Question

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is .......... × 10–2 MeV.
(Given 1 u = 931 MeV/c2 , atomic mass of helium = 4.002603 u)

Answer 1

Given reaction:

$^3_2\text{He} \longrightarrow ^{12}_6\text{C} + \gamma \text{ rays}$

Mass defect:

$\Delta m = (3m_{\text{He}} - m_{\text{C}})$

Calculating:

$\Delta m = (3 \times 4.002603 - 12) = 0.007809 \, \text{u}$

Energy released:
$\text{Energy} = 931 \Delta m \, \text{MeV}$ $= 7.27 \, \text{MeV} = 727 \times 10^{-2} \, \text{MeV}$