Paper II — Q5
(a) An X-ray beam of wavelength (λ₁) undergoes a first order Bragg reflection at a Bragg angle of 30°. X-ray of wavelength 97 nm…
An X-ray beam of wavelength (λ₁) undergoes a first order Bragg reflection at a Bragg angle of 30°. X-ray of wavelength 97 nm undergoes 3rd order reflection at a Bragg angle of 60°. Consider that the two beams are reflected from the same set of planes. Find the value of λ₁. 10 marks
Using the expression for internal energy U=3N(ℏω)/(e^ℏω/k_BT-1), show that Einstein specific heat capacity is given by; C=3R((ℏω)/(k_BT))²(e^ℏω/k_BT)/((e^ℏω/k_BT-1)²). Also show that Einstein specific heat capacity given above is proportional to e⁻ℏω/k_BT at very low temperature. 10 marks
ρ^° and K^° mesons both decay mostly to π⁺ and π⁻. Explain why the mean lifetime of ρ^° is shorter (∼10⁻²³s) compared to the mean lifetime of K^°(∼10⁻¹⁰s). 10 marks
What are the properties of the particles made up of the following quarks ? (a) ud̄ (b) ūd (c) dds (d) uss 10 marks
What are chain reactions ? What do you mean by critical size of the core in which chain reaction takes place ? 10 marks
हिंदी में प्रश्न पढ़ें
λ₁ तरंग-दैर्घ्य का एक एक्सरे किरणपुंज 30° के ब्रैग-कोण पर पहले क्रम के ब्रैग परावर्तन से गुजरता है । 97 nm तरंग-दैर्घ्य का एक्सरे 60° के ब्रैग कोण पर तृतीय क्रम के परावर्तन से गुजरता है । मान लीजिए कि दोनों किरणपुंज (बीम) एक ही तल से परावर्तित होते हैं तो λ₁ का मान ज्ञात कीजिए । (10 अंक)
आंतरिक ऊर्जा के व्यंजक U=3N(ℏω)/(e^ℏω/k_BT-1) का इस्तेमाल करते हुए दिखाइए कि आइंस्टीन की विशिष्ट ऊष्मा धारिता निम्नलिखित द्वारा निर्धारित है; C=3R((ℏω)/(k_BT))²(e^ℏω/k_BT)/((e^ℏω/k_BT-1)²)। यह भी दिखाइए कि ऊपर दी गई आइंस्टीन की विशिष्ट ऊष्मा धारिता कम तापमान पर e⁻ℏω/k_BT के समानुपाती होती है । (10 अंक)
ρ^° और K^° मेसॉन दोनों ही अधिकतर π⁺ और π⁻ में क्षय होते हैं । समझाइए कि ρ^° का औसत जीवन काल (∼10⁻²³s) K^° के औसत जीवन काल (∼10⁻¹⁰s) की तुलना में छोटा क्यों है । (10 अंक)
निम्नलिखित क्वार्कों से बने हुए कणों के गुण क्या हैं ? (a) ud̄ (b) ūd (c) dds (d) uss (10 अंक)
श्रृंखला अभिक्रियाएँ क्या होती हैं ? कोर के क्रांतिक परिमाण से, जिसके भीतर श्रृंखला अभिक्रिया होती है, आप क्या अर्थ निकालते हैं ? (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) Bragg’s law: nλ = 2d sin θ, with the same d for both beams.
First beam: λ₁ = 2d sin 30° = d. Second beam: 3 × 97 nm = 2d sin 60° = d√3, so d = 97√3 nm.
Hence λ₁ = d = 97√3 nm ≈ 168 nm.
(b) Let x = ħω/(k_BT). Then U = 3Nħω/(e^x − 1) and C = (∂U/∂T)_V.
Since dx/dT = −x/T = −ħω/(k_BT²), dU/dT = 3Nħω × e^x/(e^x − 1)² × ħω/(k_BT²) = 3Nk_B (ħω/(k_BT))² e^x/(e^x − 1)².
For one mole, Nk_B = R, so C = 3R (ħω/k_BT)² e^(ħω/k_BT)/(e^(ħω/k_BT) − 1)².
At very low temperature, x ≫ 1, so e^x − 1 ≈ e^x and C ≈ 3R x² e^(−x) = 3R (ħω/k_BT)² e^(−ħω/k_BT). Thus the exponential factor dominates and C ∝ e^(−ħω/k_BT); C falls to zero as T → 0, in agreement with the third law.
(c) ρ⁰ ≈ (uū − dd̄)/√2 has strangeness 0. The decay ρ⁰ → π⁺π⁻ conserves charge, baryon number, strangeness and isospin, so it proceeds by the strong interaction, whose characteristic time is about 10⁻²³ s.
