91Ó°ÊÓ

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.

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

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

$Z=e^{-{\mathrm{Ï\mu }}_1/kT}+e^{-{\mathrm{Ï\mu }}_2/kT}+e^{-{\mathrm{Ï\mu }}_3/kT}$

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,

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

For second state,

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

For third state,

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

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

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

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

For second state:

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

For third state

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