Paper II — Q1
(a) (i) The de Broglie wavelength of a particle in thermal equilibrium at a temperature of 27°C is λ. At what temperature will it…
The de Broglie wavelength of a particle in thermal equilibrium at a temperature of 27°C is λ. At what temperature will it be λ/2 ?
The wave function of a spin 1/2 particle is given by (−i, 2). What is the probability of finding the particle in the spin-up state ? 10 marks
The OH-radical has a moment of inertia of 1·48 × 10⁻⁴⁶ kg.m². For J = 6, calculate its angular velocity and angular momentum. Find the energy absorbed in the J = 6 → J = 7 transition. 10 marks
In a Stern-Gerlach experiment, the magnetic field varies with distance in the z-direction according to dB_z/dz = 1·5 T/mm. The speed of silver atoms coming from the oven is 725 m/s. Silver atoms travel a distance of 3 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 × 10⁻²⁵ kg and its magnetic moment is 1 Bohr magneton.] 10 marks
If the position and momentum operators in 1-D are represented by x̂ and p̂ respectively, calculate the commutator [x̂², cos (p̂)]. 10 marks
Find the eigenvalues and eigenfunctions of a system described by Hamiltonian H = ae^(bσ_x), where a, b are constants and σ_x is the x-component of Pauli matrix σ⃗. 10 marks
हिंदी में प्रश्न पढ़ें
तापक्रम 27°C पर तापीय संतुलन में एक कण की दे ब्रॉग्ली तरंगदैर्ध्य λ है। यह किस तापमान पर λ/2 हो जाएगा ?
एक प्रचक्रण 1/2 कण का तरंग फलन (−i, 2) द्वारा दिया गया है। कण के स्पिन-अप अवस्था में होने की प्रायिकता क्या है ? (10 अंक)
OH-मूलक (रैडिकल) का जड़त्व आघूर्ण 1·48 × 10⁻⁴⁶ kg.m² है। J = 6 स्तर के लिए, उसके कोणीय वेग एवं कोणीय संवेग की गणना कीजिए। J = 6 → J = 7 संक्रमण में अवशोषित ऊर्जा ज्ञात कीजिए। (10 अंक)
एक स्टर्न-गर्लैक प्रयोग में, चुम्बकीय क्षेत्र z-दिशा में दूरी के साथ dB_z/dz = 1·5 T/mm के अनुसार विचरित होता है। अवन से आने वाले सिल्वर परमाणुओं की गति 725 m/s है। सिल्वर के परमाणु 3 cm की दूरी चुम्बक से होकर तय करते हैं। जब परमाणुओं के दो पुंज चुम्बक को छोड़ते हैं, तब उनके पृथक्करण की गणना कीजिए। [दिया गया है : सिल्वर परमाणु का द्रव्यमान = 1·8 × 10⁻²⁵ kg और उसका चुम्बकीय आघूर्ण 1 बोर मैग्नेटॉन है।] (10 अंक)
यदि एक-विमीय (1-D) स्थिति एवं संवेग संकारकों को क्रमशः x̂ एवं p̂ से निरूपित किया जाता है, तो कम्यूटेटर [x̂², cos (p̂)] की गणना कीजिए। (10 अंक)
हैमिल्टोनियन H = ae^(bσ_x) द्वारा वर्णित एक निकाय के आइगेनमान एवं आइगेनफलन ज्ञात कीजिए, जहाँ a, b स्थिरांक हैं एवं σ_x पाउली मैट्रिक्स σ⃗ का x-घटक है। (10 अंक)
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) Temperature for half the de Broglie wavelength For a particle in thermal equilibrium, p = √(3mk_BT) (from ½mv²_rms = (3/2)k_BT), so λ = h/p = h/√(3mk_BT) ⇒ λ ∝ 1/√T λ₁/λ₂ = √(T₂/T₁) ⇒ 2 = √(T₂/300 K) ⇒ T₂ = 4 × 300 = 1200 K (927 °C)
(a) (ii) Probability of spin-up χ = (−i, 2)ᵀ in the (↑, ↓) basis is not normalised: ⟨χ|χ⟩ = |−i|² + |2|² = 1 + 4 = 5. Normalised χ = (1/√5)(−i, 2)ᵀ, so by the Born rule P(↑) = |−i|²/5 = 1/5 = 0.2 (and P(↓) = 4/5).
