Physics | VITEEE 2019 | Discussion Page

If the kinetic energy of a free electron doubles, it's de-Broglie wavelength changes by the factor

A) 2

$\frac{1}{2}$

$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

Option D

de-Broglie wavelength,

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2.m.(K.E)}}$

$\lambda \propto \frac{1}{\sqrt{K.E}}$

If K.E is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$

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