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Atoms & Nuclei

JEE Main level, numerical answer, medium

216 of 239

Question 216 of 239

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 $\alpha \times 10^{23} \mathrm{MeV}$ . The value of $\alpha$ is $\_\_\_\_$ .
(Avogadro Number $=6 \times 10^{23}$ per mole)

Show answer and solution

Answer: 228

The number of

^{235}\text{U}

atoms in 47 g is determined using Avogadro's number.
Molar mass of

^{235}\text{U}

is 235 g/mol, so the number of moles is

n = 47/235

mol.
The total number of atoms is

N = (47/235) \times 6 \times 10^{23}

.
Since each fission releases 190 MeV, the total energy released is:

E_{total} = N \times 190 = \frac{47 \times 6 \times 10^{23}}{235} \times 190 = \frac{47 \times 6 \times 190}{235} \times 10^{23}

MeV.
Calculating the numerator:

47 \times 6 \times 190 = 282 \times 190 = 53580

.
Therefore:

E_{total} = \frac{53580}{235} \times 10^{23} = 228 \times 10^{23}

MeV, giving

\alpha = 228

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