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Q.

An electron and a photon have the same de Broglie wavelength. Then the kinetic energy of the electron is

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a

equal to the kinetic energy of the photon

b

greater than the kinetic energy of the photon

c

zero

d

infinity

answer is B.

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Detailed Solution

$λ = \frac{h}{mv} = \frac{h}{\sqrt{(2 mE)}} Thus, m_{e} E_{e} = m_{p} E_{p} \frac{E_{e}}{E_{p}} = \frac{m_{p}}{m_{e}} > 1 (∵ m_{p} > m_{e}) or E_{e} > E_{p}$

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