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Q.

A photon and an electron possess same de-Broglie wavelength. Given that c is speed of light and v, the speed of electrons, which of the following relations is correct ? Here $E_{e}$ = K.E. of electron, $E_{ph}$ = K.E. of photon, $p_{e}$ = momentum of electron, $p_{ph}$ = momentum of photon

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a

$E_{e} / E_{ph} = \frac{2 c}{V}$

b

$E_{e} / E_{ph} = \frac{V}{2 c}$

c

$p_{e} / p_{ph} = \frac{2 c}{V}$

d

$p_{e} / p_{ph} = \frac{c}{2 v}$

answer is B.

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Detailed Solution

$E_{e} = \frac{1}{2} {mv}^{2} = \frac{1}{2} (mv) v = \frac{1}{2} \times \frac{h}{λ} \times V E_{ph} = \frac{hc}{λ} ∴ E_{e} / E_{ph} = V / 2 c$

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A photon and an electron possess same de-Broglie wavelength. Given that c is speed of light and v, the speed of electrons, which of the following relations is correct ? Here Ee = K.E. of electron, Eph = K.E. of photon, pe= momentum of electron, pph = momentum of photon