Deuterium nuclei fuse by the reaction
\({}^{2}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\rightarrow{}^{3}_{2}\mathrm{He}+\mathrm{n}+3.27\ \mathrm{MeV}\)
All the deuterium in a 1 g sample is used up in this reaction. The energy released is
(Take the molar mass of deuterium as 2 g mol−1 and \(N_A=6\times10^{23}\ \mathrm{mol^{-1}}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4.9\times10^{23}\ \mathrm{MeV}\)
Each reaction uses two deuterium nuclei and releases 3.27 MeV. So the number of reactions is half the number of nuclei in the sample.
Number of deuterium nuclei in 1 g (half a mole):
\( \begin{aligned} N &= \frac{1}{2}\times6\times10^{23} \\ &= 3\times10^{23} \end{aligned} \)
Number of reactions and the energy released:
\( \begin{aligned} n &= \frac{N}{2} \\ &= 1.5\times10^{23} \end{aligned} \)
\( \begin{aligned} E &= n\times3.27\ \mathrm{MeV} \\ &= 1.5\times10^{23}\times3.27 \\ &\approx 4.9\times10^{23}\ \mathrm{MeV} \end{aligned} \)
Hence, 1 g of deuterium releases about \( 4.9\times10^{23}\ \mathrm{MeV} \) by this reaction.
NCERT (2026-27):
Nuclear fusion – energy generation in stars – “… In reaction [13.13(b)], two deuterons combine to form the light isotope of helium. …”
\( {}^{2}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}\rightarrow{}^{3}_{2}\mathrm{He}+\mathrm{n}+3.27\ \mathrm{MeV} \qquad [13.13(b)] \)