The average energy released per fission for the nucleus of $_{92}^{235} U$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure $_{92}^{235} U$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is . . . . . . . . . . . (Avogadro Number $= 6 \times 10^{23} \text{ per mole}$)
- A$114$
- B$228$
- C$190$
- D$456$
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View SolutionOn fission,a $U^{235}$ nucleus releases $3 \times 10^{-11} \, J$ of energy. In a $1 \, GW$ nuclear reactor,$4.2 \%$ of this energy is converted to useful energy. The $U^{235}$ consumed (in grams) in half an hour is closest to (Avogadro number $N_A = 6.023 \times 10^{23}$)
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