(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 be λ/2 ?
(ii) 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)
(b) 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)
(c) 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)
(d) If the position and momentum operators in 1-D are represented by x̂ and p̂ respectively, calculate the commutator [x̂², cos (p̂)]. (10 marks)
(e) 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 पढ़ें
(a) (i) तापक्रम 27°C पर तापीय संतुलन में एक कण की दे ब्रॉग्ली तरंगदैर्ध्य λ है। यह किस तापमान पर λ/2 हो जाएगा ?
(ii) 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)
(b) 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)
(c) 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)
(d) 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)
(e) 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)
Answer approach & key points
(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. (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. (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). (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. (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).
- λ = h/√(3mk_BT) ⇒ λ ∝ 1/√T
- T₂ = 4 × 300 K = 1200 K (927 °C)
- Norm ⟨χ|χ⟩ = 5 ⇒ P(↑) = 1/5 = 0.2
- |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)
- 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
- [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
- 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
(a) A particle is in the ground state of a 1-D simple harmonic oscillator potential, V(x) = 1/2 mω²x². Calculate the position uncertainty of the particle. (15 marks)
(b) (i) A particle is in a normalized state ψ which is a superposition of the energy eigenstates ψ₁ and ψ₂ of energies 10 eV and 30 eV respectively. The average energy of the particle in the state ψ is 24 eV. Find the state ψ in terms of ψ₁ and ψ₂. (10 marks)
(ii) Obtain the allowed energy eigenvalues of a half 1-D simple harmonic oscillator defined by the Hamiltonian
H = p²/2m + V(x), where
V(x) = 1/2 mω²x² for x ≥ 0
= ∞ for x < 0
Express its eigenstates in terms of the eigenstates of a complete 1-D oscillator. (10 marks)
(c) (i) Why does the electron paramagnetic resonance (EPR) spectroscopy utilize the microwave region of the electromagnetic spectrum, while nuclear magnetic resonance (NMR) uses radio waves ? (7 marks)
(ii) Why can the experimental observation of the Lamb shift not be explained by the Dirac theory ? (8 marks)
हिंदी में पढ़ें
(a) एक कण एक-विमीय (1-D) सरल आवर्त दोलक के विभव, V(x) = 1/2 mω²x² की आद्य (निम्नतम) अवस्था में है। कण की स्थिति अनिश्चितता की गणना कीजिए। (15 अंक)
(b) (i) एक कण प्रसामान्यीकृत अवस्था ψ में है जो क्रमशः 10 eV एवं 30 eV की ऊर्जा आइगेनअवस्थाओं ψ₁ एवं ψ₂ का अध्यारोपण है। ψ अवस्था में कण की औसत ऊर्जा का मान 24 eV है। ψ अवस्था को ψ₁ एवं ψ₂ के पदों में ज्ञात कीजिए। (10 अंक)
(ii) अर्ध एक-विमीय (1-D) सरल आवर्त दोलक की अनुमत आइगेनऊर्जाओं को प्राप्त कीजिए जो हैमिल्टोनियन
H = p²/2m + V(x), जहाँ
V(x) = 1/2 mω²x² x ≥ 0 के लिए
= ∞ x < 0 के लिए
से परिभाषित है। इसकी आइगेनअवस्थाओं को पूर्ण एक-विमीय (1-D) दोलक की आइगेनअवस्थाओं के पदों में व्यक्त कीजिए। (10 अंक)
(c) (i) इलेक्ट्रॉन अनुचुंबकीय अनुनाद (EPR) स्पेक्ट्रमिकी में विद्युत-चुंबकीय स्पेक्ट्रम के सूक्ष्म तरंग क्षेत्र का उपयोग क्यों होता है, जबकि नाभिक चुंबकीय अनुनाद (NMR) में रेडियो तरंगों का उपयोग होता है ? (7 अंक)
(ii) लैम्ब शिफ्ट के प्रयोगात्मक प्रेक्षण को डिरैक सिद्धांत द्वारा क्यों व्याख्यायित नहीं किया जा सकता है ? (8 अंक)
Answer approach & key points
(a) For the ground state of the 1-D harmonic oscillator V = ½mω²x², calculate the position uncertainty Δx = √(⟨x²⟩ − ⟨x⟩²). This requires writing the normalised ground-state Gaussian (or using ladder operators), evaluating ⟨x⟩ and ⟨x²⟩ explicitly, and giving the closed-form result. (b(i)) A normalised state ψ is a superposition of two energy eigenstates (10 eV and 30 eV) with mean energy 24 eV. Determine the expansion coefficients and write ψ explicitly in terms of ψ₁ and ψ₂. (b(ii)) Obtain the allowed energy eigenvalues of the half-oscillator (V = ½mω²x² for x ≥ 0, V = ∞ for x < 0) AND express its eigenstates in terms of the eigenstates of the full 1-D oscillator. Both parts must be answered; the key argument is the boundary condition ψ(0) = 0. (c(i)) Explain, with the resonance condition and the relevant magnitudes, WHY EPR resonances fall in the microwave region while NMR resonances fall in the radiofrequency region at laboratory magnetic fields. The answer hinges on the ratio of the electron magnetic moment to the nuclear magnetic moment. (c(ii)) Explain why the observed Lamb shift lies outside the Dirac theory: state what Dirac theory predicts for 2S₁/₂ and 2P₁/₂, what is actually measured, and identify the missing physics (quantisation of the radiation field, i.e. QED).
