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 $\mathrm{E}_{\mathrm{e}}$ and $\mathrm{p}_{\mathrm{e}}$ and that of photon are $\mathrm{E}_{\mathrm{ph}}$ and $\mathrm{p}_{\mathrm{ph}}$ respectively. Which of the following is correct?

(B) $\frac{\mathrm{E}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{ph}}}=\frac{\mathrm{v}}{2 \mathrm{c}}$

(A) $\frac{E_{e}}{E_{ph}}=\frac{2 c}{v}$

(C) $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\mathrm{ph}}}=\frac{2 \mathrm{c}}{\mathrm{v}}$

(D) $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\mathrm{ph}}}=\frac{\mathrm{v}}{2 \mathrm{c}}$

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Modern Physics

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?

Answer: (B)

Since the de-Broglie wavelengths are equal:

λ_{e} = λ_{p h}

, we have

\frac{h}{m v} = \frac{h}{p _{p h}}

, so

p_{p h} = m v = p_{e}

(momenta are equal). The kinetic energy of the electron is

E_{e} = \frac{1}{2} m v^{2} = \frac{p _{e} v}{2}

. The photon energy is

E_{p h} = p c = p_{p h} c

. Therefore,

\frac{E _{e}}{E _{p h}} = \frac{p _{e} v /2}{p _{p h} c} = \frac{v}{2 c}

(since

p_{e} = p_{p h}

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option