The binding energy per nucleon in deuterium and helium nuclei are $1.1 \mathrm{MeV}$ and $7.0 \mathrm{MeV}$,…

The binding energy per nucleon in deuterium and helium nuclei are $1.1 \mathrm{MeV}$ and $7.0 \mathrm{MeV}$, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
  1. $23.6 \mathrm{MeV}$
  2. $2.2 \mathrm{MeV}$
  3. $28.0 \mathrm{MeV}$
  4. $30.2 \mathrm{MeV}$

Solution

Mass of ${ }_1 \mathrm{H}^2=2.01478 \mathrm{amu}$
Mass of ${ }_2 \mathrm{He}^4=4.00388 \mathrm{amu}$
Mass of two deuterium $=2 \times 2.01478$
$=4.02956 \mathrm{amu}$
Energy equivalent to ${ }_2 \mathrm{H}^2$
$\begin{aligned}
& =4.02956 \times 1.2 \\
& =4.4 \mathrm{MeV}
\end{aligned}$
Energy equivalent to ${ }_2 \mathrm{He}^4$
$=4.00388 \times 7=28 \mathrm{MeV}$
$\text {Energy released }=(28-4.4)$
$=23.6 \mathrm{MeV}$

Asked in: NEET 2010 (Mains)

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