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Q.
Out of a photon and an electron the equation $ E=pc, $ is valid for
ManipalManipal 2008Dual Nature of Radiation and Matter
Solution:
Relativistic energy is given by
$E =\frac{m_{0} c^{2}}{\sqrt{1-v^{2} / c^{2}}} $
$E^{2} =\frac{m_{0}^{2} c^{6}}{c^{2}-v^{2}}\ldots$(i)
Momentum is given by
$p=\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}}$
or $p^{2} c^{2}=\frac{m_{0}^{2} c^{4} v^{2}}{c^{2}-v^{2}}\ldots$(ii)
$\therefore E^{2}-p^{2} c^{2}=m_{0}^{2} c^{4}$
$E^{2}=p^{2} c^{2}+m_{0}^{2} c^{4}$
For photon, rest mass $m_{0}=0$, so $E=p c$
For electron, $m_{0} \neq 0$, so $E \neq p c$