91Ó°ÊÓ

From given frequencies, we can calculate the wavelength of each frequency by the formula $\mathbf{ν}\mathbf{=}\frac{c}{\mathrm{λ}}$. where$\mathbf{λ}$is called wavelength,$\mathbf{ν}$is called frequency, and c is called speed of light.

Since

For frequency

$\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}\mathbf{ν}\mathbf{=}\frac{c}{\mathrm{λ}}\\ \mathbf{λ}\mathbf{=}\frac{\mathbf{3}\mathbf{×}{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}}\\ \mathbf{=}\mathbf{7}\mathbf{.}\mathbf{463}\mathbf{×}{\mathbf{10}}^{-6}\mathbf{m}\\ =7463\mathrm{nm}\end{array}$

$\begin{array}{c}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{7463\mathrm{nm}}\\ \mathbf{=}\mathbf{1339}\mathbf{.}\mathbf{94}{\mathbf{cm}}^{-1}\end{array}$

For frequency $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}\mathbf{ν}\mathbf{=}\frac{c}{\mathrm{λ}}\\ \mathbf{λ}\mathbf{=}\frac{\mathbf{3}\mathbf{×}{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}}\\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{×}{\mathbf{10}}^{-5}\mathbf{m}\\ =15000\mathrm{nm}\end{array}$

$\begin{array}{c}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{15000\mathrm{nm}}\\ \mathbf{=}\mathbf{666}\mathbf{.}\mathbf{67}{\mathbf{cm}}^{-1}\end{array}$

For frequency$\text{ }\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}\mathbf{ν}\mathbf{=}\frac{c}{\mathrm{λ}}\\ \mathbf{λ}\mathbf{=}\frac{\mathbf{3}\mathbf{×}{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{255}\mathbf{×}{\mathbf{10}}^{-6}\mathbf{m}\\ =4225\mathrm{nm}\end{array}$

$\begin{array}{c}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{4225\mathrm{nm}}\\ \mathbf{=}\mathbf{2366}\mathbf{.}\mathbf{86}{\mathbf{cm}}^{-1}\end{array}$

Since, the range of IR region of electromagnetic radiation is 600- $\mathbf{4000}{\mathbf{cm}}^{-1}$. So, vibrational motions have frequencies in IR region of electromagnetic spectrum. As a result, the indicated frequencies exist in the IR area.

Calculate the energy of each frequency by using the formula$\mathbf{E}\mathbf{=}\mathbf{³óν}$

For frequency$\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}{\mathbf{E}}_1\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{×}\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}\\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{663}\mathbf{×}{\mathbf{10}}^{-20}\mathbf{J}\end{array}$

For frequency $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}{\mathbf{E}}_2\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{×}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}\\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3252}\text{ }\mathbf{×}{\mathbf{10}}^{-20}\mathbf{J}\end{array}$

For frequency$\text{ }\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}{\mathbf{E}}_3\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{×}\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{671}\text{ }\mathbf{×}{\mathbf{10}}^{-20}\mathbf{J}\end{array}$

From the calculations it is clear that $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$ has least energy.