(b) Rotation of the OH radical Rigid rotor: E_J = J(J + 1)ħ²/2I, |L| = √(J(J + 1)) ħ, L = Iω. For J = 6: L = √42 ħ = 6.481 × 1.055 × 10⁻³⁴ = 6.83 × 10⁻³⁴ J s ω = L/I = 6.83 × 10⁻³⁴/1.48 × 10⁻⁴⁶ = 4.62 × 10¹² rad s⁻¹ Energy absorbed, J = 6 → 7 (ΔJ = +1): ΔE = (ħ²/2I)(7 × 8 − 6 × 7) = 14ħ²/2I = 7ħ²/I ΔE = 7 × (1.055 × 10⁻³⁴)²/1.48 × 10⁻⁴⁶ = 5.26 × 10⁻²² J ≈ 3.28 × 10⁻³ eV (a far-infrared photon, ν ≈ 7.9 × 10¹¹ Hz).
(c) Beam separation in the Stern–Gerlach experiment Force on an atom: F_z = ±μ_B dB_z/dz = 9.274 × 10⁻²⁴ × 1.5 × 10³ = 1.39 × 10⁻²⁰ N Acceleration: a = F/m = 1.39 × 10⁻²⁰/1.8 × 10⁻²⁵ = 7.73 × 10⁴ m s⁻² Time in the magnet: t = L/v = 0.03/725 = 4.14 × 10⁻⁵ s Deflection of each beam: z = ½at² = ½ × 7.73 × 10⁴ × (4.14 × 10⁻⁵)² = 6.6 × 10⁻⁵ m The two beams deflect in opposite directions, so the separation is Δz = 2z = aL²/v² = 1.32 × 10⁻⁴ m ≈ 0.13 mm
(d) Commutator [x̂², cos p̂] Use [x̂, p̂] = iħ, the identity [x̂, f(p̂)] = iħ f′(p̂), and [ÂB̂, Ĉ] = Â[B̂, Ĉ] + [Â, Ĉ]B̂. [x̂, cos p̂] = −iħ sin p̂, [x̂, sin p̂] = iħ cos p̂ [x̂², cos p̂] = x̂[x̂, cos p̂] + [x̂, cos p̂]x̂ = −iħ(x̂ sin p̂ + sin p̂ x̂) Since sin p̂ x̂ = x̂ sin p̂ − iħ cos p̂: [x̂², cos p̂] = −iħ(x̂ sin p̂ + sin p̂ x̂) = −2iħ x̂ sin p̂ − ħ² cos p̂ (Check in momentum space, x̂ = iħ d/dp: acting on g(p) both sides give ħ²(cos p) g + 2ħ²(sin p) g′.)
(e) Eigenvalues and eigenfunctions of H = a e^bσₓ σₓ = [[0, 1], [1, 0]], σₓ² = I, so the exponential series separates into even and odd terms: e^bσₓ = I cosh b + σₓ sinh b ⇒ H = a [[cosh b, sinh b], [sinh b, cosh b]] H is a function of σₓ, so it has the eigenvectors of σₓ (eigenvalues ±1): χ₊ = (1/√2)(1, 1)ᵀ: Hχ₊ = a(cosh b + sinh b)χ₊ ⇒ E₊ = a e^b χ₋ = (1/√2)(1, −1)ᵀ: Hχ₋ = a(cosh b − sinh b)χ₋ ⇒ E₋ = a e^−b
How this answer will be evaluated
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Two independent 5-mark numericals: (i) use the thermal de Broglie wavelength λ = h/√(3mkT) to find the temperature at which λ halves, starting from 27 °C; (ii) normalise the unnormalised spinor (−i, 2) and compute the probability of a spin-up measurement. 10 marks
Must cover
- λ = h/√(3mk_BT) ⇒ λ ∝ 1/√T
- T₂ = 4 × 300 K = 1200 K (927 °C)
- Norm ⟨χ|χ⟩ = 5 ⇒ P(↑) = 1/5 = 0.2
Loses marks
- Scaling T in °C (108 °C)
- Using λ ∝ 1/T (600 K)
- Treating (−i, 2) as normalised or taking |−i|² = −1
Earns more
- Writes the normalised spinor (1/√5)(−i, 2)ᵀ and P(↓) = 4/5
- (b) A three-part rigid-rotor numerical on the OH radical with I = 1.48 × 10⁻⁴⁶ kg m²: compute (a) the angular momentum for J = 6, (b) the corresponding angular velocity, and (c) the energy absorbed in the J = 6 → J = 7 rotational transition. 10 marks
Must cover
- |L| = √(J(J + 1)) ħ = √42 ħ = 6.83 × 10⁻³⁴ J s
- ω = L/I = 4.62 × 10¹² rad s⁻¹
- ΔE = (ħ²/2I)[7·8 − 6·7] = 7ħ²/I = 5.26 × 10⁻²² J (3.28 × 10⁻³ eV)
