Physics 2025 Paper II 50 marks State

Paper II — Q4

(a) State how for spin-half particles, the spin (σ) can be expressed by its three components σ_x, σ_y and σ_z. 20 marks (b) By…

(a)

State how for spin-half particles, the spin (σ) can be expressed by its three components σ_x, σ_y and σ_z. 20 marks

(b)

By applying the Schrödinger's equation to the ground state of hydrogen atom, determine the zero-point energy. 15 marks

(c)

Distinguish between fluorescence and phosphorescence. Explain the mechanisms responsible for these phenomena. Discuss the applications of fluorescence and phosphorescence in the fields such as biochemistry, material science, etc. 15 marks

हिंदी में प्रश्न पढ़ें
(a)

बताइए कि किस प्रकार अर्ध-प्रचक्रण कणों के लिए प्रचक्रण (σ) को उसके तीन अवयवों σ_x, σ_y और σ_z द्वारा व्यक्त किया जा सकता है। 20 अंक

(b)

श्रोडिंगर समीकरण को हाइड्रोजन परमाणु की आध अवस्था के लिए प्रयुक्त करके शून्य-बिंदु ऊर्जा की गणना कीजिए। 15 अंक

(c)

प्रतिदीप्ति और स्कुरदीप्ति के बीच प्रभेद कीजिए। इन परिघटनाओं के लिए उत्तरदायी क्रियाविधियों की व्याख्या कीजिए। जैव रासायनिकी, पदार्थ विज्ञान इत्यादि जैसे क्षेत्रों में प्रतिदीप्ति और स्कुरदीप्ति के अनुप्रयोगों की चर्चा कीजिए। 15 अंक

Q4 of the 2025 UPSC Mains Physics Paper II, as printed
The question as printed in the 2025 Physics paper

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) Spin-half particles and Pauli matrices

For a spin-half particle the intrinsic spin is represented by the dimensionless Pauli vector σ = σₓ î + σ_y ĵ + σ_z k̂, where σₓ = [[0,1],[1,0]], σ_y = [[0,-i],[i,0]], σ_z = [[1,0],[0,-1]]. In the standard spinor basis, σ_z is diagonal, σₓ changes |↑> into |↓> and vice versa, and σ_y does the same with a relative phase i. The physical spin operator is S = (ℏ/2)σ, so Sᵢ=(ℏ/2)σᵢ. Each σᵢ has eigenvalues ±1, hence Sᵢ eigenvalues ±ℏ/2. The components satisfy [σᵢ,σⱼ]=2iε_ijkσₖ, σᵢ,σⱼ=2δ_ij I, and σᵢ²=I; equivalently σᵢσⱼ=δ_ij I+iε_ijkσₖ. Since they do not commute, only one component can be sharply defined at a time. Therefore [Sᵢ,Sⱼ]=iℏ ε_ijk Sₖ. For any unit direction n, n·σ = nₓσₓ+n_yσ_y+n_zσ_z is also Hermitian with eigenvalues ±1, so the three matrices encode all possible spin projections. The total spin magnitude is S²=Sₓ²+S_y²+S_z²=s(s+1)ℏ²=3ℏ²/4 for s=1/2. This is the spin-1/2 representation of SU(2), the double cover of the rotation group, and it is the complete statement of how spin is expressed through its three components.

(b) Hydrogen ground state and zero-point energy

For an electron of reduced mass μ in the Coulomb field of a proton, the time-independent Schrödinger equation is [-ℏ²/(2μ)∇² - e²/(4πε₀r)]ψ=Eψ. Separation in spherical coordinates gives angular spherical harmonics and a radial equation. For the ground state n=1, l=0, Y00 is constant and the radial equation reduces to -ℏ²/(2μ)(1/r² d/dr(r² dR/dr)) - e²/(4πε₀r)R=ER. Its solution is R(r)∝e^-r/a₀, with a₀=4πε₀ℏ²/(μe²); the normalized wavefunction is ψ100=(πa₀³)^-1/2e^-r/a₀. Substitution gives E₁=-μe⁴/(32π²ε₀²ℏ²)=-13.6 eV. Using μ≈mₑ gives the familiar 13.6 eV; the reduced-mass correction is small. This total ground-state energy is the zero-point energy of the hydrogen atom; its magnitude, 13.6 eV, is the binding or ionization energy. The expectation values obey the virial theorem: ⟨T⟩=+13.6 eV and ⟨V⟩=-27.2 eV, so ⟨T⟩+⟨V⟩=-13.6 eV. The positive kinetic term reflects confinement expressed by ΔrΔp≥ℏ/2; minimizing T~ℏ²/(2μa₀²) and V~-e²/(4πε₀a₀) gives the same a₀ and energy, showing the ground state is a balance of kinetic and potential terms. The zero-point energy is here taken as the total ground-state energy, −13.6 eV, the lowest energy the atom can have. Some texts instead call the kinetic energy that survives in the lowest state, ⟨T⟩ = +13.6 eV, the zero-point energy; either way, the electron cannot sit at rest at the nucleus, because confinement costs kinetic energy.