- Normalised ψ₀ = (mω/πħ)^{1/4} e^{−mωx²/2ħ}
- ⟨x⟩ = 0 by parity; ⟨x²⟩ = ħ/(2mω) from the Gaussian integral (or ladder operators)
- Δx = √(ħ/2mω)
- ⟨E⟩ = |c₁|²E₁ + |c₂|²E₂ with |c₁|² + |c₂|² = 1
- |c₁|² = 0.3, |c₂|² = 0.7
- ψ = √0.3 ψ₁ + √0.7 ψ₂ (up to a relative phase)
- ψ(0) = 0 from the infinite wall
- Only odd-n full-oscillator states allowed ⇒ E_k = (2k + 3/2)ħω, k = 0, 1, 2, …
- φ_k = √2 ψ_{2k+1} for x ≥ 0, 0 for x < 0
- Resonance hν = gμB for both
- μ_B/μ_N = m_p/m_e ≈ 1836 (nuclear moment far smaller)
- At lab fields: electron ≈ 28 GHz T⁻¹ (microwave, e.g. 9.5 GHz at 0.34 T); proton 42.58 MHz T⁻¹ (radio frequency)
- Dirac: energy depends on n and j only ⇒ 2S₁/₂ and 2P₁/₂ degenerate
- Lamb–Retherford (1947): 2S₁/₂ above 2P₁/₂ by ≈ 1057 MHz
- Missing physics is field quantisation (QED): self-energy from vacuum fluctuations raising s states, plus vacuum polarisation
Q3 50M calculate Quantum mechanics, NMR, hydrogen atom
(a) (i) Calculate the number of permitted electrons in a sub-shell and a shell in an atom. (ii) Explain how Pauli's exclusion principle helps in determining the electronic configuration in a many-electron system. (8+12 marks) (b) (i) Why is the relaxation process so important in nuclear magnetic resonance (NMR)? (ii) Why are 12C and 16C nuclei not suitable for the study of NMR? (10+5 marks) (c) (i) The wave function of a hydrogen atom is written as ψn, l, m (r, θ, φ). Explain the quantum numbers n, l, m and mention their ranges. Show that ψn, l, m (r, θ, φ) is n² degenerate. (ii) The electron in the hydrogen atom is found in the state ψ (r, θ, φ) = A R(r) sin θ cos θ e⁻ⁱφ. Find the z-component of angular momentum of the electron. (10+5 marks)
हिंदी में पढ़ें
(a) (i) एक परमाणु के एक उपकोश एवं कोश में अनुमत इलेक्ट्रॉनों की संख्या की गणना कीजिए। (ii) समझाइए कि पाउली का अपवर्जन सिद्धांत किस प्रकार एक बहु-इलेक्ट्रॉन निकाय में इलेक्ट्रॉनिक विन्यास को निर्धारित करने में सहायता करता है। (8+12 अंक) (b) (i) नाभिकीय चुम्बकीय अनुनाद (NMR) में विश्रांति प्रक्रिया इतनी महत्वपूर्ण क्यों है? (ii) 12C एवं 16C नाभिक एनएमआर (NMR) के अध्ययन के लिए उपयुक्त क्यों नहीं हैं? (10+5 अंक) (c) (i) एक हाइड्रोजन परमाणु के तरंग फलन को ψn, l, m (r, θ, φ) के रूप में लिखा जाता है। क्वांटम संख्याओं n, l, m की व्याख्या कीजिए एवं उनके परासों का उल्लेख कीजिए। दर्शाइए कि ψn, l, m (r, θ, φ), n² समप्रभ (डिजनरेट) है। (ii) हाइड्रोजन परमाणु में इलेक्ट्रॉन ψ (r, θ, φ) = A R(r) sin θ cos θ e⁻ⁱφ अवस्था में पाया जाता है। इलेक्ट्रॉन के कोणीय संवेग का z-घटक ज्ञात कीजिए। (10+5 अंक)
Answer approach & key points
Framework: Principle > Setup and diagram > Derivation > Result and limiting case. (a(i)) calculate: given > formula > substitution > result with units > interpretation | (a(ii)) explain: definition/context > points in order > small example > short close | (b(i)) explain: definition/context > points in order > small example > short close | (b(ii)) explain: definition/context > points in order > small example > short close | (c(i)) explain: definition/context > points in order > small example > short close | (c(ii)) calculate: given > formula > substitution > result with units > interpretation Full marks: Rigorous derivations, correct quantum number logic, clear NMR physics, and accurate calculation of Lz.
