Question

The wavelength of the first overtone of the $N-H$stretching vibration is about $1.5$ µm. What is the approximate wavenumber and wavelength of the fundamental band for the $N-H$ stretch?

Short answer

The wave number is$3,300{\mathrm{cm}}^{-1}$and wavelength is$3\mu m$for fundamental stretch.

Step-by-step solution

Step 1. Given information

Stretching vibration is given as$1.5\mu m$

Step 2. Formula used

For calculating part, the formula is

$\overline{v}\left({\mathrm{cm}}^{-1}\right)=\frac{1}{\lambda \left(\mu \mathrm{m}\right)}\times {10}^4(\mu \mathrm{m}/\mathrm{cm})$

Where $\overline{V}$ is wave number and $\lambda$ is wavelength

Step 3. Solution

Substitute the value in formula,

$$\begin{gathered} \overline{v}\left({\mathrm{cm}}^{-1}\right)=\frac{1}{\lambda \left(\mu \mathrm{m}\right)}\times {10}^4(\mu \mathrm{m}/\mathrm{cm}) \\ =\frac{1}{1.5\mu \mathrm{m}}\times {10}^4(\mu \mathrm{m}/\mathrm{cm}) \\ =0.66\times {10}^4{\mathrm{cm}}^{-1} \end{gathered}$$

This is the initial overtone's wavenumber. However, we require the fundamental wavenumber. Use the following relationship to do this:

$$\begin{gathered} \text{wave number of first overtone}\left(\overline{v}\right)=2\text{(wave number of fundamental)} \\ \text{wave number of fundamental}=\frac{\text{wave number of first overtone}}{2} \\ =\frac{0.66\times {10}^4{\mathrm{cm}}^{-1}}{2} \\ =0.33\times {10}^4{\mathrm{cm}}^{-1} \\ =3300{\mathrm{cm}}^{-1} \end{gathered}$$

Step 4. Calculating wavelength

Now the fundamental wavelength,

$$\begin{gathered} \text{wave length of fundamental}=\frac{1}{\text{wavenumber of fundamental}} \\ =\frac{1}{3300{\mathrm{cm}}^{-1}} \\ =3.0\times {10}^{-4}\mathrm{cm} \end{gathered}$$