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 $n$ .
$\left( m _{ e }=9.1 \times 10^{-31} kg , \quad c =3 \times 10^{8} \quad m \quad s ^{-1}, \quad e =1.6 \times 10^{-19}\right.$

Answer 1

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