The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n is _______ MeV. Given atomic masses of ²³⁵U: 235.0439u; ¹⁴⁰Ce: 139.9054u, ⁹⁴Zr: 93.9063u; n: 1.0086u, Value of c² = 931 MeV/u. Numerical answer type. Work it out, then open the answer.

Answer: 208. Q = [mU - (mCe + mZr + mn)] c². Products: 139.9054 + 93.9063 + 1.0086 = 234.8203 u, so Δ m = 235.0439 - 234.8203 = 0.2236 u. Q = 0.2236×931 ≈ 208 MeV.

The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce +… | JEE Main 2024 (4 Apr, Shift 2)

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Nuclear Fission & Fusion

The disintegration energy $Q$ for the nuclear fission of ${}^{235}\text{U} \rightarrow {}^{140}\text{Ce} + {}^{94}\text{Zr} + n$ is _______ $\text{MeV}$ . Given atomic masses of ${}^{235}\text{U} : 235.0439u;\ {}^{140}\text{Ce} : 139.9054u,\ {}^{94}\text{Zr} : 93.9063u;\ n : 1.0086u$ , Value of $c^{2} = 931\text{ MeV/u}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 208

$Q = [m_U - (m_{Ce} + m_{Zr} + m_n)]\,c^2$ .
Products: $139.9054 + 93.9063 + 1.0086 = 234.8203\text{ u}$ , so $\Delta m = 235.0439 - 234.8203 = 0.2236\text{ u}$ .
$Q = 0.2236\times931 \approx 208\text{ MeV}$ .

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