Question

The mass of the deuterium molecule (${\mathit{D}}_{\mathbf{2}}$) is twice that of the hydrogen molecule (${\mathit{H}}_{\mathbf{2}}$). If the vibrational frequency of ${\mathit{H}}_{\mathbf{2}}$is $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{ }\mathbf{\text{Hz}}$, what is the vibrational frequency of ${\mathit{D}}_{\mathbf{2}}$? Assume the “spring constant” of attracting forces is the same for the two molecules.

Short answer

The vibrational frequency of$D_2$ is$f_D=9.19\times {10}^{13} \text{Hz}$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The vibrational frequency of$H_2$is,$1.30\times {10}^{14} \text{Hz}$.
• The mass of the deuterium molecule ($D_2$) is twice that of the hydrogen molecule ($H_2$).

Significance of the period of the oscillation of the particle

The period of the oscillation of the particle is expressed as,

$\mathit{T}\mathbf{=}\mathbf{2}\mathit{\pi }\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

From here we can write the angular frequency:

${\mathit{\omega }}_{\mathbf{n}}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

And the inverse of the period is the frequency

Determination of the vibrational frequency of D2

Let mass of hydrogen molecules is m.

Then according to given Mass of deuterium molecule is 2m

By using the concept and from step (1), we have

${\omega }_n=\sqrt{\frac{k}{m}}$

But we know that:

$$\begin{gathered} f_H=\frac{{\omega }_n}{2\pi } \\ {f}_H=\frac{1}{2\pi }\sqrt{\frac{k}{m}} \end{gathered}$$

And,

$$\begin{gathered} f_D=\frac{{\omega }_n}{2\pi } \\ {f}_D=\frac{1}{2\pi }\sqrt{\frac{k}{2m}} \end{gathered}$$

We take ratio of both, we get

$$\begin{gathered} \frac{f_D}{f_H}=\frac{\sqrt{\frac{k}{2m}}}{\sqrt{\frac{k}{m}}} \\ {f}_D=\frac{f_H}{\sqrt{2}} \end{gathered}$$

Substitute all the value in the above equation,

$$\begin{gathered} f_D=\frac{1.3\times {10}^{14} \text{Hz}}{\sqrt{2}} \\ {f}_D=9.19\times {10}^{13} \text{Hz} \end{gathered}$$

Hence the vibrational frequency ofdeuteriumis, $f_D=9.19\times {10}^{13} \text{Hz}$.