Question

The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is ( $c =$ velocity of light, $h =$ Planck's constant)
A) $h$
B) $c$
C) $\dfrac{1}{c}$
D) None of these

Answer 1

In this solution, we will use the relation of de-Broglie wavelength with the momentum of the particle in question that is inverse in nature. A photon always travels at the speed of light.

Formula used: In this solution, we will use the following formula
$\lambda = \dfrac{h}{p}$ where $\lambda$ is the de Broglie wavelength of a particle travelling with momentum $p$
The energy of the photon ${E_p} = \dfrac{{hc}}{{{\lambda _p}}}$ where $h$ is the Planck's constant, $c$ is the speed of light, and ${\lambda _p}$ is the wavelength of the photon

Complete step by step answer
We’ve been given that the de-Broglie wavelength of an electron and the photon is the same. So, we can say that
${\lambda _e} = {\lambda _p}$
From the de-Broglie relation, we can relate the wavelength of the electron and its momentum as:
${\lambda _e} = \dfrac{h}{{{p_e}}}$
$\Rightarrow {p_e} = \dfrac{h}{{{\lambda _e}}}$
The energy of the photon is defined using the relation
${E_p} = \dfrac{{hc}}{{{\lambda _p}}}$ where ${\lambda _p}$ is the wavelength of the photon.
So, the ratio of the energy of that photon and the momentum of that electron is
$\dfrac{{{E_p}}}{{{p_e}}} = \dfrac{{\dfrac{{hc}}{{{\lambda _p}}}}}{{\dfrac{h}{{{\lambda _e}}}}}$
However since ${\lambda _e} = {\lambda _p}$, we can cancel the terms from above and write
$\dfrac{{{E_p}}}{{{p_e}}} = c$

Hence the ratio of the energy of that photon and the momentum of that electron is $c$ which corresponds to option (B).

Note
We must be careful in not using the de-Broglie relation for photons are photons are massless particles that travel at the speed of light and hence the de-Broglie relation for them involves the energy of the photon i.e. ${\lambda _p} = \dfrac{h}{{{p_p}}}$ where ${p_p} = \dfrac{E}{c}$. Despite being massless, the photons have a non-zero momentum which is associated with the particle behaviour of light where the photon has nonzero momentum due to its non-zero energy.