The energy released per fission of nucleus of ^240 X is 200 MeV. The energy released if all the atoms in 120g of pure ^240 X undergo fission is \underline\hspace2cm\times 10 ^25 MeV. (Given N_A=6\times10^23 )

Atoms and Nuclei PYQ — JEE Main 2023 · 24 Jan · Shift 2 · Q66 with Solution

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Correct answer: 6

Since the molar mass of $^{240}X$ is $240\text{ g/mol}$ ,

\text{moles} = \frac{120}{240} = 0.5

Using Avogadro's number $N_A = 6\times 10^{23}$ ,

N = 0.5 \times 6\times 10^{23} = 3\times 10^{23}

Given energy per fission = $200\,\text{MeV}$ .

So total energy released is

E = (3\times 10^{23})(200)

E = 6\times 10^{25}\,\text{MeV}

The question asks for

\underline{\hspace{2cm}} \times 10^{25}\,\text{MeV}

Thus, the required integer is

6

\boxed{6}

Stored correct answer = $6$

Our derived answer = $6$

So, the answer agrees with the stored correct answer.