An electron with speed v and a photon with speed c have the same...

An electron with speed

v

and a photon with speed

c

have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are

{E}_{e}

and

{P}_{e}

and that of photon are

{E}_{ph}

and

{P}_{ph}

respectively. Which of the following is correct?

Options

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 $\dfrac{{p}_{e}}{{p}_{ph}} = \dfrac{v}{2c}$

Correct answer

B. $\dfrac{{E}_{e}}{{E}_{ph}} = \dfrac{v}{2c}$

Step-by-step solution

We have, ${\lambda}_{p} = \dfrac{h}{{p}_{p}}$ and ${\lambda}_{e} = \dfrac{h}{{p}_{e}}$

Given here, ${\lambda}_{e} = {\lambda}_{p}$ , then ${p}_{p} = {p}_{e}$

Ratio of their energy is $\dfrac{{E}_{ph}}{{E}_{e}} = \dfrac{\dfrac{hc}{{\lambda}_{P}}}{{p}_{e}^{2}2m} = \dfrac{2mc}{{p}_{e}} = \dfrac{2mc}{mv} = \dfrac{2c}{v}$

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