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}].

Consider a system of N particles with only 3 possible energy levels separated by ε. Let the ground state energy be zero. The system occupies a fixed volume V and is in thermal equilibrium with a reservoir at temperature T. Ignore interactions between particles and assume Boltzmann statistics apply.

(a) What is the partition function for a single particle in the system?

(b) What is the average energy per particle?

(c) What is the probability that the 2ε level is occupied in the
high-temperature limit k_{B}T >> ε?

Explain your answer on physical grounds.

(d) What is the average energy per particle in the high-temperature limit k_{B}T
>> ε?

(e) At what energy is the ground state 1.1 times as likely to be occupied
as the 2ε level?

(f) Find the heat capacity C_{V} of the system, analyze the low-T
(k_{B}T >> ε) and high-T (k_{B}T >> ε) limit, and sketch CV as a
function of T.

Solution:

- Concepts:

Boltzmann statistics - Reasoning:

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:

(a) P(E) = Cexp(-E/(k_{B}T)).

Σ_{j}P(E_{j}) = CΣ_{j}exp(-E_{j}/(k_{B}T)) = 1, C = 1/Σ_{j}exp(-E_{j}/(k_{B}T)) = 1/Z.

Let β = 1/(k_{B}T)).

Z = 1 + exp(-βε) + exp(-2βε).

(b) <ε> = (ε exp(-βε) + 2ε exp(-2βε))/Z

= ε(exp(-βε) + 2exp(-2βε))/(1 + exp(-βε) + exp(-2βε)).

(c) P = exp(-2βε)/(1 + exp(-βε) + exp(-2βε)) ~ (1 - 2βε)/( 1 + 1 - βε + 1 - 2βε) = 1/3.

In the high energy limit the probability is the same for all three levels.

(d) <ε> = 0/3 + ε/3 + 2ε/3 = ε.

(e) exp(-2βε) = 1/1.1, 2βε = ln(1.1), T = 2ε/(k_{B}*ln(1.1)).

(f) C_{V}= dU/dT = Nd<ε>/dT = Nd<ε>/dβ)(dβ/dT).

C_{V}= (Nε^{2}/(k_{B}T^{2}))[ exp(-βε) + 4 exp(-2βε) + exp(-3βε)]/[1 + exp(-βε) + exp(-2βε)]^{2}.^{ }Low temperature limit: C_{V}= (Nε^{2}/(k_{B}T^{2})) exp(-ε/(k_{B}T)). (C_{V}--> 0 as T --> 0)

High temperature limit: C_{V}= (2/3)(Nε^{2}/(k_{B}T^{2})). ( C_{V}--> 0 as T --> ∞)