Physics 2024 Paper II 50 marks Discuss

Paper II — Q7

(a) (i) What is the origin of paramagnetism? Discuss the temperature dependence of paramagnetic susceptibility. (ii) Calculate…

(a)
(i)

What is the origin of paramagnetism? Discuss the temperature dependence of paramagnetic susceptibility.

(ii)

Calculate the diamagnetic susceptibility of atomic Hydrogen in the ground state at STP. Assume the mean square distance of electronic charge distribution of atomic Hydrogen from the nucleus r² = 3a₀², a₀ being the radius of the first Bohr orbit of Hydrogen. (8+7=15 marks)

(b)
(i)

What is the concept of effective mass of the electron?

(ii)

The energy near a valence band edge is given by E(k) = −1 × 10⁻²⁶ k² ergs. An electron is removed from the orbital k = 1 × 10⁷ kₓ cm⁻¹. Find the sign and magnitude of effective mass of the hole. (10+10=20 marks)

(c)

An RC-coupled transistor amplifier has mid-frequency gain A_vm = 100. The values of the lower and upper cut-off frequencies are f₁ = 20 Hz and f₂ = 50 kHz. Find the frequencies at which the gain is reduced to 80. 15 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

अनुचुंबकत्व का मूल क्या है ? अनुचुंबकीय प्रवृत्ति की तापमान निर्भरता की विवेचना कीजिए।

(ii)

परमाण्विक हाइड्रोजन की मानक ताप दाब पर आद्य (निम्नतम) अवस्था में प्रतिचुंबकीय प्रवृत्ति की गणना कीजिए। मान लीजिए कि परमाण्विक हाइड्रोजन के इलेक्ट्रॉनिक आवेश वितरण की नाभिक से माध्य वर्ग दूरी r² = 3a₀² है, जहाँ a₀ हाइड्रोजन की प्रथम बोर कक्ष की त्रिज्या है। (8+7=15 अंक)

(b)
(i)

इलेक्ट्रॉन के प्रभावी द्रव्यमान की क्या अवधारणा है ?

(ii)

संयोजकता बैंड के किनारे पर ऊर्जा स्तर E(k) = −1 × 10⁻²⁶ k² ergs दिया गया है। एक इलेक्ट्रॉन ऑर्बिटल k = 1 × 10⁷ kₓ cm⁻¹ से निकाल लिया जाता है। होल के प्रभावी द्रव्यमान के चिह्न एवं परिमाण ज्ञात कीजिए। (10+10=20 अंक)

(c)

एक RC-युग्मित ट्रांज़िस्टर ऐम्प्लीफायर का मध्य-आवृत्ति लाभांक (लब्धि) A_vm = 100 है। नीचे और ऊपर की अंतक (कट-ऑफ) आवृत्तियाँ f₁ = 20 Hz और f₂ = 50 kHz हैं। उस आवृत्ति को ज्ञात कीजिए जिस पर लाभांक (लब्धि) घट कर 80 हो जाता है। (15 अंक)

Q7 of the 2024 UPSC Mains Physics Paper II, as printed
The question as printed in the 2024 Physics paper

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a)(i) Origin of paramagnetism and its temperature dependence

Paramagnetism arises in materials whose atoms, ions or molecules carry a permanent magnetic dipole moment, which requires an unpaired electron: the moment is the vector sum of the spin and orbital moments of the partly filled shell, μ = g μB √(J(J+1)) for a free ion with total angular momentum J. Transition-metal and rare-earth ions, free radicals, O₂ and NO are paramagnetic. In zero field the moments point at random and the net magnetisation is zero; an applied field B tends to align them, since a moment has energy −μ·B, while thermal agitation tends to randomise them. The magnetisation is therefore small, positive and field-dependent, and the susceptibility is positive but much smaller than one.

