A nuclear reactor generates power by fission | vedclass

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Question

(A) The fission reaction is: ${}_{92}U^{235} ightarrow 2 	imes {}_{46}Pd^{116} + 3{}_{0}n^{1}$.Binding energy of parent nucleus ${}_{92}U^{235} = 2

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The fission reaction is: ${}_{92}U^{235} ightarrow 2 	imes {}_{46}Pd^{116} + 3{}_{0}n^{1}$.Binding energy of parent nucleus ${}_{92}U^{235} = 235 	imes 7.2 	ext{ MeV} = 1692 	ext{ MeV}$.Binding energy of two fragments ${}_{46}Pd^{116} = 2 	imes (116 	imes 8.2 	ext{ MeV}) = 2 	imes 951.2 	ext{ MeV} = 1902.4 	ext{ MeV}$.Total energy released in fission $Q = (B.E._{	ext{products}} - B.E._{	ext{reactants}}) = 1902.4 - 1692 = 210.4 	ext{ MeV}$.Energy carried away by two $\gamma$-rays $= 2 	imes 5.2 	ext{ MeV} = 10.4 	ext{ MeV}$.Usable energy per fission event $= Q - E_{\gamma} = 210.4 	ext{ MeV} - 10.4 	ext{ MeV} = 200 	ext{ MeV}$.

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