KCET · Physics · Nuclear 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 \(25.8 \mathrm{MeV}\)
- C \(23.6 \mathrm{MeV}\)
- D \(12.9 \mathrm{MeV}\)
Answer & Solution
Correct Answer
(C) \(23.6 \mathrm{MeV}\)
Step-by-step Solution
Detailed explanation
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, \(\therefore\)
\(\begin{aligned}
&{ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{Q} \\
&Q=\mathrm{BE}_{P}-\mathrm{BE}_{R} \\
&\quad=4 \times 7-4 \times 1.1=23.6 \mathrm{MeV}
\end{aligned}\)
\(\begin{aligned}
&{ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{Q} \\
&Q=\mathrm{BE}_{P}-\mathrm{BE}_{R} \\
&\quad=4 \times 7-4 \times 1.1=23.6 \mathrm{MeV}
\end{aligned}\)
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