Paper II — Q1
(a) Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400 nm wavelength when a…
Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400 nm wavelength when a spectrometer whose resolution is 0.010 nm is used. Write the answer in the nearest high integer. 10 marks
1.(b) Normalised wave function of hydrogen atom for 1s state is ψ₁₀₀ = (1/√(πa₀³))e⁻ʳ/ᵃ⁰, where a₀ = ℏ²/me² being the Bohr radius. Calculate the expectation value of potential energy in this state. 10 marks
1.(c) A beam of 12 eV electron is incident on a potential barrier of height 25 eV and width 0.05 nm. Calculate the transmission coefficient. 10 marks
1.(d) Calculate the Larmor precessional frequency for a magnetic induction field of 0.5 T. Hence calculate the splitting in wave numbers of a spectral line due to normal Zeeman effect for the same field. 10 marks
1.(e) The first line in the pure rotational spectrum of HCl appears at 21.18 cm⁻¹. Calculate bond length of the molecule. Given atomic masses of H and Cl are 1.008 and 35.45 amu, respectively. 10 marks
हिंदी में प्रश्न पढ़ें
400 nm तरंग-दैर्ध्य की स्पेक्ट्रम रेखा में देखे जाने के लिए जीमैन प्रभाव के लिए आवश्यक न्यूनतम चुंबकीय क्षेत्र का पता लगाएँ जब कि एक स्पेक्ट्रोमीटर जिसका विभेदन 0.010 nm है, का उपयोग किया जाता है। उत्तर को निकटतम उच्च पूर्णक में लिखें। (10 अंक)
1.(b) हाइड्रोजन परमाणु की 1s अवस्था में इलेक्ट्रॉन के लिए सामान्यीकृत तरंग फलन निम्नलिखित है: ψ₁₀₀ = (1/√(πa₀³))e⁻ʳ/ᵃ⁰, जहाँ a₀ = ℏ²/me² बोहर त्रिज्या है। इस अवस्था में स्थितिज ऊर्जा के अपेक्षित मान की गणना कीजिए। (10 अंक)
1.(c) 25 eV ऊँचाई और 0.05 nm चौड़ाई के विभव रोध पर 12 eV इलेक्ट्रॉन का एक किरण-पुंज आपतित होता है। संचरण गुणांक की गणना कीजिए। (10 अंक)
1.(d) 0.5 T के चुंबकीय प्रेरण क्षेत्र के लिए लार्मर पुत्रस्तरण आवृत्ति की गणना कीजिए। समान क्षेत्र के लिए सामान्य जीमैन प्रभाव के कारण स्पेक्ट्रम रेखाओं की तरंग संख्याओं में विभाजन की गणना कीजिए। (10 अंक)
1.(e) HCl के शुद्ध घूर्णी वर्णक्रम (स्पेक्ट्रम) में पहली पट्टी 21.18 cm⁻¹ पर दिखाई देती है। अणु की बंधन लंबाई की गणना कीजिए। हाइड्रोजन परमाणु का द्रव्यमान 1.008 और क्लोरिन परमाणु का द्रव्यमान 35.45 amu है जहाँ amu परमाणुिक द्रव्यमान इकाई 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) Minimum field for the Zeeman effect to be resolved at 400 nm
In the normal Zeeman effect each component of the line is shifted by Δν = eB/(4πm), which in wavelength is
Δλ = (λ²/c)Δν = eBλ²/(4πmc).
The splitting is observable only if it is at least the spectrometer's resolution, Δλ = 0.010 nm = 1.0×10⁻¹¹ m. Hence
B_min = 4πmcΔλ/(eλ²) = 4π × 9.109×10⁻³¹ × 2.998×10⁸ × 1.0×10⁻¹¹/(1.602×10⁻¹⁹ × (4.00×10⁻⁷)²) = 1.34 T.
To the nearest higher integer, B_min = 2 T.
(b) Expectation value of potential energy in the hydrogen 1s state
With V = −e²/r (Gaussian units, matching a₀ = ħ²/me²) and |ψ₁₀₀|² = e^(−2r/a₀)/(πa₀³):
⟨V⟩ = −e² ∫ (1/r)|ψ|² 4πr² dr = −(4e²/a₀³) ∫₀^∞ r e^(−2r/a₀) dr.
Using ∫₀^∞ r e^(−αr) dr = 1/α² with α = 2/a₀, the integral is a₀²/4, so
⟨V⟩ = −e²/a₀ = −me⁴/ħ² = −27.2 eV (= −4.36×10⁻¹⁸ J).
This is twice the binding energy, as the virial theorem requires (⟨K⟩ = +13.6 eV and E = −13.6 eV).
(c) Transmission through a 25 eV barrier
For E < V₀, the transmission coefficient of a rectangular barrier of width a is
T = 1/[1 + V₀² sinh²(κa)/(4E(V₀ − E))], κ = √(2m(V₀ − E))/ħ.
Here E = 12 eV, V₀ = 25 eV, V₀ − E = 13 eV and, taking the English text's width a = 0.05 nm:
κ = √(2 × 9.109×10⁻³¹ × 13 × 1.602×10⁻¹⁹)/1.055×10⁻³⁴ = 1.85×10¹⁰ m⁻¹, κa = 0.923.
sinh²(0.923) = 1.125 and V₀²/(4E(V₀ − E)) = 625/624 ≈ 1.002, so
T = 1/(1 + 1.002 × 1.125) = 1/2.127 ≈ 0.47.
