The energy released in the fission of ${\rm{1}}\,{\rm{Kg}}$ of _92 $_{92}{{\rm{U}}^{235}}$ is (energy per fission ${\rm{ = 200 }}\,{\rm{MeV}}$)

$8.2\,\, \times \,\,{10^{13}}\,\,{\rm{J}}$

$5.1\,\, \times \,\,{10^{26}}\,\,{\rm{eV}}$

$5.1\,\, \times \,\,{10^{26}}\,\,{\rm{J}}$

$8.2\,\, \times \,\,{10^{13}}\,\,{\rm{MeV}}$

JEE PHYSICS MCQ: NUCLEI

1 Medium

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Application +4 | -1

The energy released in the fission of

\mathrm{1}\,\mathrm{Kg}

of _92

_{92}{\mathrm{U}^{235}}

is (energy per fission

\mathrm{ = 200 }\,\mathrm{MeV}

)

View answer and explanation

Correct option: C

8.2\,\, \times \,\,{10^{13}}\,\,\mathrm{J}

Option C (

8.2 \times 10^{13} J

) is correct because it accurately calculates the energy released from 1 kg of Uranium-235 using the energy per fission and Avogadro's number.