The average energy released per fission for the nucleus of 235U is 190 MeV when all the atoms of 47 g pure 235U undergo fission process
Q. The average energy released per fission for the nucleus of ${}^{235}_{92}\mathrm{U}$ is $190\,\text{MeV}$. When all the atoms of $47\,\text{g}$ pure ${}^{235}_{92}\mathrm{U}$ undergo fission process, the energy released is $\alpha \times 10^{23}\,\text{MeV}$. The value of $\alpha$ is ____.

(Avogadro Number $= 6 \times 10^{23}$ per mole)
Correct Answer: 228

Explanation

The number of moles of uranium is calculated using:

$$ n = \frac{\text{Given mass}}{\text{Molar mass}} $$

For ${}^{235}\mathrm{U}$:

$$ n = \frac{47}{235} = \frac{1}{5} = 0.2\ \text{mole} $$

The number of uranium nuclei present is:

$$ N = n \times N_A $$

$$ N = 0.2 \times (6 \times 10^{23}) = 1.2 \times 10^{23} $$

Each nucleus releases $190\,\text{MeV}$ of energy during fission. Therefore, total energy released is:

$$ E = N \times \text{Energy per fission} $$

$$ E = (1.2 \times 10^{23}) \times 190 $$

$$ E = 228 \times 10^{23}\ \text{MeV} $$

Comparing with $\alpha \times 10^{23}\,\text{MeV}$:

$$ \alpha = 228 $$

Final Answer: $$ \boxed{228} $$

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