Photoelectric Effect and WaveParticle Duality — Physics STD 12 Science — Question

$\text{E}=\text{mc}^2$

^{$\Rightarrow\text{E}=\frac{\text{m}_0\text{c}^2}{\sqrt{1\frac{\text{v}^2}{\text{c}^2}}}$}

Relativistic momentum,

$\text{p}=\text{mv}$

$\Rightarrow\text{p}=\frac{\text{m}_0\text{c}^2}{\sqrt{1\frac{\text{v}^2}{\text{c}^2}}}$

Combining the above equations, we get:

$\text{E}^2=\text{m}_0^2\text{c}^ 4+\text{p}^2\text{c}^2$

From the above equation, it is clear that for $\text{p}=\frac{\text{E}}{\text{c}}$ to be valid, the rest mass of the body should be zero. As electrons do not have zero rest mass, this equation is not valid for electrons.