In the sun, four hydrogen atoms combine to form one helium atom with the release of 26.7 MeV. The energy released when 1 g of hydrogen is fully changed into helium in this way is
(Take the molar mass of hydrogen as 1 g mol−1, \(N_A=6\times10^{23}\ \mathrm{mol^{-1}}\) and \(1\ \mathrm{MeV}=1.6\times10^{-13}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(6.4\times10^{11}\ \mathrm{J}\)
One gram of hydrogen is one mole, so it has \( 6\times10^{23} \) atoms. Every four atoms form one helium atom and give 26.7 MeV.
\( \begin{aligned} n &= \frac{6\times10^{23}}{4} \\ &= 1.5\times10^{23} \end{aligned} \)
\( \begin{aligned} E &= 1.5\times10^{23}\times26.7\ \mathrm{MeV} \\ &\approx 4.0\times10^{24}\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} E &= 4.0\times10^{24}\times1.6\times10^{-13}\ \mathrm{J} \\ &\approx 6.4\times10^{11}\ \mathrm{J} \end{aligned} \)
Hence, about \( 6.4\times10^{11}\ \mathrm{J} \) is released when 1 g of hydrogen turns into helium.
NCERT (2026-27):
Nuclear fusion – energy generation in stars – “Thus, four hydrogen atoms combine to form an 42He atom with a release of 26.7 MeV of energy.”