As noted in example 10.2, the HD molecule differs from ${\mathbf{H}}_{\mathbf{2}}$in that a deuterium atom replaces a hydrogen atom (a) What effect, if any, does the replacement have on the bond length and force constant? Explain. (b) What effect does it have on the rotational energy levels ? (c) And what effect does it have on the vibrational energy levels.?

Consider the formula for the reduced mass as:

$\mathbf{\mu }\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots ..\left(1\right)$

Consider the relation between reduced mass constant and rotational energy as:

$$\begin{gathered} {\mathbf{E}}_{\mathbf{rot}}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}\mathbf{l}\left(\mathbf{l}\mathbf{+}\mathbf{1}\right)}{\mathbf{2}{\mathbf{\mu a}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ {\mathbf{E}}_{\mathbf{rot}}\mathbf{\mu }\frac{\mathbf{1}}{\mathbf{\mu }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

(a)

Determine the effect of the bond length and the force constant on the mass of the Hydrogen as:

$$\begin{gathered} m_1=1.007\text{u} \\ {m}_1=1.007\times 1.66\times {10}^{-27}\text{kg} \end{gathered}$$

Solve for the mass of the Deuterium as:

$$\begin{gathered} m_2=2.013\text{u} \\ {m}_2=2.013\times 1.66\times {10}^{-27}\text{kg} \end{gathered}$$

Substitute the values in the equation (1) and solve for the reduced mass as:

For hydrogen:

$$\begin{gathered} {\mu }_{H2}=\frac{{\displaystyle 1.007\times 1.007}}{{\displaystyle 1.007+1.007}} \\ =0.503\mu \end{gathered}$$

For Deuterium:

$$\begin{gathered} {\mu }_{HD}=\frac{1.007\times 2.013}{1.007+2.013} \\ =0.671\mu \end{gathered}$$

Therefore, the deuterium differs from the Hydrogen by one neutron and do not contribute to the potential as the reduced mass do not alter with the a or k and remain significantly same for both $H_2$and $HD$.

(b)

From equation (2) determine the effect on the rotational energy as:

$$\begin{gathered} E_{rot}\propto \frac{1}{\mu }\backslash \mathrm{hfill} \\ \frac{1}{{\mu }_{HD}}<\frac{1}{{\mu }_{H2}}\left\{{\mu }_{HD}>{\mu }_{H2}\right\} \end{gathered}$$

(c)

From equation (2) determine the effect on the rotational energy as:

$$\begin{gathered} E_{vib}=\left(n+\frac{1}{2}\right)h\sqrt{\frac{\kappa }{2}} \\ {\mu }_{HD}>{\mu }_{H2}\backslash \mathrm{hfill} \\ \sqrt{{\mu }_{HD}}>\sqrt{{\mu }_{H2}} \\ \frac{1}{\sqrt{{\mu }_{HD}}}<\frac{1}{\sqrt{{\mu }_{H2}}} \end{gathered}$$

Therefore, the required relation is $E_{vibHD}<E_{VIBH2}$.