(The Hindi text prints the width as 0.5 nm; then κa = 9.23 and T ≈ 4×10⁻⁸.)
(d) Larmor frequency and normal Zeeman splitting at 0.5 T
Larmor frequency: ν_L = eB/(4πm) = 1.602×10⁻¹⁹ × 0.5/(4π × 9.109×10⁻³¹) = 7.0×10⁹ Hz (angular, ω_L = eB/2m = 4.4×10¹⁰ rad/s).
In the normal Zeeman effect, the line splits into three components displaced by ±ν_L, so the shift in wave number is
Δν̃ = ν_L/c = eB/(4πmc) = 7.0×10⁹/(3×10¹⁰ cm/s) = 0.233 cm⁻¹ (about 0.467 cm⁻¹ per tesla).
The outer components are therefore 0.467 cm⁻¹ apart.
(e) Bond length of HCl from the first rotational line
For a rigid rotor, the J = 0 → 1 line lies at ν̃ = 2B, so B = 21.18/2 = 10.59 cm⁻¹. With B = h/(8π²cI):
I = 6.626×10⁻³⁴/(8π² × 2.998×10¹⁰ cm/s × 10.59 cm⁻¹) = 2.64×10⁻⁴⁷ kg m².
Reduced mass μ = (1.008 × 35.45)/(36.458) = 0.980 amu = 1.628×10⁻²⁷ kg.
r = √(I/μ) = √(2.64×10⁻⁴⁷/1.628×10⁻²⁷) = 1.27×10⁻¹⁰ m = 1.27 Å.
How this answer will be evaluated
Framework: Principle > Setup and diagram > Derivation > Result and limiting case
What earns full marks: Complete derivations with correct units, physical interpretation, and accurate numerical results.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find the minimum magnetic field for the Zeeman splitting at 400 nm to be resolved by a 0.010 nm spectrometer. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Δλ = eBλ²/(4πmc) for the normal Zeeman shift
- Set Δλ = 0.010 nm and solve B_min = 4πmcΔλ/(eλ²)
- B_min = 1.34 T, so 2 T to the nearest higher integer
Loses marks
- Not rounding up as asked
- Using total splitting without stating it
Earns more
- Correct unit conversion of λ and Δλ
- Δν = eB/4πm
Extra mark
- Cross-check via wave number 0.467 cm⁻¹/T
- (b) Calculate the expectation value of potential energy in the hydrogen 1s state. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- ⟨V⟩ = -e²∫(1/r)|ψ|²d³r with |ψ|² = e^(-2r/a₀)/(πa₀³)
- Evaluates ∫r e^(-2r/a₀)dr = a₀²/4
- ⟨V⟩ = -e²/a₀ = -27.2 eV
Loses marks
- Giving -13.6 eV
- Missing the r² in the volume element
Earns more
- Angular integral 4π
- Link e²/a₀ = 2 × 13.6 eV
Extra mark
- Virial theorem consistency
- (c) Calculate the transmission coefficient of a 12 eV electron through a 25 eV, 0.05 nm barrier. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- κ = √(2m(V₀-E))/ħ = 1.85e10 m⁻¹, κa = 0.923
- T = 1/[1 + V₀²sinh²(κa)/(4E(V₀-E))]
- T ≈ 0.47
Loses marks
- Using the thick-barrier approximation for κa < 1
- Mixing eV and joules in κ
Earns more
- sinh²(κa) = 1.125
- States the width taken
Extra mark
- Notes the Hindi text's 0.5 nm giving T ≈ 4e-8
- (d) Find the Larmor frequency at 0.5 T and the normal Zeeman splitting in wave numbers. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- ν_L = eB/(4πm) = 7.0 GHz
- Normal Zeeman components at ±ν_L
- Δν̃ = ν_L/c = 0.233 cm⁻¹
Loses marks
- Using ω instead of ν
- Wrong power of ten in cm⁻¹
Earns more
- Larmor angular frequency ω_L = eB/2m
- Three-component pattern
Extra mark
- Notes 0.467 cm⁻¹ between outer components
- (e) Calculate the bond length of HCl from the first rotational line at 21.18 cm⁻¹. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- First line J = 0 → 1 at 2B, so B = 10.59 cm⁻¹
- I = h/(8π²cB) = 2.64e-47 kg m²
- μ = 0.980 amu; r = √(I/μ) ≈ 1.27 Å
Loses marks
- Using the atomic mass of Cl alone instead of reduced mass
- Taking B = 21.18 cm⁻¹
Earns more
- Reduced mass computed
- Correct unit use of c in cm/s
Extra mark
- Notes the rigid-rotor assumption
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
Evaluate my answer →More from Physics 2021 Paper II
- Q1 (a) Find the minimum magnetic field needed for the Zeeman effect to be observed in a spec…
- Q2 (a) Using Pauli spin matrices prove that, (i) σₓσᵧ + σᵧσₓ = 0; σᵧσᵤ + σᵤσᵧ = 0; σₓσᵤ + σᵤ…
- Q3 (a) In observing the Raman spectrum of a sample using 3637Å as the exciting line, one get…
- Q4 (a) A particle of rest mass m₀ has a kinetic energy K, show that its de Broglie wavelengt…