: Align fibrous molecules to increase the number of planes parallel to the fiber axis.
• Temperature control: Adjust temperature to reduce mosaicity and enhance coherent scattering.

(c)(ii) Principle and working of Laue’s diffraction method

Laue’s method uses a stationary single crystal and a continuous (white) X-ray spectrum. Each set of planes selects its characteristic wavelength from the spectrum that satisfies Bragg’s law 2d sinθ = nλ. The diffracted beams form discrete spots—Laue spots—on a photographic plate. The origin of Laue spots lies in constructive interference from planes whose spacing and orientation match available wavelengths. The pattern directly maps crystal symmetry: spot positions reveal plane spacings and angles, while intensities encode atomic form factors. Laue patterns are indispensable for quick symmetry determination and orientation of single crystals before more precise monochromatic methods.

Q7. (i) What is diamagnetism ? Why do diamagnetic materials have negative magnetic susceptibility ? (ii) Discuss the limitations of the quantum theory of diamagnetism. (iii) Draw a diagram of magnetic susceptibility as a function of temperature. Pay attention to the relative scales in your diagram so that the relative strengths of the magnetic phenomena are presented. 5+5+10 (b)   (I) (II) The input signal as shown in Figure (I) is applied to the comparator circuit given in Figure (II). Make a sketch of the output signal showing its proper relationship to the input signal. Assume that the maximum output levels of the op-amp are ± 12 V.   (I) (II) (c) What is the criticality of a self-sustained nuclear reactor ? Write the basic processes affecting the nuclear chain reactions of a finite-size nuclear reactor. Also, write the critical size of reactors of different shapes in terms of geometrical buckling constant. (15 marks)

How to approach this question

The directive word “discuss” demands a layered explanation: definition, microscopic origin, and limitations, followed by a quantitative diagram. Examiners test conceptual clarity on diamagnetism, quantum underpinnings, and ability to sketch temperature dependence with correct relative scales. The common mistake is to confuse diamagnetism with paramagnetism or to draw the susceptibility-temperature plot without marking the tiny diamagnetic values relative to paramagnetic/ferromagnetic ranges.

Model answer

Diamagnetism is a form of magnetism that appears in all materials when subjected to an external magnetic field, resulting in a weak repulsion. It arises from the induction of orbital currents in closed electron shells that oppose the applied field, in accordance with Lenz’s law. Because the induced magnetic moment is antiparallel to the field, diamagnetic susceptibility χ is negative and typically of order 10⁻⁵ to 10⁻⁶.

The negative susceptibility follows directly from the Langevin theory of induced magnetic moments. In a classical picture, an electron orbiting with angular frequency ω in a field B experiences an additional Lorentz force that shifts its orbital frequency by Δω = eB/(2m), producing an induced dipole moment μ_ind = −(e²r²/4m)B. Summing over all electrons yields a net magnetization M = −(n e² Z ⟨r²⟩/6m) B, giving χ = M/H = −μ₀ n e² Z ⟨r²⟩/6m, which is inherently negative.

Quantum theory refines this picture by quantizing orbital motion and including spin-orbit coupling. The Landau levels and the van Vleck contribution modify the induced moment, but the net result remains a small, negative χ independent of temperature for closed-shell atoms. However, the theory assumes perfect spherical symmetry and neglects exchange interactions and spin fluctuations—limitations that become evident in transition metals and superconductors where diamagnetism can be anomalously large or temperature dependent.

Limitations include: (i) inability to predict temperature dependence in metals where Landau quantization competes with Pauli paramagnetism, (ii) omission of many-body effects that enhance diamagnetism in high-Tc superconductors, and (iii) failure to account for core diamagnetism in ions with unfilled d/f shells.

Susceptibility vs. temperature diagram:

• Diamagnetic χ_dia (≈ −10⁻⁵) is a horizontal line near zero, independent of T.
• Paramagnetic χ_para (≈ +10⁻³ to +10⁻²) follows a 1/T Curie law, diverging as T→0.
• Ferromagnetic χ_ferro diverges at the Curie temperature Tc and flips sign below it.

The sketch must show χ_dia as a nearly flat, slightly negative line, χ_para rising steeply as T decreases, and χ_ferro showing a sharp peak at Tc, with vertical arrows marking Tc and horizontal arrows marking relative magnitudes.

In conclusion, diamagnetism is a universal, temperature-independent, negative susceptibility arising from induced orbital currents. While quantum theory explains its sign and order of magnitude, it cannot capture temperature anomalies in real materials, underscoring the need for many-body corrections in advanced applications such as superconductivity and topological insulators.

