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

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Modern Physics

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

α \times 1 0^{23}

MeV. The value of

α

____

.
(Avogadro Number

= 6 \times 1 0^{23}

per mole)

Answer: 228

Solution

The number of moles in

47 g

_{92}^{235} U

\frac{47}{235} = \frac{1}{5}

, so the number of atoms is

\frac{1}{5} \times 6 \times 1 0^{23} = 1.2 \times 1 0^{23}

. Each fission releases

190 MeV

, so the total energy released is

1.2 \times 1 0^{23} \times 190 = 228 \times 1 0^{23} MeV

. Thus

α = 228

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Integer answer

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