Question

A photon has the same wavelength as the de Broglie wavelength of electrons. Given $C=\text{speed of light}$, $v=\text{speed of electron}$. Which of the following relations is correct?
(Here ${{E}_{e}}=$ kinetic energy of electron, ${{E}_{ph}}=$ energy of photon, ${{P}_{e}}=$ momentum of electron and ${{P}_{ph}}=$ momentum of photon)
A. $\dfrac{{{E}_{e}}}{{{E}_{ph}}}=\dfrac{2C}{v}$
B. $\dfrac{{{E}_{e}}}{{{E}_{ph}}}=\dfrac{v}{2C}$
C. $\dfrac{{{p}_{e}}}{{{p}_{ph}}}=\dfrac{2C}{v}$
D. $\dfrac{{{p}_{e}}}{{{p}_{ph}}}=\dfrac{C}{v}$

Answer 1

Hint: To solve the given problem, we must know about the formulae for the energy of a photon and momentum of a photon. Also the formulae for the kinetic energy and wavelength of an electron. Use this formulae and the given data to find the relation between the energies and the momentums.

Formula used:
${{E}_{ph}}=\dfrac{hC}{{{\lambda }_{ph}}}$
${{P}_{ph}}=\dfrac{h}{{{\lambda }_{ph}}}$
${{\lambda }_{e}}=\dfrac{h}{{{P}_{e}}}$
$P=mv$
$E=\dfrac{1}{2}m{{v}^{2}}$

Complete step by step answer:
Let us first the data for the photon. The energy of a photon is given as ${{E}_{ph}}=\dfrac{hC}{{{\lambda }_{ph}}}$ ..… (i),
where h is the Planck’s constant, C is the speed of the photon (equal to speed of light) and ${{\lambda }_{ph}}$ is the wavelength of the light.
The momentum of a photon is given as ${{P}_{ph}}=\dfrac{h}{{{\lambda }_{ph}}}$ …. (ii).
Let us now discuss the electron. It is found that an electron also shows wave nature. Therefore, it has a wavelength just like light. This wavelength of the electron is called as de Broglie wavelength. The de Broglie wavelength of an electron is given as ${{\lambda }_{e}}=\dfrac{h}{{{P}_{e}}}$ ….. (iii),
where ${{P}_{e}}$ is the momentum of the electron.
From mechanics, we know that the momentum of a particle is given as $P=mv$, where m is its mass and v is its velocity.
This means that ${{P}_{e}}=mv$ ….. (iv).
The kinetic energy of a particle is given as $E=\dfrac{1}{2}m{{v}^{2}}$. Therefore, the kinetic energy of the electron is ${{E}_{e}}=\dfrac{1}{2}m{{v}^{2}}$ ….. (v).
Now, divide equation (v) by (i).
$\Rightarrow \dfrac{{{E}_{e}}}{{{E}_{ph}}}=\dfrac{\dfrac{1}{2}m{{v}^{2}}}{\dfrac{hC}{{{\lambda }_{ph}}}}$
$\Rightarrow \dfrac{{{E}_{e}}}{{{E}_{ph}}}=\dfrac{m{{v}^{2}}{{\lambda }_{ph}}}{2hC}$ ….. (vi).
It is given that the wavelength of the photon is equal to the de Broglie wavelength of the electron.
$\Rightarrow {{\lambda }_{ph}}={{\lambda }_{e}}$.
Substitute this value in (vi).
$\Rightarrow \dfrac{{{E}_{e}}}{{{E}_{ph}}}=\dfrac{m{{v}^{2}}{{\lambda }_{e}}}{2hC}=\dfrac{(mv)v{{\lambda }_{e}}}{2hC}$.
Substitute the values of (mv) and ${{\lambda }_{e}}$ from (iv) and (iii) respectively.
$\Rightarrow \dfrac{{{E}_{e}}}{{{E}_{ph}}}=\dfrac{{{P}_{e}}vh}{2hC{{P}_{e}}}=\dfrac{v}{2C}$.

So, the correct answer is “Option B”.

Note:
Let us find the relation between the momentums of the photon and the electron.
For this divide equation (ii) by equation (iii).
$\Rightarrow \dfrac{{{P}_{ph}}}{{{\lambda }_{e}}}=\dfrac{\dfrac{h}{{{\lambda }_{ph}}}}{\dfrac{h}{{{P}_{e}}}}$
$\Rightarrow \dfrac{{{P}_{ph}}}{{{\lambda }_{e}}}=\dfrac{{{P}_{e}}}{{{\lambda }_{ph}}}$
However, ${{\lambda }_{ph}}={{\lambda }_{e}}$.
This means that ${{P}_{ph}}={{P}_{e}}$.
Therefore, the momentum of the photon is equal to the momentum of the electron.