Question

Assume that an electron and a positron pair is formed from a gamma ray photon having energy $3.0MeV.$ If the total kinetic energy of the positron and the electron that is formed is $n\times 10^{-2}MeV$ . Find the value of $n$ .
$(m_e=9.1\times 10^{-31}kg,c=3\times 10^{8}m{s}^{-1},e=1.6\times 10^{-19}$

Answer 1

Answer: 198

$\gamma \to \_{-1}^{}{e}^{\circ }+\_{+1}^{}{e}^{\circ }$
From energy-mass equivalence
$E=m_e{C}^2+m_e{C}^2$
$=2m_e{C}^2=\frac{2\times 9.1\times 10^{-31}\times {\left(3\times 10^8\right)}^2}{1.6\times 10^{-19}\times 10^6}$
$=1.02MeV$
$\therefore K.E.=3-1.02=1.98MeV=198\times 10^{-2}MeV$