An electron with speed V and a photon with speed c have the same de Broglie wavelength. If the kinetic energy and momentum of electrons is E_e and P_e and that of photon is E_ph and P_ph respectively, then the correct statement is

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$\frac{E_{e}}{E_{p h}} = \frac{2 c}{V}$

$\frac{E_{e}}{E_{p h}} = \frac{V}{2 c}$

$\frac{P_{e}}{P_{p h}} = \frac{2 c}{V}$

$\frac{P_{e}}{P_{p h}} = \frac{V}{2 c}$

answer is B.

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Detailed Solution

$E_{p h o t o n} = \frac{h c}{λ} E_{e} = \frac{h}{m V}$

$\begin{array}{l} λ_{e} = λ_{ph} \\ \frac{h}{p_{e}} = \frac{h}{p_{ph}} \\ \sqrt{2 mE} = \frac{E_{ph}}{c} \\ 2 {mEE}_{e} = \frac{{E_{ph}}^{2}}{c^{2}} \\ \frac{E_{e}}{E_{ph}} = \frac{E_{ph}}{c^{2}} \frac{1}{2 m} = \frac{p_{ph}}{c} \frac{1}{2 m} = \frac{p_{e}}{c} \frac{1}{2 m} = \frac{mv}{c} \frac{1}{2 m} = \frac{v}{2 c} \end{array}$