Treating the moments classically, the average moment along the field is μL(x), L the Langevin function, x = μB/kT; for x ≪ 1, L(x) ≈ x/3, so M = Nμ²B/(3kT) and χ = μ₀Nμ²/(3kT) = C/T. Quantum mechanically the 2J + 1 orientations give the Brillouin function and the same low-field result with μ² replaced by g²μB²J(J+1). This inverse dependence on temperature is Curie's law, χ = C/T: raising the temperature increases the thermal disorder and lowers the susceptibility, so a plot of 1/χ against T is a straight line through the origin. In materials whose moments interact weakly the line is shifted, χ = C/(T − θ), the Curie–Weiss law, θ positive for ferromagnetic and negative for antiferromagnetic coupling. Two cases do not obey Curie's law: the Pauli paramagnetism of conduction electrons in metals, where only the electrons within kT of the Fermi level can respond, giving a small susceptibility nearly independent of temperature, and Van Vleck paramagnetism from field-induced mixing of excited states, also temperature-independent.

(a)(ii) Diamagnetic susceptibility of atomic hydrogen at STP

The Langevin diamagnetic susceptibility of N atoms per unit volume with Z electrons each is

χ = −μ₀ N Z e² ⟨r²⟩/(6m),

where ⟨r²⟩ is the mean square distance of the electrons from the nucleus. Here Z = 1 and ⟨r²⟩ = 3a₀² = 3 × (0.529 × 10⁻¹⁰)² = 8.39 × 10⁻²¹ m². At STP an ideal gas contains N = 2.69 × 10²⁵ atoms/m³ (Loschmidt's number).

χ = −(4π × 10⁻⁷ × 2.69 × 10²⁵ × (1.6 × 10⁻¹⁹)² × 8.39 × 10⁻²¹)/(6 × 9.11 × 10⁻³¹) = −(7.3 × 10⁻³⁹)/(5.47 × 10⁻³⁰) = −1.3 × 10⁻⁹.

The volume susceptibility of atomic hydrogen at STP is about −1.3 × 10⁻⁹ (SI, dimensionless), negative and tiny, as expected for an induced moment in a dilute gas; per mole it is about −2.4 × 10⁻⁶ cm³/mol in CGS units.

(b)(i) Effective mass of an electron

An electron in a crystal moves under the periodic potential of the lattice as well as any applied force. Its response to an external force F can still be written in Newton's form, dv/dt = F/m*, provided the free-electron mass is replaced by the effective mass

1/m* = (1/ħ²) d²E/dk²,

which follows from the group velocity v = (1/ħ) dE/dk and ħ dk/dt = F. The effective mass is a property of the band structure: it is small where the band is sharply curved, large where the band is flat, and negative near the top of a band where d²E/dk² < 0, meaning the electron accelerates opposite to the force because the lattice pushes back more than the field pushes forward. In a real crystal m* is a tensor. Near the top of a nearly full band the collective behaviour of the electrons is described more simply by the missing electron, the hole, which has positive charge and positive effective mass −m*e.

(b)(ii) Effective mass of the hole

E(k) = −1 × 10⁻²⁶ k² erg (k in cm⁻¹), so d²E/dk² = −2 × 10⁻²⁶ erg cm², and for an electron in this band

m*e = ħ²/(d²E/dk²) = (1.055 × 10⁻²⁷)²/(−2 × 10⁻²⁶) = 1.113 × 10⁻⁵⁴/(−2 × 10⁻²⁶) = −5.6 × 10⁻²⁹ g.

The electron near the top of the valence band has negative effective mass, as expected for a band curving downwards. Removing it from the orbital k = 10⁷ k̂ₓ cm⁻¹ leaves a hole with wave vector −10⁷ k̂ₓ cm⁻¹, positive charge and effective mass

m*h = −m*e = +5.6 × 10⁻²⁹ g ≈ 0.061 mₑ.

The sign is positive and the magnitude 5.6 × 10⁻²⁹ g, about six per cent of the free-electron mass; because the band is parabolic the value does not depend on which k the electron was removed from.

(c) Frequencies at which the gain falls to 80

For an RC-coupled amplifier the gain at low frequency is A = Am/√(1 + (f₁/f)²) and at high frequency A = Am/√(1 + (f/f₂)²), where f₁ = 20 Hz and f₂ = 50 kHz are the lower and upper half-power (cut-off) frequencies. Gain 80 means A/Am = 0.8, so √(1 + x²) = 1.25, x² = 0.5625, x = 0.75.