- 2l + 1 values of m_l, two spins each ⇒ 2(2l + 1) per sub-shell
- Σ_{l=0}^{n−1} 2(2l + 1) = 2n² per shell
- Basis: Pauli exclusion principle
- Statement with all four quantum numbers (or antisymmetry of the wavefunction)
- Limits: 2 per orbital, 2(2l + 1) per sub-shell, 2n² per shell ⇒ electrons fill successive levels (Aufbau)
- Hund's rule for partly filled sub-shells, with an example configuration
- Relaxation restores the Boltzmann population difference after excitation
- T₁ (spin–lattice) prevents saturation and sets the repetition rate
- T₂ (spin–spin) governs FID decay and line width Δν ≈ 1/πT₂
- NMR needs nuclear spin I ≠ 0 (magnetic moment)
- ¹²C and ¹⁶C are even–even nuclei with I = 0, so no Zeeman splitting
- n = 1, 2, …; l = 0 … n − 1; m = −l … +l, with physical meaning of each
- Energy depends only on n for hydrogen
- Σ_{l=0}^{n−1}(2l + 1) = n² ⇒ n²-fold degeneracy
- L̂_z = −iħ ∂/∂φ applied to e^{−iφ}
- L_z = −ħ (m = −1)
(a) (i) What are hot bands in vibrational spectroscopy ? Why are they called so ? (8 marks)
(ii) The fundamental and 2nd harmonic transitions of ¹⁴N¹⁶O are recorded at 1876·06 cm⁻¹ and 3724·2 cm⁻¹, respectively. Calculate the equilibrium vibrational frequency, the anharmonicity constant and the zero-point energy of the molecule. [Given : Mass of ¹⁴N = 23·25 × 10⁻²⁷ kg, Mass of ¹⁶O = 26·56 × 10⁻²⁷ kg] (7 marks)
(b) A particle of mass m is in the ground state of a 1-D box of infinite potential of length a. If the length of the box is changed to 2a symmetrically without disturbing the wave function of the particle, find the probability that the particle will be in the ground state of the new box. (20 marks)
(c) (i) Explain why the rotational transition for J = 0 → J = 1 is usually not of maximum intensity. (8 marks)
(ii) Determine the positions of the first four rotational Raman lines in the spectrum of hydrogen (H₂), given that its bond length is 0·742 Å. What effect will the nuclear spin have on the spectrum ? [Given : Mass of ¹H = 1·673 × 10⁻²⁷ kg] (7 marks)
हिंदी में पढ़ें
(a) (i) कंपनिक स्पेक्ट्रमिकी में तप्त बैंड्स (हॉट बैंड्स) क्या हैं ? इन्हें ऐसा क्यों कहा जाता है ? (8 अंक)
(ii) ¹⁴N¹⁶O के मूल एवं द्वितीय हार्मोनिक संक्रमणों को क्रमश: 1876·06 cm⁻¹ एवं 3724·2 cm⁻¹ पर अभिलेखित (रिकॉर्डिंग) किया जाता है । अणु की साम्यावस्था कंपन आवृत्ति, अप्रसंवादिता स्थिरांक एवं शून्य-बिंदु ऊर्जा की गणना कीजिए । [दिया गया है : ¹⁴N का द्रव्यमान = 23·25 × 10⁻²⁷ kg, ¹⁶O का द्रव्यमान = 26·56 × 10⁻²⁷ kg] (7 अंक)
(b) एक m द्रव्यमान का कण, अनंत विभव के 1-D बॉक्स, जिसकी लम्बाई a है, की आद्य (निम्नतम) अवस्था में है । यदि बॉक्स की लम्बाई, कण के तरंग फलन में विचलन के बिना, सममितीय रूप से 2a कर दी जाती है, तो नए बॉक्स की आद्य अवस्था में कण के होने की प्रायिकता ज्ञात कीजिए । (20 अंक)
