Question

Out of a photon and an electron the equation $E=pc,$ is valid for

• A. both
• B. neither
• C. photon only
• D. electron only

Answer 1

Answer: (C)

Relativistic energy is given by
$E=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}$
$E^2=\frac{m_0^2{c}^6}{c^2-v^2}\dots$(i)
Momentum is given by
$p=\frac{m_{0}v}{\sqrt{1-v^2/c^2}}$
or $p^2{c}^2=\frac{m_0^2{c}^4{v}^2}{c^2-v^2}\dots$(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 $E=pc$
For electron, $m_0\ne 0$, so $E\ne pc$