Question

The nuclear reaction ${ }_{1} H^{1}+{ }_{1} H^{1} \rightarrow{ }_{2} H e^{4}$ (mass of deuteron $=2.0141\, amu$ and of $He =4.0024$ amu) is

• A. fusion reaction releasing 24 MeV energy
• B. fusion reaction absorbing 24 MeV energy
• C. fission reaction releasing 0.0258 MeV energy
• D. fission reaction absorbing 0.0258 MeV energy

Answer 1

Answer: (A)

${ }_{1} H^{1}+{ }_{1} H^{1} \rightarrow{ }_{2} H e^{4}$
Mass defect
$\Delta m =2 \times \text { mass of }{ }_{1} H^{1}-\text { mass of }{ }_{2} H e^{4}$
$=2 \times 2.0141-4.0024=0.0258$
Energy released $=\Delta m \times 931\, MeV =0.0258 \times 931=24\, MeV$