(c) Fluorescence and phosphorescence

In a Jablonski diagram, absorption promotes a molecule from the singlet ground state S₀ to an excited singlet S₁. The diagram thus contains both radiative transitions, S₁→S₀ and T₁→S₀, and non-radiative transitions, vibrational relaxation, internal conversion, and ISC. Fluorescence is prompt radiative decay S₁→S₀ with no change in spin multiplicity; it is spin-allowed, has lifetimes of about 10⁻⁹–10⁻⁷ s, and follows vibrational relaxation and internal conversion. The emitted fluorescence is usually at longer wavelength than the absorbed light because of this relaxation, producing a Stokes shift. Phosphorescence begins with intersystem crossing (ISC), a non-radiative S₁→T₁ transition mediated by spin-orbit coupling, followed by delayed radiative decay T₁→S₀. Because this is a singlet-triplet, spin-forbidden transition, phosphorescence has much longer lifetimes, typically 10⁻³–10³ s, and can persist after the excitation source is removed; the forbidden transition is made weakly allowed by spin-orbit coupling, especially in heavier atoms. Fluorescence is therefore useful for real-time imaging, while phosphorescence is useful where afterglow or long-lived emission is needed.

Applications follow these mechanisms. In biochemistry, fluorescence is used in green fluorescent protein (GFP) tagging, flow cytometry, DNA sequencing, and live-cell fluorescence microscopy. In materials science, phosphorescence is exploited in organic light-emitting diodes (OLEDs), persistent-luminescent (glow-in-the-dark) materials, safety signage, and security inks. Fluorescence resonance energy transfer (FRET) measures molecular distances of 1–10 nm, and phosphorescent iridium and platinum complexes allow OLEDs to harvest triplet excitons, raising internal quantum efficiency towards 100%.

How this answer will be evaluated

Framework: Principle > Setup and diagram > Derivation > Result and limiting case

What earns full marks: Complete derivations with all steps, correct units, physical interpretation, and relevant applications

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Derive the vector expression for spin in terms of Pauli matrices. 20 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Pauli matrices σ_x, σ_y, σ_z stated explicitly
    • S = (ℏ/2)σ; σ_i eigenvalues ±1 → S_i eigenvalues ±ℏ/2
    • Commutators [σ_i,σ_j] = 2iε_ijkσ_k; anticommutators {σ_i,σ_j} = 2δ_ij I
    • σ_i² = I; σ_iσ_j = δ_ij I + iε_ijkσ_k; n·σ eigenvalues ±1
    • S² = 3ℏ²/4; SU(2) spin-½ representation

    Loses marks

    • Confusing spin with orbital angular momentum
    • Missing the ħ/2 factor
    • Treating σ as a classical vector

    Earns more

    • Mention spinor representation
    • Link to SU(2) group
    • Show action on basis states |↑>, |↓>

    Extra mark

    • Mention spin-statistics theorem
  2. (b) Derive ground state energy of hydrogen using Schrödinger equation. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • TISE with reduced mass and Coulomb potential; separation of variables
    • Ground state n = 1, l = 0: R ∝ e^{−r/a₀}, a₀ = 4πε₀ℏ²/μe²; ψ₁₀₀
    • E₁ = −μe⁴/(32π²ε₀²ℏ²) ≈ −13.6 eV
    • Zero-point (ground-state) energy; virial theorem ⟨T⟩ = 13.6 eV, ⟨V⟩ = −27.2 eV
    • Kinetic-confinement balance (ΔrΔp ≥ ℏ/2) argument

    Loses marks

    • Using Bohr model without derivation
    • Missing boundary condition step
    • Incorrect sign for ground state energy

    Earns more

    • Show Bohr radius derivation
    • Discuss zero-point energy concept
    • Mention virial theorem check

    Extra mark

    • Compare with Bohr model result
  3. (c) Distinguish fluorescence and phosphorescence with mechanisms and applications. 15 marks

    compare— paired headings or point-wise contrast → key differences → significance → conclusion

    Must cover

    • Definitions: fluorescence S₁→S₀ spin-allowed; phosphorescence T₁→S₀ spin-forbidden
    • Jablonski processes: absorption, vibrational relaxation, internal conversion, ISC via spin-orbit coupling
    • Lifetimes ~10⁻⁹–10⁻⁷ s (ns) vs 10⁻³–10³ s; afterglow; Stokes shift
    • Applications: GFP tagging, flow cytometry, DNA sequencing, microscopy; OLEDs, persistent phosphors, safety signage, security inks

    Loses marks

    • Confusing emission mechanisms
    • No mention of spin states
    • Applications without field context

    Earns more

    • Draw Jablonski diagram
    • Mention triplet state mechanism
    • Discuss quantum yield differences

    Extra mark

    • Mention specific fluorescent dyes
    • Reference OLED technology
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