In a reactor, each fission of a \( {}^{235}_{92}\mathrm{U} \) nucleus releases 200 MeV of energy. The energy released if all the nuclei in 2.35 g of \( {}^{235}_{92}\mathrm{U} \) undergo fission is (Avogadro number \( N_A=6\times10^{23}\,\mathrm{mol^{-1}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.2\times10^{24}\,\mathrm{MeV} \)
One mole of \( {}^{235}_{92}\mathrm{U} \) has a mass of 235 g and contains \( N_A \) nuclei. So the number of nuclei in a mass \( m \) is \( N=(m/M)\,N_A \). Each of them gives 200 MeV when it undergoes fission.
The number of nuclei in 2.35 g of uranium-235 is:
\( \begin{aligned} N &= \frac{m}{M}\,N_A \\ &= \frac{2.35\,\mathrm{g}}{235\,\mathrm{g\,mol^{-1}}}\times6\times10^{23}\,\mathrm{mol^{-1}} \\ &= 6\times10^{21} \end{aligned} \)
The total energy released when all of them undergo fission is:
\( \begin{aligned} E &= N\times200\,\mathrm{MeV} \\ &= 6\times10^{21}\times200\,\mathrm{MeV} \\ &= 1.2\times10^{24}\,\mathrm{MeV} \end{aligned} \)