The disintegration energy Q for the nuclear f | vedclass

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Question

(A) The nuclear fission reaction is: ${ }^{235} U ightarrow{ }^{140} Ce+{ }^{94} Zr+n$.The disintegration energy $Q$ is given by $Q = (m_{	ext{reac

Vedclass · Accepted Answer

$208$

Vedclass · Answer

(A) The nuclear fission reaction is: ${ }^{235} U ightarrow{ }^{140} Ce+{ }^{94} Zr+n$.The disintegration energy $Q$ is given by $Q = (m_{	ext{reactants}} - m_{	ext{products}}) c^2$.Mass of reactants $(m_{	ext{reactants}})$ = $m({ }^{235} U) = 235.0439 	ext{ u}$.Mass of products $(m_{	ext{products}})$ = $m({ }^{140} Ce) + m({ }^{94} Zr) + m(n) = 139.9054 	ext{ u} + 93.9063 	ext{ u} + 1.0086 	ext{ u} = 234.8203 	ext{ u}$.Mass defect $\Delta m = m_{	ext{reactants}} - m_{	ext{products}} = 235.0439 	ext{ u} - 234.8203 	ext{ u} = 0.2236 	ext{ u}$.Disintegration energy $Q = \Delta m 	imes 931 	ext{ MeV/u} = 0.2236 	imes 931 	ext{ MeV} = 208.1716 	ext{ MeV}$.Rounding to the nearest integer,$Q \approx 208 	ext{ MeV}$.

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