Due to the presence everywhere of the cosmic background radiation the minimum possible temperature of a gas in interstellar space is 2.7 K. This implies that a significant fraction of molecules in space may be in low-level excited states. Consider a hypothetical molecule with one possible excited state. What would the excitation energy have to be for 20% of the molecules to be in the excited state?

Solution:

- Concepts:

Boltzmann statistics - Reasoning:

Boltzmann statistics tells us that N_{excited}/N_{ground}= e^{-ΔE/kT}, if the degeneracy of both states is 1. - Details of the calculation:

0.2/0.8 = exp(-ΔE/(kT))

ln(0.25) = -ΔE/(kT)

ΔE = -ln(0.25)*(kT) = -ln(0.25)*8.617*10^{-5}*2.7 eV = 3.2*10^{-4}eV = 0.32 meV

Consider a particle in a simple two level system with eigenstates ψ_{1}
and ψ_{2} such that

Hψ_{1} = E_{1}ψ_{1} and Hψ_{2} = E_{2}ψ_{2}.
Derive an expression for the specific heat per particle c_{V} = ∂<E>/∂T of the
system as a function of temperature.

Solution:

- Concepts:

Boltzmann statistics - Reasoning:

The system consisting of many particles is in a statistical mixture of states.

In thermodynamic equilibrium P(E_{i}) ∝ g_{i}exp(-E_{i}/(kT)

Here the degeneracy g_{i}is 1 for all i. - Details of the calculation:

Average energy: <E> = (E_{1}exp(-E_{1}/(kT)) + E_{2}exp(-E_{2}/(kT)))/(exp(-E_{1}/(kT)) + exp(-E_{2}/(kT)))

<E> = (E_{1}exp(-βE_{1}) + E_{2}exp(-βE_{2}))/(exp(-βE_{1}) + exp(-βE_{2}))

= (E_{1}+ E_{2}exp(-β∆))/(1 + exp(-β∆)),

with β = 1/(kT) and ∆ = E_{2}– E_{1}.

c_{V}= ∂<E>/∂T. c_{V}= (∂<E>/∂β)(∂β/∂T) = -(1/(kT^{2}))∂<E>/∂β.

c_{V}= [1/(kT^{2})][(E_{2}∆exp(-β∆))/(1 + exp(-β∆)) - (E_{1}+ E_{2}exp(-β∆))(∆exp(-β∆)))/(1 + exp(-β∆))^{2}].