K⁰ (d s̄) has strangeness S = +1, while π⁺π⁻ has S = 0. The decay K⁰ → π⁺π⁻ therefore changes strangeness by one unit (ΔS = 1). Strong and electromagnetic interactions conserve strangeness, so the decay must go through the weak interaction, which is far slower. Hence the lifetime is about 10⁻¹⁰ s (the short-lived K⁰ₛ component).
(d)
- u d̄: Q = 2/3 + 1/3 = +1, B = 0, S = 0. This is the π⁺ meson, J^P = 0⁻, I = 1, I₃ = +1, mass ≈ 139.6 MeV/c², lifetime ≈ 2.6 × 10⁻⁸ s (weak decay).
- ū d: Q = −1, B = 0, S = 0. This is the π⁻ meson, the antiparticle of π⁺, with the same J^P, mass and lifetime, and I₃ = −1.
- d d s: Q = −1/3 − 1/3 − 1/3 = −1, B = 1, S = −1. This is the Σ⁻ baryon, J^P = 1/2⁺, I = 1, I₃ = −1, mass ≈ 1197 MeV/c², lifetime ≈ 1.5 × 10⁻¹⁰ s (weak decay).
- u s s: Q = 2/3 − 1/3 − 1/3 = 0, B = 1, S = −2. This is the Ξ⁰ baryon, J^P = 1/2⁺, I = 1/2, I₃ = +1/2, mass ≈ 1315 MeV/c², lifetime ≈ 2.9 × 10⁻¹⁰ s (weak decay).
(e) A chain reaction is a self-sustaining series of fissions in which neutrons released by one fission induce further fissions. For example, a slow neutron absorbed by ²³⁵U gives fission fragments, 2–3 neutrons and about 200 MeV.
The multiplication factor k is the ratio of fission-causing neutrons in one generation to those in the previous generation. If k < 1 the reaction dies out (subcritical), if k = 1 it is steady (critical), and if k > 1 it grows (supercritical).
The critical size is the minimum size of the fissile core for which k = 1. In a smaller core, neutrons leak through the surface and are lost to parasitic absorption faster than they are produced, so the chain cannot be sustained. The critical mass is M_c = ρV_c. For a bare homogeneous sphere in one-speed diffusion theory, R_c = π√[D/(νΣ_f − Σ_a)]. The critical size is lowered by a spherical shape, higher enrichment or density, and a neutron reflector.
How this answer will be evaluated
What earns full marks: Complete derivations with correct limits; precise particle identification; clear physical reasoning for lifetimes.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find λ1 using Bragg's law for two orders from the same planes. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- nλ = 2d sin θ with the same d
- d = 97√3 nm from the second beam
- λ1 = 97√3 ≈ 168 nm
Loses marks
- Treating n = 1 for both beams
- Arithmetic error in d
Earns more
- Correct use of sin 30° and sin 60°
- (b) Derive Einstein's specific heat and show C ∝ e^(−ħω/kT) at low T. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Differentiate U with respect to T using x = ħω/kT
- Result C = 3R x² e^x/(e^x−1)²
- Low T: e^x − 1 ≈ e^x gives C ∝ e^(−x)
Loses marks
- Sign or chain-rule error in dx/dT
- Missing factor x²
Earns more
- Use Nk_B = R for one mole
Extra mark
- Note C → 0 as T → 0
- (c) Explain why ρ0 lives far shorter than K0 though both decay to π+π−. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- ρ0 decay is strong, conserving strangeness; time about 1e-23 s
- K0 has S = +1, pions S = 0, so ΔS = 1
- Hence K0 decays weakly with lifetime about 1e-10 s
Loses marks
- Attributing the difference to mass alone
Earns more
- Quark content of ρ0 and K0
Extra mark
- Mention K0_S component
- (d) Identify the particles for the four quark combinations and give their properties. 10 marks
describe— define → structure or process in order → labelled diagram → significance
Must cover
- ud̄ = π+ and ūd = π−, Q = ±1, B = 0, S = 0
- dds = Σ−, Q = −1, B = 1, S = −1
- uss = Ξ0, Q = 0, B = 1, S = −2
Loses marks
- Wrong charge or strangeness
- Naming dds as Σ+
Earns more
- Spin-parity, isospin, mass or lifetime values
Extra mark
- Note weak decays for all four
- (e) Explain chain reaction and the critical size of the core. 10 marks
define— precise definition → the distinguishing feature → one example
Must cover
- Self-sustaining fission by released neutrons, e.g. U-235
- Multiplication factor k with k < 1, = 1, > 1 cases
- Critical size = minimum core size for k = 1; leakage argument
Loses marks
- Confusing critical size with critical mass without explanation
- Equating chain reaction with fusion
Earns more
- Critical mass and dependence on shape, enrichment, reflector
- Diffusion-theory expression for R_c
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
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