Imagine a particle that can be in only three states, with energies -0.05 eV, 0, and 0.05 eV. This particle is in equilibrium with a reservoir at 300 K.

(a) Calculate the partition function for this particle.

(b) Calculate the probability for this particle to be in each of the three states.

(c) Because the zero point for measuring energies is arbitrary, we could just as well say that the energies of the three states are 0, +0.05 eV, and +0.10 eV, respectively. Repeat parts (a) and (b) using these numbers. Explain what changes and what doesn't.

A particle that can be in only three states, with energies -0.05 eV, 0, and 0.05 eV. This particle is in equilibrium with a reservoir at 300 K.

(a) The partition function is equal to the total of all Boltzmann factors, i.e.

$Z=\sum _s{e}^{-E\left(s\right)/kT}$

Consider a particle that can exist in three states, each with an energy of ${ϵ}_1$=-0.05 eV, ${ϵ}_2$= 0 eV, and ${ϵ}_3$= 0.05 eV; the partition function is:

$Z=e^{-{ϵ}_1/kT}+e^{-{ϵ}_2/kT}+e^{-{ϵ}_3/kT}$

Substitute the energies and Boltzmann's constant

$$\begin{gathered} Z=e^{0.05\mathrm{eV}/0.025851\mathrm{eV}}+e^0+e^{-0.05\mathrm{eV}/0.025851\mathrm{eV}} \\ Z=8.063 \end{gathered}$$

(b)The probability is given by:

$P=\frac{1}{Z}{e}^{-E\left(s\right)/kT}$

For first state,

$$\begin{gathered} P_1=\frac{1}{Z}{e}^{-{ϵ}_1/kT} \\ =\left(\frac{1}{8.063}\right)e^{0.05\mathrm{eV}/0.025851\mathrm{eV}} \\ {P}_1=0.858 \end{gathered}$$

For second state,

$$\begin{gathered} P_2=\frac{1}{Z}{e}^{-{ϵ}_2/kT} \\ =\frac{1}{8.063} \\ {P}_2=0.124 \end{gathered}$$

For third state,

$$\begin{gathered} P_3=\frac{1}{Z}{e}^{-{ϵ}_3/kT} \\ =\left(\frac{1}{8.063}\right)e^{-0.05\mathrm{eV}/0.025851\mathrm{eV}} \\ {P}_3=0.018 \end{gathered}$$

(c) Now imagine a particle that can exist in three states with energies of ${ϵ}_1$= 0 eV, ${ϵ}_2$= 0.05 eV, and ${ϵ}_3$= 0.10 eV; the partition function is:$Z=e^{-{ϵ}_1/kT}+e^{-{ϵ}_2/kT}+e^{-{ϵ}_3/kT}$

Substitute with energies and Boltzmann's constant:

role="math" localid="1647298423935" $$\begin{gathered} Z=e^0+e^{-0.05\mathrm{eV}/0.025851\mathrm{eV}}+e^{-0.10\mathrm{eV}/0.025851\mathrm{eV}} \\ Z=1.165 \end{gathered}$$

For first state

$$\begin{gathered} P_1=\frac{1}{Z}{e}^{-{ϵ}_1/kT} \\ =\frac{1}{1.165} \\ {P}_1=0.858 \end{gathered}$$

For second state:

$$\begin{gathered} P_2=\frac{1}{Z}{e}^{-{ϵ}_2/kT} \\ =\left(\frac{1}{1.165}\right)e^{-0.05\mathrm{eV}/0.025851\mathrm{eV}} \\ {P}_2=0.124 \end{gathered}$$

For third state

$$\begin{gathered} P_3=\frac{1}{Z}{e}^{-{ϵ}_3/kT} \\ =\left(\frac{1}{1.165}\right)e^{-0.10\mathrm{eV}/0.025851\mathrm{eV}} \\ {P}_3=0.018 \end{gathered}$$