Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ 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)}

Hint:

Total Energy = Number of Nuclei $\times$ Energy per Nucleus. Remember $N = \frac{m}{M} N_A$.

Answer 1

Correct Answer: 228

First, calculate the number of moles of Uranium-235: $n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{47}{235} = 0.2 \text{ mol}$.
Calculate the number of atoms $N = n \times N_A = 0.2 \times 6 \times 10^{23} = 1.2 \times 10^{23}$.
Total energy released $E = N \times (\text{Energy per fission})$.
$E = 1.2 \times 10^{23} \times 190 \text{ MeV}$.
$E = (1.2 \times 190) \times 10^{23} = 228 \times 10^{23} \text{ MeV}$.
Comparing with $\alpha \times 10^{23}$, we find $\alpha = 228$.