Question

Calculate the energy released when an electron annihilates a positron.

Answer 1

The equation is given by: ${}_{-1}{\beta }^0{+}_{+1}{\beta }^0\to hv+hv$
$\Delta m=2\times 9\times {10}^{-31}$
$=18\times {10}^{-31}$kg
$\Delta E=\Delta m{c}^2$
$=18\times {10}^{-31}\times (3\times {10}^8{)}^2$
$=18\times {10}^{-31}\times 9\times {10}^{16}$
$=162\times {10}^{-15}$
$=1.62\times {10}^{-13}$J
$=\frac{1.62\times {10}^{-13}}{1.6\times {10}^{-19}}$eV
$=1.0125\times {10}^{-6}$eV
$=1.0125$MeV.