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Q.
A particle with rest mass zero is moving with speed $c$. The de-Broglie wavelength associated with it
J & K CETJ & K CET 2004
Solution:
The de-Broglie wavelength $(\lambda)$ is given by
$\lambda=\frac{h}{p}$ ... (i)
where $h$ is Planck's constant, $p$ the momentum.
Also, $p=m v=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{C^{2}}}}$ ... (ii)
where $m_{0}$ is rest mass, $v$ the velocity and $c$ the speed of light.
From Eqs. (i) and (ii),
we get $\lambda=\frac{h \sqrt{1-\frac{v^{2}}{c^{2}}}}{m_{0} v}$
Given: $v=c$
$\therefore \lambda=0$