Q136 / 170 · Manipal (MET) 2019 · Physics
The nuclear reaction ${ }_1 \mathrm{H}^1+{ }_1 \mathrm{H}^1 \rightarrow{ }_2 \mathrm{He}^4$ (mass of deuteron $=2.0141 \mathrm{amu}$ and of $\mathrm{He}=4.0024 \mathrm{amu})$ is
- Correct answerAfusion reaction releasing 24 MeV energy
- Bfusion reaction absorbing 24 MeV energy
- Cfission reaction releasing 0.0258 MeV energy
- Dfission reaction absorbing 0.0258 MeV energy
Solution
${ }_1 \mathrm{H}^1+{ }_1 \mathrm{H}^1 \longrightarrow{ }_2 \mathrm{He}^4$
Mass defect $\begin{aligned} \Delta m & =2 \times \text { mass of } 1_1 \mathrm{H}^1-\text { mass of }{ }_2 \mathrm{He}^4 \\ & =2 \times 2.0141-4.0024 \\ & =0.0258 \end{aligned}$ $\begin{aligned} \text { Energy released } & =\Delta m \times 931 \mathrm{MeV} \\ & =0.0258 \times 931=24 \mathrm{MeV} \end{aligned}$
Mass defect $\begin{aligned} \Delta m & =2 \times \text { mass of } 1_1 \mathrm{H}^1-\text { mass of }{ }_2 \mathrm{He}^4 \\ & =2 \times 2.0141-4.0024 \\ & =0.0258 \end{aligned}$ $\begin{aligned} \text { Energy released } & =\Delta m \times 931 \mathrm{MeV} \\ & =0.0258 \times 931=24 \mathrm{MeV} \end{aligned}$
Hence, the correct answer is option: 'A'.