Question

Calculate the wavenumber of a photon whose wavelength is 400 nm.
a) $0.001n{{m}^{-1}}$
b) $0.002n{{m}^{-1}}$
c) $0.0025n{{m}^{-1}}$
d) $0.0045n{{m}^{-1}}$

Answer 1

Hint: Since, kinetic energy of a photon is given by the formula: $E=hf=\dfrac{hc}{\lambda }$where f is frequency, c is speed of light and $\lambda $ is wavelength of photon.
From this equation, we get: $f=\dfrac{c}{\lambda }$
If we divide the whole equation by c, we get: $\dfrac{f}{c}=\dfrac{1}{\lambda }$
The ratio of frequency and speed of light is termed as wave number and is given as \[\overline{\nu }=\dfrac{1}{\lambda }\]

Complete step by step answer:
So, we are given $\lambda =400nm$
By using the formula: \[\overline{\nu }=\dfrac{1}{\lambda }\]
We get:
\[\begin{align}
  & \overline{\nu }=\dfrac{1}{400} \\
 & =0.0025n{{m}^{-1}}
\end{align}\]

So, the correct answer is “Option C”.

Additional Information:
The wavelength is the distance between identical points in a cycle of a wave. The wavenumber is the reciprocal of the wavelength. Therefore, it is the number of waves or cycles per unit distance. So, the unit of wavenumber is $n{{m}^{-1}}$.

Note:
Another way to solve for a wavenumber is to find the frequency of the photon and divide it by speed of light as follows:
For the given value of wavelength $\lambda =400nm$; by using the frequency – wavelength relation $f=\dfrac{c}{\lambda }$, we get frequency of the photon as:
$f=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{400nm}$
Since, 1 nm = ${{10}^{-9}}$m
So, frequency is: $\begin{align}
  & f=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{400\times {{10}^{-9}}m} \\
 & =7.5\times {{10}^{14}}{{s}^{-1}}
\end{align}$
Now, divide frequency by speed of light to get the wave number:
\[\begin{align}
  & \overline{\nu }=\dfrac{7.5\times {{10}^{14}}{{s}^{-1}}}{3\times {{10}^{8}}m{{s}^{-1}}} \\
 & =2.5\times {{10}^{6}}{{m}^{-1}} \\
 & =0.0025n{{m}^{-1}}
\end{align}\]
But this method is a bit lengthy because we are multiplying and dividing by a similar quantity in two different steps. So, to avoid this, we may use the direct method, i.e. reciprocal of wavelength. So, we might not require speed of light to calculate the wavenumber of a photon