Question

A system consists of N particles that can occupy two energy levels: a nondegenerate ground state and a three-fold degenerate excited state, which lies at an energy of 0.25 eV above the ground state. At a temperature of 960 K, find the ratio of the number of particles in the excited state to the number in the ground state.

Answer 1

Ng = 0.893 N, Ne = 0.107N

Step-by-step explanation:

Number of particles in Ground state = Ng

Number of particles in Excited state = Ne

Ne/Ng = e^{(-ΔE)/kt}

Since excited state is 3 fold degenerate

Ne/Ng =3 x e^{(-ΔE)/kt}

ΔE = Energy difference between ground and excited states = 0.25eV

T = 960 K

Constant k = 8.617 x 10^-5 eV/K

Ne/Ng = 3 x e^{-0.25/(8.617x10^-5) x 960}

= 3 x e^(-3.188645)

= 3 x 0.0412 = 0.1237 ≅ 0.12

Ne = 0.12 Ng

but Ne + Ng = N, where is N is total number of particles, substituting Ne into equation we get,

Ng(1 + 0.12) = N

Ng = N/1.12 = 0.893N

and Ne = 0.12 x 0.893 N = 0.107 N