Q8. (i) Discuss the Debye model of lattice specific heat. What are the limitations of the Debye model ? (ii) At very low temperatures, the specific heat of rock salt varies with temperature according to the Debye T$^{3}$ law. The Debye temperature for rock salt is 281 K. How much heat will be required to raise the temperature of 2 kilo-mol of rock salt from 10 K to 50 K ? 10+5 (b) What is the photovoltaic effect ? What are the processes that are required to obtain a useful power output from photon interactions in a semiconductor ? List the factors that affect the efficiency of a solar cell. (c) What do you understand by the SU(3) symmetry for the classification of baryons and mesons ? Draw the octets for baryons and mesons along with their quarks and antiquarks. (15 marks)

How to approach this question

The directive word “Discuss” requires a balanced synthesis of theory, critique, and application. Examiners test (i) conceptual clarity of the Debye model, (ii) critical awareness of its limitations, and (iii) numerical problem-solving under low-temperature conditions. The most common mistake is to state limitations without connecting them to physical assumptions (e.g., dispersionless phonons, isotropic continuum). For part (b), clarity on the photovoltaic effect and the full chain from photon absorption to power delivery is essential; listing efficiency factors without categorising them (material, device, environmental) is a frequent pitfall. Part (c) demands precise group-theoretic understanding and graphical octet construction; aspirants often mislabel quark content or omit antiquark partners.

Model answer

(i) Debye model of lattice specific heat and its limitations

The Debye model treats lattice vibrations as phonons in a continuous elastic medium with a maximum cutoff frequency ω_D determined by the total number of vibrational modes. The model predicts the specific heat C_V ∝ T³ at low T and approaches the Dulong–Petit law at high T. Mathematically, C_V = 9Nk_B(T/θ_D)³ ∫₀^θ_D/T x⁴ e^x(e^x−1)⁻² dx, where θ_D is the Debye temperature.

The limitations are: (1) it assumes an isotropic, dispersionless continuum, which breaks down near the Brillouin zone boundary; (2) it overestimates the density of states at high frequencies; (3) it neglects optical phonon branches, crucial in ionic crystals; and (4) it cannot account for anharmonic effects responsible for thermal expansion and phonon–phonon scattering.

(ii) Heat required to raise temperature of rock salt

At T ≪ θ_D (10–50 K ≪ 281 K), the Debye T³ law applies. The molar specific heat is C_V = (12π⁴/5) n R (T/θ_D)³. For 2 kilo-mol, n = 2000. Integrating C_V dT from 10 K to 50 K gives

Q = ∫₁₀⁵⁰ n C_V(T) dT = n (12π⁴/5) R (1/θ_D)³ ∫₁₀⁵⁰ T³ dT

= 2000 × (12π⁴/5) × 8.314 J mol⁻¹ K⁻¹ × (1/281)³ × [T⁴/4]₁₀⁵⁰

≈ 2000 × 1944 × 8.314 × 4.53×10⁻⁸ × (6.25×10⁶ − 2.5×10⁴)

≈ 9.2 × 10⁶ J.

(b) Photovoltaic effect and power generation in semiconductors

The photovoltaic effect is the generation of a voltage and electric current in a material upon absorption of photons with energy greater than the band gap, creating electron–hole pairs. To obtain useful power, four processes are required: (1) photon absorption with energy ≥ E_g; (2) efficient charge separation via a built-in electric field (p–n junction); (3) collection of carriers at selective contacts without recombination; and (4) external circuit extraction with minimal resistive losses.

Efficiency is affected by material factors (band gap, minority carrier lifetime, defect density), device factors (series/shunt resistance, anti-reflection coating, passivation), and environmental factors (spectrum, temperature, angle of incidence).

(c) SU(3) symmetry and quark-model octets

SU(3) flavour symmetry classifies hadrons by treating the three light quarks (u, d, s) as a fundamental triplet. Baryons (qqq) and mesons (q q̄) transform under the 8-dimensional adjoint and 8+1-dimensional representations, respectively. The baryon octet contains spin-½ states: p (uud), n (udd), Σ⁺ (uus), Σ⁰ (uds), Σ⁻ (dds), Ξ⁰ (uss), Ξ⁻ (dss), and Λ⁰ (uds). The meson octet includes π⁺ (u d̄), π⁰ (u ū − d d̄)/√2, π⁻ (d ū), K⁺ (u s̄), K⁰ (d s̄), K̄⁰ (s d̄), K⁻ (s ū), and η (mixed flavour singlet).

Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.

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