(c) (i) समझाइए कि क्यों J = 0 → J = 1 के लिए घूर्णी संक्रमण प्राय: अधिकतम तीव्रता का नहीं होता है । (8 अंक)
(ii) हाइड्रोजन (H₂) के स्पेक्ट्रम में प्रथम चार घूर्णी रमन रेखाओं की स्थितियों को ज्ञात कीजिए, यदि उसके आबंध की लम्बाई 0·742 Å है । नाभिकीय प्रचक्रण का स्पेक्ट्रम पर क्या प्रभाव पड़ेगा ? [दिया गया है : ¹H का द्रव्यमान = 1·673 × 10⁻²⁷ kg] (7 अंक)
Answer approach & key points
(a(i)) Define hot bands in vibrational (IR/Raman) spectroscopy and explain the origin of the name — both halves must be answered, the second requiring the Boltzmann population argument. (a(ii)) From the fundamental (1876.06 cm⁻¹) and the '2nd harmonic' (first overtone, 3724.2 cm⁻¹) of ¹⁴N¹⁶O, calculate THREE quantities: the equilibrium vibrational frequency ω̃_e, the anharmonicity constant, and the zero-point energy of the molecule. (b) A particle sits in the ground state of an infinite well of width a; the well is suddenly widened SYMMETRICALLY to width 2a with the wave function unchanged. Compute the probability of finding the particle in the ground state of the new well — i.e. apply the sudden approximation and evaluate the overlap integral. (c(i)) Explain why the J = 0 → J = 1 line is usually NOT the most intense line in a rotational (microwave) spectrum: the intensity follows the thermal population of the lower level, which is the product of the degeneracy (2J+1) and the Boltzmann factor, and peaks at an intermediate J. (c(ii)) Two demands: (a) compute the positions (Raman shifts) of the first four rotational Raman lines of H₂ given r = 0.742 Å and m(¹H) = 1.673 × 10⁻²⁷ kg; (b) state the effect of nuclear spin (ortho/para hydrogen) on the spectrum.
- Transition from a thermally populated excited level (v = 1 → 2), not from v = 0
- Named 'hot' because intensity ∝ Boltzmann population of v = 1, rising with temperature
- Anharmonicity puts it ≈ 2ω̃_e x_e below the fundamental
- ω̃_e − 2ω̃_e x_e = 1876.06 and 2ω̃_e − 6ω̃_e x_e = 3724.2 (0 → 2 overtone)
- ω̃_e x_e = 13.96 cm⁻¹, x_e = 7.33 × 10⁻³
- ω̃_e = 1903.98 cm⁻¹
- ZPE = ½ω̃_e − ¼ω̃_e x_e = 948.5 cm⁻¹
- Sudden approximation: P = |⟨φ₁|ψ⟩|²
- ψ = √(2/a) cos(πx/a) on |x| ≤ a/2; φ₁ = √(1/a) cos(πx/2a) on |x| ≤ a
- Overlap integral over |x| ≤ a/2 only: c = 8/(3π) = 0.849
- P = 64/(9π²) ≈ 0.72
- Line intensity follows the population of the lower level J
- N_J ∝ (2J + 1) e^{−hcBJ(J + 1)/k_BT}: degeneracy rises, Boltzmann factor falls
- Maximum at J_max = √(k_BT/2hcB) − ½, not at J = 0
- μ = m_H/2, I = 4.61 × 10⁻⁴⁸ kg m², B = 60.8 cm⁻¹
- ΔJ = ±2, Δν̃ = B(4J + 6): 364.7, 607.8, 850.9, 1094.1 cm⁻¹
- Nuclear spin: ortho (odd J, weight 3) and para (even J, weight 1) ⇒ 1 : 3 intensity alternation, starting weak