Physics | MHT CET 2018 | Discussion Page

The energy of an electron having de-Broglie wavelength $\lambda$ is

(where, h= planck's constant, m= mass of electron)

$\frac{h}{2m\lambda}$

$\frac{h^{2}}{2m\lambda^{2}}$

$\frac{h^{2}}{2m^{2}\lambda^{2}}$

$\frac{h^{2}}{2m^{2}\lambda^{}}$

Option B

We know that ,

$\lambda= \frac{h}{\sqrt{2m(KE)}}$

on taking square on both sides,we get

$\lambda^{2}= \frac{h^{2}}{2m(KE)}$

$KE = \frac{h^{2}}{2m\lambda^{2}}$

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