JEE Physics 2026 Question with Solution

Given: Energy released per fission is $190 \, \text{MeV}$. Mass of pure $^{235}_{92}U$ is $47 \, \text{g}$. Avogadro number is $6 \times 10^{23}$ per mole.

Find: The value of $\alpha$ in total energy $= \alpha \times 10^{23} \, \text{MeV}$.

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 get $\alpha = 228$.

Therefore, the required numerical value is $228$.

Given: Total Energy = Number of Nuclei $\times$ Energy per Nucleus.

Find: The numerical coefficient multiplying $10^{23}$.

Use

$$N = \frac{m}{M} N_A$$

Substitute the given values:

$$N = \frac{47}{235} \times 6 \times 10^{23}$$ $$N = 0.2 \times 6 \times 10^{23} = 1.2 \times 10^{23}$$

Now multiply by energy released per fission:

$$E = 1.2 \times 10^{23} \times 190 \, \text{MeV}$$ $$E = 228 \times 10^{23} \, \text{MeV}$$

Hence, in the form $\alpha \times 10^{23} \, \text{MeV}$, the value of $\alpha$ is $228$.