A) both
B) neither
C) photon only
D) electron only
Correct Answer: C
Solution :
Relativistic energy is given by \[E=\frac{{{m}_{0}}{{c}^{2}}}{\sqrt{1-{{v}^{2}}/{{c}^{2}}}}\] Or \[{{E}^{2}}=\frac{m_{0}^{2}{{c}^{6}}}{{{c}^{2}}-{{v}^{2}}}\] ...(i) Momentum is given by \[P=\frac{{{m}_{0}}v}{\sqrt{1-{{v}^{2}}/{{c}^{2}}}}\] Or \[{{p}^{2}}{{c}^{2}}=\frac{m_{0}^{2}{{c}^{4}}{{v}^{2}}}{{{c}^{2}}-{{v}^{2}}}\] ?(ii) \[\therefore \] \[{{E}^{2}}-{{p}^{2}}{{c}^{2}}=m_{0}^{2}{{c}^{4}}\] Or \[{{E}^{2}}={{p}^{2}}{{c}^{2}}+m_{0}^{2}{{c}^{4}}\] For photon, rest mass\[{{m}_{0}}=0,\]so \[E=pc\] For electron, \[{{m}_{0}}\ne 0,\] so \[E\ne pc\]
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