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Q.

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 $E_{p h}$ and $P_{p h}$ respectively. Which of the following is correct?

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a

$\frac{p_{e}}{p_{p h}} = \frac{2 c}{v}$

b

$\frac{p_{e}}{p_{p h}} = \frac{v}{2 c}$

c

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

d

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

answer is B.

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Detailed Solution

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$λ_{e} = λ_{p h o t o n}$

$\frac{h}{m v} = \frac{h}{P_{p h o t o n}} \Rightarrow P_{p h o t o n} = m v$

$\frac{E_{e}}{E_{p h}} = \frac{\frac{1}{2} m v^{2}}{\frac{h c}{λ}} = \frac{1}{2} \frac{m v}{p_{p h} C} \times v = \frac{v}{2 C}$

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