Loses marks
- L = Jħ = 6ħ
- ΔE with factor J = 6 instead of J + 1 = 7
- Using h instead of ħ
Earns more
- E_J = J(J + 1)ħ²/2I and selection rule ΔJ = ±1 stated
- (c) Compute the beam separation in a Stern–Gerlach apparatus: silver atoms (m = 1.8 × 10⁻²⁵ kg, μ = 1 Bohr magneton) at v = 725 m s⁻¹ traverse 3 cm of a field gradient dB_z/dz = 1.5 T mm⁻¹. The answer required is the separation of the TWO beams at the exit of the magnet (not the deflection of one beam). 10 marks
Must cover
- F = μ_B dB_z/dz with 1.5 T/mm = 1500 T/m ⇒ 1.39 × 10⁻²⁰ N
- a = 7.73 × 10⁴ m s⁻², t = L/v = 4.14 × 10⁻⁵ s, z = ½at² = 6.6 × 10⁻⁵ m
- Separation 2z = 1.32 × 10⁻⁴ m
Loses marks
- Giving one beam's deflection as the separation
- No T/mm to T/m conversion
- Using z = at² or vt
Earns more
- Compact form Δz = (μ_B dB/dz)L²/(mv²)
- (d) Evaluate the operator commutator [x̂², cos p̂] in one dimension. The candidate must use the canonical relation [x̂, p̂] = iħ, the identity [x̂, f(p̂)] = iħ ∂f/∂p̂, and the product rule for commutators, and present the ordered (non-commuting) result. 10 marks
Must cover
- [x̂, f(p̂)] = iħ f′(p̂) ⇒ [x̂, cos p̂] = −iħ sin p̂
- Product rule [x̂², C] = x̂[x̂, C] + [x̂, C]x̂
- [x̂², cos p̂] = −iħ(x̂ sin p̂ + sin p̂ x̂) = −2iħ x̂ sin p̂ − ħ² cos p̂
Loses marks
- Sign error: [x̂, cos p̂] = +iħ sin p̂
- Answer −2iħ x̂ sin p̂ without the −ħ² cos p̂ term (unless written in the symmetric form)
- Concluding the commutator is zero
Earns more
- Proof of [x̂, f(p̂)] = iħ f′(p̂) via series or momentum representation
- Verification by acting on a test function
- (e) Find BOTH the eigenvalues and the eigenfunctions (eigenspinors) of H = a e^(b σ_x). The key idea is that any function of σ_x is diagonal in the eigenbasis of σ_x, so the eigenvectors are those of σ_x and the eigenvalues are a e^(±b). 10 marks
Must cover
- σ_x² = I ⇒ e^{bσ_x} = I cosh b + σ_x sinh b
- Eigenvalues E± = a e^{±b}
- Eigenvectors (1/√2)(1, ±1)ᵀ, the eigenvectors of σ_x
Loses marks
- Eigenvalues ±a e^{b}
- Using σ_z eigenvectors (1, 0), (0, 1)
- Giving eigenvalues without eigenfunctions
Earns more
- Explicit matrix H = a[[cosh b, sinh b], [sinh b, cosh b]] and direct verification
What "State" is asking you to do
Give the formulation itself — the theorem, rule, statutory provision or position — worded accurately. Precision of wording is the whole of the mark; no background or justification is being asked for.
Structure that answers it
The statement in full, complete in itself → the conditions under which it holds → an illustration only where the stem asks for one
Where marks are lost
Approximating the wording. A theorem or a provision stated loosely forfeits the mark that exact statement would have carried.
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