Question

The wavelength ${\lambda }_e$ of an electron and ${\lambda }_p$ of a photon of same energy $E$ are related by:
$A){\lambda }_p\propto {{\lambda }_e}^2$
$B){\lambda }_p\propto {\lambda }_e$
$C){\lambda }_p\propto \sqrt{{\lambda }_e}$
$D){\lambda }_p\propto {\displaystyle \frac{1}{\sqrt{{\lambda }_e}}}$

Answer 1

Hint: This problem can be solved by using the direct formula for the wavelength of a body in terms of its momentum (which can be related to the energy) and the wavelength of a photon in terms of its energy. Hence, we can get the required relation between the wavelength of an electron and a photon.

Formula used:
$\lambda ={\displaystyle \frac{h}{p}}$
$E={\displaystyle \frac{hc}{\lambda }}$
$p=\sqrt{2mKE}$

Complete step-by-step answer:
We will use the formula for the wavelength of a photon of a certain energy and the wavelength of a particle of certain momentum and relate the momentum with the kinetic energy.
Now, let us analyze the question.
The wavelength of the photon is ${\lambda }_p$.
The wavelength of the electron is ${\lambda }_e$.
Let the momentum of the electron be $p$.
The energy of the photon and the electron both are $E$.
Now, the wavelength $\lambda$ of a photon is related to its energy $E$ as
$E={\displaystyle \frac{hc}{\lambda }}$ --(1)
Where $h=6.636\times {10}^{-34}J.s$ is the Planck’s constant and $c=3\times {10}^{8}m/s$ is the speed of light in vacuum.
Therefore, using (1), we get
$E={\displaystyle \frac{hc}{{\lambda }_p}}$ --(2)
The wavelength $\lambda$ of a body having momentum $p$ is given by
$\lambda ={\displaystyle \frac{h}{p}}$ --(3)
Where $h=6.636\times {10}^{-34}J.s$ is the Planck’s constant.
Using (3), we get
${\lambda }_e={\displaystyle \frac{h}{p}}$ --(4)
Now, the momentum $p$ and kinetic energy $KE$ of a body of mass $m$ is related as
$p=\sqrt{2mKE}$ --(5)
Therefore, using (5), we get
$p=\sqrt{2m_{e}E}$ --(6)
Where $m_e$ is the mass of the electron.
Putting (6) in (4), we get
${\lambda }_e={\displaystyle \frac{h}{\sqrt{2m_{e}E}}}$
Squaring both sides we get
${\left({\lambda }_e\right)}^2={\left({\displaystyle \frac{h}{\sqrt{2m_{e}E}}}\right)}^2$
$\therefore {{\lambda }_e}^2={\displaystyle \frac{h^2}{2m_{e}E}}$
$\therefore E={\displaystyle \frac{h^2}{2m_e{{\lambda }_e}^2}}$ --(7)
Equating (2) and (7), we get
${\displaystyle \frac{hc}{{\lambda }_p}}={\displaystyle \frac{h^2}{2m_e{{\lambda }_e}^2}}$
$\therefore {\lambda }_p=\left({\displaystyle \frac{2m_{e}c}{h}}\right){{\lambda }_e}^2$
$\therefore {\lambda }_p=K{{\lambda }_e}^2$
Where $K={\displaystyle \frac{2m_{e}c}{h}}$ is a constant.
$\therefore {\lambda }_p\propto {{\lambda }_e}^2$
Hence, we have got the required relation as ${\lambda }_p\propto {{\lambda }_e}^2$.
Therefore, the correct option is $A){\lambda }_p\propto {{\lambda }_e}^2$.

Note: We could have also solved this problem by determining the proportionalities between the energy and the wavelengths of the electron and the photon separately as two equations and then combine them to get the relation of the proportionality of between the wavelength of the photon. This method would have removed the unnecessary variables from the calculation. However, it is difficult to always recall the proportionality relations and hence, it is best to go step by step and derive the result to avoid mistakes that can happen while writing the direct proportionality relations from memory.