Question

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.?

Short answer

(a)Thedeuterium differs from the Hydrogen by one neutron and do not contribute to the potential as the reduced mass do not alter with the or $\kappa$and remain significantly same for both $H_2$and $HD$.

(b) The relation is $\frac{1}{{\mu }_{HD}}<\frac{1}{{\mu }_{H2}}$.

(c) The required relation is $E_{vibHD}<E_{VIBH2}$.

Step-by-step solution

Determine the formulas:

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:

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

Determine the effect of the replacement on the bond length and the force constants:

(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 $HD$.

Determine the effect on the rotational energy

(b)

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

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

Determine the effect on the vibrational energy levels

(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 $E_{vibHD}<E_{VIBH2}$.