The average energy released per fission for the nucleus of \({ }_{92}^{235} U\) is 190 MeV . When all the atoms of 47 g pure \({ }_{92}^{235} U\) undergo fission process, the energy released is \(\alpha \times 10^{23} MeV\). The value of \(\alpha\) is ___________.
(Avogadro Number \(=6 \times 10^{23}\) per mole)

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Correct Answer

(B) 228

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Detailed explanation

Total numbers of U-235 atom is
\(47 g=\frac{47}{235}\) moles \(=\frac{1}{5}\) moles
∴ Total energy released \(=\frac{1}{5} \times 6 \times 10^{23} \times 190 MeV\)
\(=228 \times 10^{23} MeV\)

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