(a) Find out the difference in frequencies of Lyman-alpha line in hydrogen and deuterium atoms. 15 marks (b) The Stern-Gerlach…

(a)

The Lyman-alpha transition is n = 2 → 1. For a hydrogen-like atom, isotope shift arises because the Rydberg constant depends on the reduced mass μ. ν = (3/4)cR∞ / (1 + mₑ/M), where M is the nuclear mass. Thus ν_H = (3/4)cR∞ / (1 + mₑ/M_p), ν_D = (3/4)cR∞ / (1 + mₑ/M_d). Therefore, Δν = ν_D − ν_H = (3/4)cR∞ [1/(1 + mₑ/M_d) − 1/(1 + mₑ/M_p)]. Using R∞ = 1.0973731568×10⁷ m⁻¹, c = 2.99792458×10⁸ m s⁻¹, mₑ/M_p = 1/1836.15 = 5.44617×10⁻⁴, mₑ/M_d = 1/3670.48 = 2.72444×10⁻⁴, the bracket is approximately 2.71951×10⁻⁴. Hence Δν = (3/4)(2.99792458×10⁸)(1.0973731568×10⁷)(2.71951×10⁻⁴) = 6.71×10¹¹ Hz. The deuterium Lyman-alpha frequency is higher than that of hydrogen by this amount.

(b)

The Stern-Gerlach experiment passed a beam of silver atoms through a strongly inhomogeneous magnetic field. Classically, randomly oriented atomic magnetic moments should give a continuous spread on the plate. Instead, the beam split into two distinct traces. This established space quantization: the component of angular momentum along the field direction takes discrete values.

For silver, the ground-state valence electron has zero orbital angular momentum, so the observed magnetic moment could not arise from orbital motion. The two beams correspond to the two spin orientations of the electron: m_s = +1/2 and m_s = −1/2. Thus the experiment gave direct evidence for electron spin and spin magnetic moment.

The force on an atom is F_z = μ_z ∂B_z/∂z = −m_s g_s μ_B ∂B_z/∂z. For an electron, g_s ≈ 2, so two equal and opposite deflections occur. The experiment also showed that only certain spin components can be measured. Sequential Stern-Gerlach experiments along different axes demonstrate that spin components along different directions do not commute, so they cannot be simultaneously definite. This became the basis of spinor quantum mechanics, Pauli matrices, quantum state preparation, and qubits.

(c)(i)

For a rigid diatomic rotor, rotational term is F(J) = B J(J+1), where B is the rotational constant in wavenumber units. Selection rule: ΔJ = ±1. Absorption line: ν̄(J → J+1) = 2B(J+1). So the difference between consecutive rotational lines is Δν̄ = 2B. Given Δν̄ = 4050 m⁻¹, (1) B = 4050/2 = 2025 m⁻¹ = 20.25 cm⁻¹.

(2) The moment of inertia is I = h/(8π²cB). Using h = 6.62607015×10⁻³⁴ J s, c = 2.99792458×10⁸ m s⁻¹, B = 2025 m⁻¹, I = 6.62607015×10⁻³⁴/(8π²×2.99792458×10⁸×2025) = 1.38236×10⁻⁴⁷ kg m².

(3) Reduced mass of HF: μ = M_H M_F/(M_H + M_F) = (1×19)/20 u = 0.95 u = 0.95×1.660539×10⁻²⁷ kg = 1.57751×10⁻²⁷ kg. Now I = μr², so r = √(I/μ) = √(1.38236×10⁻⁴⁷/1.57751×10⁻²⁷) = √(8.7629×10⁻²¹) = 9.361×10⁻¹¹ m = 0.9361 Å. Valid under rigid-rotor approximation, neglecting centrifugal distortion.

(c)(ii)

For a harmonic oscillator, vibrational wavenumber is ν̄_vib = (1/(2πc))√(k/μ). Given k = 4.8×10⁵ dyne/cm = 4.8×10⁵ g s⁻², μ = 1.61×10⁻²⁴ g, c = 3.0×10¹⁰ cm s⁻¹. k/μ = 4.8×10⁵/1.61×10⁻²⁴ = 2.98137×10²⁹ s⁻². √(k/μ) = 5.46019×10¹⁴ s⁻¹. Thus ν̄_vib = 5.46019×10¹⁴/(2π×3.0×10¹⁰) = 2.8967×10³ cm⁻¹.

Incident radiation: λ = 4358 Å = 4.358×10⁻⁵ cm. ν̄_ex = 1/λ = 1/(4.358×10⁻⁵) = 2.29463×10⁴ cm⁻¹.

Stokes line: ν̄_S = ν̄_ex − ν̄_vib = 2.29463×10⁴ − 2.8967×10³ = 2.00496×10⁴ cm⁻¹.

Anti-Stokes line: ν̄_A = ν̄_ex + ν̄_vib = 2.29463×10⁴ + 2.8967×10³ = 2.58430×10⁴ cm⁻¹. Result valid in harmonic approximation.