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

Hint: 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.