Question

A material particle with a rest mass ${}^{\prime }{m}_0{}^{\prime }$ is moving with speed of light 'c'. The de-Broglie wavelength associated is given by

• A. $\frac{h}{m_{0}c}$
• B. $\frac{m_{0}c}{h}$
• C. Zero
• D. $\infty$

Answer 1

The correct option is C Zero

Theory of special relativity

$m^{\prime }=\gamma m=\frac{m}{\sqrt{1-\frac{V^2}{c^2}}}$

Debroglie wavelength

$\lambda =\frac{h}{m^{\prime }v}=\frac{h\sqrt{1-\frac{V^2}{c^2}}}{mv}$

If $v=c$
then $\lambda =0$