Question

Determine the mean energy at 400 K of a two-level system. The energy separation is 600 /cm.

Answer 1

Given data:

The temperature of the two-level system (T) = 400 K

Energy separation (B) = 600 Cm^-1.

β = 1/kT, where k = boltzmann's constant, i.e. 1.38*10^-23 J/K.

Therefore, β = 1/(1.38*10^-23 * 400), i.e. 1.81 * 10²⁰ / J

E = hcB, where h = Planck's constant, i.e. 6.625*10^-34 J.s

c = velocity of light, i.e. 3*10¹⁰ Cm.s^-1

Therefore, E = 1.9875*10^-23 J

The rotational partition function can be calculated as shown below.

q^r = 1 + e^-2β​E

Therefore, q^r = 1 + e^{-3.975*10^(-23)β}

The average energy can be calculated as shown below.

<E> = -{ln(q^r)}/β

= -(-3.975*10^-23β)/β

= 3.975*10^-23 J/molecule

and U = N<E>

= 6.023*10²³ * 3.975*10^-23 J/mol

~ 24 J /mol