Explanation

(a) The de-Broglie wavelength ($\lambda)$ is given by
$\lambda = \frac{h}{p}$ …(i)
Where $h$ is Planck’s constant, $p$ the momentum.
Also, $p = mv = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ …(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, $v = c\ \ \ \ \ \ \ \therefore\lambda = 0$