Physics 2024 Paper II 50 marks Explain

Q6

(a) Does the nucleus possess magnetic moment ? Justify your answer. Define nuclear magneton (μN) and Bohr magneton (μB). Calculate their values. 7+8=15 (b) (i) Write semi-empirical mass formula. Calculate the atomic number (Z) of most stable nucleus for given mass number (A) using the above formula. (Use the value of fitted coefficients for Coulomb energy a3 = 0·711 MeV and that for asymmetry energy a4 = 23·702 MeV). (ii) Calculate the Q-value of the following nuclear reaction : 4Be9 + 2He4 = 6C12 + 0n1 Given : the mass of neutral atoms of Be, He and C are 9·015060, 4·003874 and 12·003815 amu, respectively. The mass of neutron is 1·008986 amu. 15+5=20 (c) What are the various conservation laws for elementary particles ? Apply these conservation laws to confirm whether the following reactions are possible or not : (i) π+ + n0 → K0 + K+ (ii) ν̄e + p+ → n0 + e− 15

हिंदी में प्रश्न पढ़ें

(a) क्या नाभिक का चुंबकीय आघूर्ण होता है ? अपने उत्तर का औचित्य दीजिए । नाभिकीय मैग्नेटॉन (μN) और बोर मैग्नेटॉन (μB) को परिभाषित कीजिए । इनके मानों की गणना कीजिए । 7+8=15 (b) (i) अर्ध-अनुभविक द्रव्यमान सूत्र लिखिए । उपर्युक्त सूत्र का उपयोग करके दी गई द्रव्यमान संख्या (A) के लिए सबसे स्थिर नाभिक के परमाणु क्रमांक (Z) की गणना कीजिए । (आसंजित गुणांकों के मान कूलॉम ऊर्जा के लिए a3 = 0·711 MeV और असममिति ऊर्जा के लिए a4 = 23·702 MeV का उपयोग कीजिए) (ii) निम्नलिखित नाभिकीय अभिक्रिया के Q-मान की गणना कीजिए : 4Be9 + 2He4 = 6C12 + 0n1 दिया गया है : Be का अनाविष्ट परमाणु द्रव्यमान = 9·015060 amu, He का अनाविष्ट परमाणु द्रव्यमान = 4·003874 amu और C का अनाविष्ट परमाणु द्रव्यमान = 12·003815 amu. न्यूट्रॉन का द्रव्यमान = 1·008986 amu है । 15+5=20 (c) मूल कणों के लिए विभिन्न संरक्षण नियम क्या हैं ? उन संरक्षण नियमों को लागू कर सत्यापित कीजिए कि क्या निम्नलिखित अभिक्रियाएँ संभव हैं या नहीं : (i) π+ + n0 → K0 + K+ (ii) ν̄e + p+ → n0 + e− 15

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How this answer will be evaluated

Approach

Explain the nuclear magnetic moment origin and define both magnetons with calculations in part (a) (~15 marks, 30% time). For part (b), derive the semi-empirical mass formula, optimize for Z using given coefficients, then calculate Q-value with proper mass-energy conversion (~20 marks, 40% time). Conclude with part (c) enumerating conservation laws and applying them systematically to verify both reactions (~15 marks, 30% time). Structure: direct definitions → derivations → numerical work → conservation analysis.

