(a) How does Planck's law resolve the ultraviolet catastrophe predicted by classical physics? Calculate the average energy ε̄ of an oscillator of frequency 0·60×10¹⁴ s⁻¹ at T = 1800 K, treating it as (i) classical oscillator and (ii) Planck's oscilla

Question

(a) How does Planck's law resolve the ultraviolet catastrophe predicted by classical physics? Calculate the average energy ε̄ of an oscillator of frequency 0·60×10¹⁴ s⁻¹ at T = 1800 K, treating it as (i) classical oscillator and (ii) Planck's oscillator. (15 marks)

(b) (i) What do you understand by macrostates and microstates? Briefly explain. (5 marks)
(ii) A three-level laser system emits laser light at a wavelength of 550 nm. If the population of the upper level exceeds that of the lower level by 25%, determine the negative temperature characterizing the system. (10 marks)

(c) Consider a situation shown in the figure below. The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance l. The resistance of rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At t = 0, the wire PQ is pushed towards right with a speed V₀. Find (i) the current in the loop at an instant when the speed of the wire PQ is V and (ii) the acceleration of the wire at this instant. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Classical physics treats each oscillator as having continuous energy and, by equipartition, average energy ε̄ = kT. The number of radiation modes per unit volume in the frequency interval ν to ν+dν is proportional to ν², so the Rayleigh–Jeans energy density becomes u(ν)dν = (8πν²/c³) kT dν. This increases without limit as ν → ∞, predicting infinite high-frequency energy emission—the ultraviolet catastrophe.

Planck resolved this by assuming that an oscillator of frequency ν can have only discrete energies E_n = n hν, n = 0, 1, 2, … . Using Boltzmann weighting, the partition function is
Z = Σ e^(−n hν/kT) = 1/[1 − e^(−hν/kT)].
The average energy is
ε̄ = Σ n hν e^(−n hν/kT) / Z = hν/[e^(hν/kT) − 1].
The resulting Planck spectral energy density is
u(ν)dν = (8πhν³/c³) [1/(e^(hν/kT) − 1)] dν.
For large ν, e^(hν/kT) grows rapidly, so high-frequency oscillators have very small average energy. Hence u(ν) → 0 as ν → ∞, the spectral distribution has a finite peak, and the ultraviolet catastrophe disappears. At low ν it reduces to the Rayleigh–Jeans law.

Given ν = 0.60×10¹⁴ s⁻¹ = 6.0×10¹³ Hz, T = 1800 K. Take h = 6.626×10⁻³⁴ J s, k = 1.381×10⁻²³ J K⁻¹.

(i) Classical oscillator:
ε̄_cl = kT = 1.381×10⁻²³ × 1800 = 2.486×10⁻²⁰ J = 0.155 eV.

(ii) Planck oscillator:
hν = 6.626×10⁻³⁴ × 6.0×10¹³ = 3.976×10⁻²⁰ J.
kT = 2.486×10⁻²⁰ J.
Thus hν/kT = 3.976×10⁻²⁰ / 2.486×10⁻²⁰ = 1.600.
ε̄_P = hν/[e^(hν/kT) − 1]
= 3.976×10⁻²⁰/[e^(1.600) − 1].
Since e^(1.600) = 4.953,
ε̄_P = 3.976×10⁻²⁰/(4.953 − 1)
= 1.006×10⁻²⁰ J = 0.0628 eV.
So ε̄_cl ≈ 2.49×10⁻²⁰ J and ε̄_P ≈ 1.01×10⁻²⁰ J.

(b)(i) A macrostate specifies the observable macroscopic properties of a system, such as total energy U, volume V, number of particles N, pressure P and temperature T. It does not specify which particle is in which state. A microstate specifies the complete microscopic description of every particle, such as all positions and momenta in classical mechanics or all quantum numbers in quantum mechanics. Many microstates correspond to one macrostate. If Ω is the number of microstates belonging to a macrostate, the entropy is S = k ln Ω. Equilibrium corresponds to the macrostate with the largest Ω.

(b)(ii) For two levels with energies E₂ > E₁, the Boltzmann population ratio is
N₂/N₁ = e^[−(E₂ − E₁)/kT].
For population inversion, N₂ > N₁, so the effective temperature T is negative. The emitted photon energy is
ΔE = hν = hc/λ.
Given λ = 550 nm = 550×10⁻⁹ m,
ΔE = (6.626×10⁻³⁴ × 3.00×10⁸)/(550×10⁻⁹)
= 3.61×10⁻¹⁹ J.
The condition is N₂ = 1.25 N₁, so
ln(N₂/N₁) = ln(1.25) = 0.2231.
Therefore
T = −ΔE/[k ln(N₂/N₁)]
= −3.61×10⁻¹⁹/[1.381×10⁻²³ × 0.2231]
= −1.17×10⁵ K.
So T ≈ −1.17×10⁵ K. A negative temperature here does not mean colder than absolute zero; it means a population-inverted state, hotter than infinite temperature.

(c)(i) Let the rails be horizontal, P on the top rail and Q on the bottom rail, separated by distance l. The magnetic field B is into the page. When the rod PQ moves to the right with speed V, the magnetic flux through the loop changes because the area of the loop lying in the field changes at rate lV. By Faraday’s law,
ε = −dΦ/dt = B l V.
The total resistance of the loop is R + r, since the rails have negligible resistance. Hence the induced current is
I = ε/(R + r) = B l V/(R + r).
By Lenz’s law, the current opposes the change in flux. For the given field into the page and motion to the right, the current is clockwise in the loop; in the rod PQ it flows from Q to P. Thus
I = B l V/(R + r), clockwise, i.e. Q → P in the rod.

(c)(ii) The current-carrying rod of length l in the magnetic field B experiences a force
F = I l B.
Using I = B l V/(R + r),
F = (B l V/(R + r)) l B = B² l² V/(R + r).
By Fleming’s left-hand rule, this force is directed to the left, opposite to the velocity. Therefore the acceleration is
a = F/m = B² l² V/[m(R + r)].
So
a = B² l² V/[m(R + r)] directed opposite to V, i.e. to the left.
This result is valid at instants when the rod PQ is still inside the magnetic-field region and self-inductance is neglected.

(a) How does Planck's law resolve the ultraviolet catastrophe predicted by classical physics? Calculate the average energy ε̄ of…

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