Question

The wave nature of particles was studied using diffraction of particle beams by crystal lattices. The wavelength of the waves associated with fast moving particles was found to be in agreement with the de Broglie relation.

For a particle of mass $m$ moving with kinetic energy $E$, the de Broglie wavelength is

• A. $\frac{h}{2mE}$
• B. $h\sqrt{2mE}$
• C. $\frac{h}{\sqrt{2mE}}$
• D. $h\sqrt{\frac{2}{mE}}$

Answer 1

Answer: (C)

$\frac{h}{\sqrt{2mE}}$

The energy of photon is
$E=h\nu$
According to Einstein's energy mass relation
$E=m{c}^2$
From above equations
$h\nu =m{c}^2$

$\frac{hc}{\lambda }=pc$ ..................(since, $p=mc$)

$\lambda =\frac{h}{p}$

This is the De-Broglie's equation for wavelength for the photon.
For the particle having mass m and velocity v, De-Broglie's equation for wavelength becomes

$\lambda =\frac{h}{mv}$ .................(1)

The energy of the particle is

$E=\frac{1}{2}m{v}^2$

$E=\frac{p^2}{2m}$

$p=\sqrt{2mE}$
using this value in (1), we get

$\lambda =\frac{h}{\sqrt{2mE}}$

So, the answer is an option (C).