The disintegration energy Q for the nuclear fission of ²³⁵ U → ¹⁴⁰ Ce+ ⁹⁴ Zr+n is _ _ _ M eV. Given atomic masses of ²³⁵ U: 235.0439 u; ¹⁴⁰ Ce: 139.9054 u, ⁹⁴ Zr: 93.9063 u; n: 1.0086 u, Value of c²=931 ~MeV / u.

Atoms and Nuclei: JEE Main 2024 Physics question

JEE Main 2024 · 4 Apr, Evening shift · Q56

JEE Main 20244 Apr, Evening shiftPhysicsHardAtoms and Nuclei183× since 2002

The disintegration energy

Q

for the nuclear fission of

{ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+n

is _______

\mathrm{MeV}

. Given atomic masses of

{ }^{235} \mathrm{U}: 235.0439 u ;{ }^{140} \mathrm{Ce}: 139.9054 u, { }^{94} \mathrm{Zr}: 93.9063 u ; n: 1.0086 u

, Value of

c^2=931 \mathrm{~MeV} / \mathrm{u}

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2024 Atoms and Nuclei Question with Answer (4 Apr Evening shift, Q56) | JEE Arc