Explanation

(a) Einstein’s mass energy relation , the energy released is
$\Delta E = \Delta mc^{2}$
Where, $c$ is speed of light and $\ \Delta m$ is mass.
Given,
$\Delta m = 1.67\ \times \ 10^{- 27}\ kg\ ,\ c\ = \ 3\ \times \ 10^{8}\ ms^{- 1}$
$\therefore\ \ \ \Delta E = 1.67\ \times \ 10^{- 27}\ \times \ \left( 3\ \times \ 10^{8} \right)^{2}$
$\Delta E \approx \ 1.5\ \times \ 10^{- 10}\ J$