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QJEEM23Phee7e: JEE Main 2023 Physics - Class 12 JEEM23Phee7e Solution: The energy released per fission of nucleus of X is 200 MeV. The …

The energy released per fission of nucleus of 240^{240} X is 200 MeV. The energy released if all the atoms in 120g of pure 240^{240} X undergo fission is ____________ ×\times 10 25^{25} MeV.

(Given NA=6×1023\mathrm{N_A=6\times10^{23}} )

Step-by-Step Verified Solution

120 g120 \mathrm{~g} of 240X^{240}X will have 12\frac{1}{2} mole of XX

Number of atom of X=12×NA=3×1023X=\frac{1}{2} \times N_{A}=3 \times 10^{23} atom

Energy released =3×1023×200 MeV=3 \times 10^{23} \times 200 ~ \mathrm{MeV}

=6×1025 MeV=6 \times 10^{25} ~\mathrm{MeV}

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