91Ó°ÊÓ

The given molecule is $H_2$.

Heat capacity:

role="math" localid="1647281996407" $C=\frac{{\mathrm{Î\mu }}^{2}N{e}^{\frac{\mathrm{Î\mu }}{kT}}}{k{T}^2{\left(e^{\frac{\mathrm{Î\mu }}{kT}}-1\right)}^2}$

The heat capacity is given as:

$C=\frac{{\mathrm{Î\mu }}^{2}N{e}^{\frac{\mathrm{Î\mu }}{kT}}}{k{T}^2{\left(e^{\frac{\mathrm{Î\mu }}{kT}}-1\right)}^2}$

Let $t=\frac{kT}{\mathrm{Î\mu }}$,

Hence, the above equation becomes,

$$\begin{gathered} C=\frac{{\mathrm{Î\mu }}^{2}N{e}^{\frac{\mathrm{Î\mu }}{kT}}}{k{T}^2{\left(e^{\frac{\mathrm{Î\mu }}{kT}}-1\right)}^2} \\ C=\frac{Nk{e}^{\frac{1}{t}}}{t^2{\left(e^{\frac{1}{t}}-1\right)}^2} \\ \frac{C}{Nk}=\frac{e^{\frac{1}{t}}}{t^2{\left(e^{\frac{1}{t}}-1\right)}^2} \end{gathered}$$

The value of $t$at which the heat capacity is half of its maximum value is:

$C=\frac{1}{2}Nk$

Hence, the above equation becomes,

$$\begin{gathered} \frac{C}{Nk}=\frac{e^{\frac{1}{t}}}{t^2{\left(e^{\frac{1}{t}}-1\right)}^2} \\ \frac{1}{2}{t}^2{\left(e^{\frac{1}{t}}-1\right)}^2=e^{\frac{1}{t}} \\ t=0.335 \end{gathered}$$

Now, by resubstituting the value of $t$, we get,

role="math" localid="1647282058861" $\frac{kT}{\mathrm{Î\mu }}=0.335$

For vibrational motion of ${\mathrm{H}}_2$gas molecules at its heat capacity, half of its maximum value, the temperature is, $T=1700\mathrm{K}$.

Hence, $\mathrm{Î\mu }$can be calculated as:

$$\begin{gathered} \mathrm{Ï\mu }=\frac{kT}{0.335} \\ \mathrm{Î\mu }=\frac{8.62×{10}^{-5}×1700}{0.335} \\ \mathrm{Î\mu }=0.44eV \end{gathered}$$

Hence, the value of$\mathrm{Î\mu }$can be calculated as$0.44eV$.