Low-frequency side: f₁/f = 0.75, f = 20/0.75 = 26.7 Hz.

High-frequency side: f/f₂ = 0.75, f = 0.75 × 50 kHz = 37.5 kHz.

The gain is 80 or more (within 2 dB of the mid-band value) between 26.7 Hz and 37.5 kHz; outside this band it falls further, reaching 70.7 at the two cut-off frequencies.

How this answer will be evaluated

Framework: Principle > Setup and diagram > Derivation > Result and limiting case

What earns full marks: Complete derivations, correct units, physical interpretation, all steps shown

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a)(i) (a)(i) Origin of paramagnetism; temperature dependence of paramagnetic susceptibility. (8) 8 marks

    discuss— intro → 3-4 dimensions → example → balanced close

    Must cover

    • Permanent moments from unpaired electrons (spin + orbital); μ = gμ_B√(J(J+1))
    • Competition between alignment (−μ·B) and thermal agitation
    • Langevin/Brillouin: χ = Nμ₀μ²/3kT = C/T (Curie law); Curie–Weiss χ = C/(T − θ)
    • Exceptions: Pauli (temperature-independent) and Van Vleck paramagnetism

    Loses marks

    • Negative susceptibility for paramagnets

    Earns more

    • 1/χ vs T plot

    Extra mark

    • Saturation at high B/T
  2. (a)(ii) (a)(ii) Diamagnetic susceptibility of atomic hydrogen at STP with ⟨r²⟩ = 3a₀². (7) 7 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Langevin formula χ = −μ₀NZe²⟨r²⟩/6m
    • ⟨r²⟩ = 8.39×10⁻²¹ m²; N = 2.69×10²⁵ m⁻³ (Loschmidt)
    • χ ≈ −1.3×10⁻⁹ (SI volume susceptibility)
    • Molar value ≈ −2.4×10⁻⁶ cm³/mol (CGS); sign and smallness explained

    Loses marks

    • Positive sign

    Earns more

    • Derivation via Larmor precession

    Extra mark

    • Comparison with measured values of H₂
  3. (b)(i) (b)(i) Concept of effective mass of the electron. (10) 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Band electron under external force obeys ħ dk/dt = F; v_g = (1/ħ) dE/dk
    • 1/m* = (1/ħ²) d²E/dk²
    • m* depends on band curvature; negative near band top
    • Tensor nature; hole concept with positive mass

    Loses marks

    • Treating m* as the rest mass

    Earns more

    • E–k sketch

    Extra mark

    • Typical values in Si/GaAs
  4. (b)(ii) (b)(ii) E(k) = −1×10⁻²⁶ k² erg near valence-band edge; electron removed from k = 10⁷ kₓ cm⁻¹: sign and magnitude of hole effective mass. (10) 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • d²E/dk² = −2×10⁻²⁶ erg cm²
    • m*_e = ħ²/(d²E/dk²) ≈ −5.6×10⁻²⁹ g
    • m*_h = −m*_e = +5.6×10⁻²⁹ g ≈ 0.061 mₑ (positive)
    • Hole wave vector −10⁷ kₓ cm⁻¹; result independent of k for parabolic band

    Loses marks

    • Negative hole mass

    Earns more

    • CGS ħ = 1.055×10⁻²⁷ erg s

    Extra mark

    • Hole velocity equals electron velocity at k
  5. (c) (c) RC-coupled amplifier A_vm = 100, f₁ = 20 Hz, f₂ = 50 kHz: frequencies where gain falls to 80. (15) 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Low-frequency gain A = A_m/√(1 + (f₁/f)²)
    • High-frequency gain A = A_m/√(1 + (f/f₂)²)
    • A/A_m = 0.8 ⇒ ratio 0.75
    • f ≈ 26.7 Hz and 37.5 kHz

    Loses marks

    • Using 0.707 relation for gain 80

    Earns more

    • Frequency-response sketch

    Extra mark

    • dB interpretation (−1.94 dB)
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