The disintegration energy Q for the nuclear fission of is _________ MeV.

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Solution

Q.
The disintegration energy Q for the nuclear fission of ${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$ is _________ MeV.

Given atomic masses of ${}^{235}U : 235.0439 u; C e : 139.905 u {}^{94}Z r : 93.9063 u; n : 1.0086 u,$ Value of $c^{2} = 931 M e V .$ [2024]

Ans.
(208)   ${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$

Disintegration energy $Q = Δ m c^{2}$

   $Q = (m_{R} - m_{P}) \cdot c^{2}$

   $m_{R} = 235.0439 u$

$m_{P} = 139.9054 u + 93.9063 u + 1.0086 u = 234.8203 u$

$Q = (235.0439 u - 234.8203 u) c^{2} = 0.2236 \times 931 MeV$

   $\approx 208.17 MeV \approx 208 MeV$

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