(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

(b) Normalised w

Question

(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

(b) Normalised wave function of hydrogen atom for 1s state is
ψ₁₀₀ = 1/(√(π a₀³)) e^-r/a₀, where a₀ = (ℏ²)/(me²)
being the Bohr radius. Calculate the expectation value of potential energy in this state. 10 marks

(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

(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

(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

UPSC Answer Check · Accepted Answer

(a) For the normal Zeeman effect, the energy of a magnetic substate is ΔE = m_l μ_B B, where μ_B = eħ/(2m). For an allowed transition with Δm_l = ±1, the frequency displacement is

Δν = ΔE/h = eB/(4πm).

Using λ = c/ν, the corresponding wavelength displacement is

Δλ = (λ²/c) Δν = eλ²B/(4πmc).

The spectrometer can just resolve a wavelength difference equal to its resolution, so the minimum field satisfies

Δλ = 0.010 nm = 1.0×10⁻¹¹ m.

Thus

B_min = 4πmc Δλ/(eλ²).

Substitute m = 9.109×10⁻³¹ kg, c = 2.998×10⁸ m s⁻¹, e = 1.602×10⁻¹⁹ C, λ = 400 nm = 4.00×10⁻⁷ m:

B_min = [4π(9.109×10⁻³¹)(2.998×10⁸)(1.0×10⁻¹¹)] / [(1.602×10⁻¹⁹)(4.00×10⁻⁷)²]
= 3.43×10⁻³² / 2.56×10⁻³² T
= 1.34 T.

Since the question asks for the nearest higher integer, the minimum required field is

B_min ≈ 2 T.

(b) The potential energy of the electron in the hydrogen atom is V(r) = −e²/r, in the units implicit in a₀ = ħ²/(m e²). Therefore

⟨V⟩ = ∫ ψ₁₀₀* (−e²/r) ψ₁₀₀ dτ.

With

ψ₁₀₀ = (πa₀³)^(−1/2) e^(−r/a₀),

we have

|ψ₁₀₀|² = (1/(πa₀³)) e^(−2r/a₀).

The volume element is dτ = r² sinθ dr dθ dφ, and the angular integral gives 4π. Hence

⟨V⟩ = −(4πe²/(πa₀³)) ∫₀∞ e^(−2r/a₀) r dr
= −(4e²/a₀³) ∫₀∞ r e^(−2r/a₀) dr.

Use

∫₀∞ r e^(−αr) dr = 1/α².

Here α = 2/a₀, so the integral equals a₀²/4. Therefore

⟨V⟩ = −(4e²/a₀³)(a₀²/4) = −e²/a₀.

Using a₀ = ħ²/(m e²),

e²/a₀ = m e⁴/ħ² = 2 × 13.6 eV = 27.2 eV.

Thus

⟨V⟩ = −27.2 eV = −4.36×10⁻¹⁸ J.

(c) For a rectangular potential barrier of height V₀ and width a, with E < V₀, define

κ = √(2m(V₀ − E))/ħ.

The exact transmission coefficient is

T = 1 / [1 + (V₀²/(4E(V₀ − E))) sinh²(κa)].

Here

E = 12 eV, V₀ = 25 eV, V₀ − E = 13 eV = 13 × 1.602×10⁻¹⁹ J = 2.083×10⁻¹⁸ J,

a = 0.05 nm = 5.0×10⁻¹¹ m.

Therefore

κ = √(2 × 9.109×10⁻³¹ × 2.083×10⁻¹⁸) / (1.055×10⁻³⁴)
= 1.847×10¹⁰ m⁻¹.

Then

κa = 1.847×10¹⁰ × 5.0×10⁻¹¹ = 0.9235.

So

sinh(0.9235) = 1.0605,

and

sinh²(κa) = 1.1247.

The prefactor is

V₀²/[4E(V₀ − E)] = 25²/[4 × 12 × 13] = 625/624 = 1.0016.

Hence

T = 1 / [1 + 1.0016 × 1.1247]
= 1 / 2.1265
= 0.470.

Thus

T ≈ 0.470 = 47.0%.

This is valid for a rectangular barrier, non-relativistic electrons, and no image-potential corrections.

(d)(i) The Larmor angular frequency for an electron in a magnetic field B is

ω_L = eB/(2m).

The ordinary precessional frequency is therefore

ν_L = ω_L/(2π) = eB/(4πm).

For B = 0.5 T,

ν_L = (1.602×10⁻¹⁹ × 0.5) / (4π × 9.109×10⁻³¹)
= 6.998×10⁹ s⁻¹
≈ 7.0×10⁹ Hz
= 7.0 GHz.

So

ν_L ≈ 7.0×10⁹ Hz = 7.0 GHz.

(d)(ii) In the normal Zeeman effect, the energy shift is

ΔE = μ_B B = eħB/(2m).

The corresponding frequency shift is

Δν = ΔE/h = eB/(4πm),

which is the same as the Larmor precessional frequency. The splitting in wave numbers is

Δν~ = Δν/c = eB/(4πmc).

Using c = 2.998×10¹⁰ cm s⁻¹,

e/(4πmc) = (1.602×10⁻¹⁹) / [4π × 9.109×10⁻³¹ × 2.998×10¹⁰]
= 0.4669 cm⁻¹ T⁻¹.

For B = 0.5 T,

Δν~ = 0.4669 × 0.5 = 0.2334 cm⁻¹.

Thus the adjacent Zeeman components are separated by

Δν~ ≈ 0.233 cm⁻¹.

If one defines the separation between the two extreme σ-components, it is twice this value:

2Δν~ ≈ 0.467 cm⁻¹.

(e) Within the rigid-rotor approximation for a diatomic molecule, the rotational term is

F(J) = B J(J + 1),

where

B = h/(8π²cI).

The first pure rotational line corresponds to J = 0 → 1. Its wave number is

Δν~ = F(1) − F(0) = 2B.

Given Δν~ = 21.18 cm⁻¹,

B = 21.18/2 = 10.59 cm⁻¹.

Therefore the moment of inertia is

I = h/(8π²cB).

Using B in cm⁻¹ and c = 2.998×10¹⁰ cm s⁻¹,

I = 6.626×10⁻³⁴ / [8π² × 2.998×10¹⁰ × 10.59]
= 2.643×10⁻⁴⁷ kg m².

The reduced mass is

μ = m_H m_Cl/(m_H + m_Cl).

With m_H = 1.008 amu and m_Cl = 35.45 amu,

μ = (1.008 × 35.45)/(1.008 + 35.45) amu
= 35.7336/36.458 amu
= 0.98013 amu.

Using 1 amu = 1.66054×10⁻²⁷ kg,

μ = 0.98013 × 1.66054×10⁻²⁷
= 1.6275×10⁻²⁷ kg.

The bond length r is

r = √(I/μ)
= √(2.643×10⁻⁴⁷ / 1.6275×10⁻²⁷)
= √(1.624×10⁻²⁰)
= 1.274×10⁻¹⁰ m.

Thus

r ≈ 1.274×10⁻¹⁰ m = 0.1274 nm = 1.274 Å.

(a) Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400 nm wavelength when a…

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