Answer
Wavelength: $4$ $\mu m$
Frequency: $7.5\times 10^{13}$ $Hz$
Work Step by Step
The relation between wavenumber $(\tilde v)$ and wavelength $(\lambda)$ is given by
$\tilde v=\frac{1}{\lambda}$ ...............$(1)$
And the relation between wavelength $(\lambda)$ and frequency $(v)$ is given by
$V=v\lambda$
or, $v=\frac{V}{\lambda}$ ...................$(2)$
where $V$ is the speed of the wave.
The wave number of a typical vibrational transition of a hydrocarbon is given
$\tilde v=2500$ $cm^{−1}$
Using eq. $1$, we find the corresponding wavelength:
$\lambda=\frac{1}{\tilde v}$
or, $\lambda=\frac{1}{2500}$ $cm$
or, $\lambda=4\times 10^{-4}$ $cm$
or, $\lambda=4\times 10^{-6}$ $m$
or, $\lambda=4$ $\mu m$
The wavelength $\lambda=4$ $\mu m$ corresponds to the wave in near-infrared region.
The speed of near-infrared wave is equals to the speed of light in vacuum.
Thus, $V=3\times 10^{8}$ $m/s$
Now, using eq. $2$, we find the corresponding frequency:
$v=\frac{3\times 10^{8}}{4\times 10^{-6}}$
$v=7.5\times 10^{13}$ $Hz$
$\therefore$ The wavelength and frequency of a typical vibrational transition of a hydrocarbon are $4$ $\mu m$ and $7.5\times 10^{13}$ $Hz$ respectively.
