Question

What is the energy released in this nuclear reaction 42 19 K ? 42 20 C a + 0 ? 1 e ? (The atomic mass of 42 K is 41.962403 u and that of 42 C a is 41.958618 u) (MeV)

Answer 1

amu of Potassium = 41.962403 u

amu of Calcium = 41.958618 u

amu of electron = 0.000549 u

Mass defect,

?m = 41.962403 u -  41.958618 u -  0.000549 u

?m = 0.003236 u

But 1 u = 1.66 x 10^-27 kg

thus 0.003236 u = 0.00537176 x 10 ^-27 kg

The energy released, ?E = ?m c² , where c = 2.99 x 10⁸ m s^-1

Therefore ?E = 0.00537176 X 10^-27 x (3 x 10⁸)²

Thus E = 0.04834584 x 10^-11 J

We know that 1 MeV = 1.602 x 10^-13 J  

Therefore energy released E = 3.01784 MeV.