Question

Consider blackbody radiation at a temperature T. Show that for an energy threshold E₀ ≫ kT, the fraction of the blackbody photons that have energy hf > E₀ is

a) 0
b) 1
c) e^(-E₀/kT)
d) 1 - e^(-E₀/kT)

Answer 1

The correct answer is b) 1.

Step-by-step explanation:

To determine the fraction of blackbody photons with energy greater than E₀, we need to consider the energy distribution of blackbody radiation. According to the Planck's Law, the energy density per unit frequency of blackbody radiation is given by:

ρ(ƒ) = (8πƒ²/c³) × (hf/(exp(hf/kT) - 1))

Where ρ(ƒ) is the energy density, ƒ is the frequency, c is the speed of light, k is the Boltzmann constant, and T is the temperature. To find the fraction of photons with energy greater than E₀, we integrate this energy density over a range of frequencies where the energy is greater than E₀:

Fraction = ∫(ƒ > E₀/h) ρ(ƒ) dƒ

This integral can be complex and difficult to solve analytically, but in the limit where E₀ ≫ kT, we can approximate the integral as:

Fraction ≈ ∫(ƒ > 0) ρ(ƒ) dƒ

Since E₀ ≫ kT, the exponential term in the denominator becomes negligible and the fraction of photons with energy greater than E₀ is approximately 1.