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. $\frac{1}{c}$
• D. None of these

Answer 1

The correct option is B $c$

We have ${\lambda }_e=\frac{h}{mv}$

and ${\lambda }_p=\frac{h}{mc}$

According to the question,

${\lambda }_e={\lambda }_p$

$\therefore v=c$

$\frac{E_p}{P_e}=\frac{m{c}^2}{mv}=\frac{c^2}{c}=c$