Q163 / 180 · KCET 2013 · Physics

The binding energy/nucleon of deuteron $\left({ }_{1} \mathrm{H}^{2}\right)$ and the helium atom $\left({ }_{2} \mathrm{He}^{4}\right)$ are $1.1 \mathrm{MeV}$ and $7 \mathrm{MeV}$ respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is

  • A
    $26.9 \mathrm{MeV}$
  • B
    $25.8 \mathrm{MeV}$
  • Correct answer
    C
    $23.6 \mathrm{MeV}$
  • D
    $12.9 \mathrm{MeV}$

Solution

Given, BE/nucleon for ${ }_{1} \mathrm{H}^{2}=1.1 \mathrm{MeV}$ and $\quad \mathrm{BE} /$ nucleon for ${ }_{2} \mathrm{He}^{4}=7 \mathrm{MeV}$ Also, $\therefore$ $\begin{aligned} &{ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{Q} \\ &Q=\mathrm{BE}_{P}-\mathrm{BE}_{R} \\ &\quad=4 \times 7-4 \times 1.1=23.6 \mathrm{MeV} \end{aligned}$

Hence, the correct answer is option: 'C'.

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