Key points expected

  • Part (a): Nuclear magnetic moment arises from unpaired nucleons (proton spin + orbital motion, neutron spin only); justification via Schmidt limits or odd-A nuclei data; definitions μN = eℏ/2mp and μB = eℏ/2me with calculated values ~5.05×10⁻²⁷ J/T and ~9.27×10⁻²⁴ J/T
  • Part (b)(i): Semi-empirical mass formula with volume, surface, Coulomb, asymmetry and pairing terms; derivation of Zmin = A/[2 + (2a3A²ᐟ³)/a4] using given a3=0.711 MeV, a4=23.702 MeV
  • Part (b)(ii): Q-value calculation using atomic masses with electron cancellation: Q = [m(Be) + m(He) - m(C) - m(n)] × 931.5 MeV/u, proper handling of neutral atom masses
  • Part (c): Conservation laws—energy-momentum, charge, baryon number, lepton number (separate families), strangeness, parity (strong/EM), isospin (strong); systematic table for each reaction
  • Reaction (i) analysis: π⁺(Q=+1, S=0, B=0) + n(Q=0, S=0, B=1) → K⁰(Q=0, S=+1, B=0) + K⁺(Q=+1, S=+1, B=0); check charge OK, baryon OK, but ΔS=+2 violates strong interaction strangeness conservation → forbidden or weak interaction only
  • Reaction (ii) analysis: ν̄e(Q=0, Le=−1) + p(Q=+1, B=1) → n(Q=0, B=1) + e⁻(Q=−1, Le=+1); charge conserved, baryon conserved, lepton: −1+0 → 0+1 gives ΔLe=0; this is inverse beta decay, allowed via weak interaction

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Precisely identifies nuclear magnetic moment sources (proton orbital + spin, neutron spin only), correctly states both magneton definitions with proper mass dependence (mp vs me), and accurately enumerates all five conservation laws with their applicability domains (strong/EM/weak)Identifies that nuclei have magnetic moments and gives basic definitions, but confuses proton/neutron contributions or omits one conservation law; minor errors in stating which laws apply to which interactionIncorrectly attributes nuclear magnetism to electrons, swaps μN and μB definitions, or misses 2+ conservation laws; fundamental misunderstanding of reaction classification
Derivation rigour20%10Complete SEMF with all five terms clearly written; explicit differentiation ∂M/∂Z=0 leading to Zmin formula with proper algebraic steps; systematic conservation law application with quantum numbers tabulated for both reactionsWrites SEMF with most terms but skips derivation steps for Zmin or makes minor algebraic errors; applies conservation laws verbally without systematic tabulationWrites only volume and Coulomb terms in SEMF, no derivation attempt, or applies conservation laws inconsistently without checking all quantum numbers
Diagram / FBD10%5Clear diagram showing nucleon magnetic moment vectors (spin and orbital) for odd-A nucleus in (a); binding energy per nucleon vs A curve sketch with SEMF contributions labeled in (b); reaction diagrams showing particle flow and quantum number accounting in (c)Basic sketch of magnetic moment orientation or generic nuclear level diagram; minimal or no reaction diagramsNo diagrams where helpful, or completely incorrect sketches that misrepresent physical situations
Numerical accuracy25%12.5μN = 5.05078×10⁻²⁷ J/T (or 3.152×10⁻⁸ eV/T), μB = 9.274×10⁻²⁴ J/T; Zmin calculated correctly with given coefficients (~Z≈A/2 for light, reduced for heavy); Q-value = [9.015060+4.003874−12.003815−1.008986]×931.5 ≈ 5.7 MeV with proper sign and unitsCorrect order of magnitude for magnetons; Zmin formula correct but arithmetic error; Q-value calculation with wrong mass excess or unit conversion error (e.g., using amu directly as MeV)Wrong formulas leading to incorrect orders of magnitude; fundamental error like using electron mass for μN or omitting 931.5 conversion entirely
Physical interpretation25%12.5Explains why μN << μB (nuclear magneton smaller by mp/me ≈ 1836); interprets Zmin result as stability valley and neutron-rich trend for heavy nuclei; identifies reaction (i) as strongly forbidden (ΔS=2) and (ii) as weak-allowed inverse beta decay with physical context (neutrino detection, DAE's India-based Neutrino Observatory relevance)States that nuclear moments are smaller but without mass ratio explanation; notes stability trend qualitatively; identifies reactions as allowed/forbidden without interaction-type specificityNo physical interpretation of numerical results; cannot explain why reactions are allowed or forbidden beyond guessing; misses connection to weak vs strong interactions

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