The energy liberated on complete fission of 1 | vedclass

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Question

(C) The number of atoms in $1 \, kg$ of $_{92}U^{235}$ is given by $N = \frac{m}{M} 	imes N_A$,where $m = 1000 \, g$,$M = 235 \, g/mol$,and $N_A = 6.

Vedclass · Accepted Answer

$8.2 	imes 10^{13} \, J$

Vedclass · Answer

(C) The number of atoms in $1 \, kg$ of $_{92}U^{235}$ is given by $N = \frac{m}{M} 	imes N_A$,where $m = 1000 \, g$,$M = 235 \, g/mol$,and $N_A = 6.023 	imes 10^{23} \, atoms/mol$.$N = \frac{1000}{235} 	imes 6.023 	imes 10^{23} \approx 2.563 	imes 10^{24} \, nuclei$.Energy released per nucleus is $200 \, MeV = 200 	imes 1.602 	imes 10^{-13} \, J = 3.204 	imes 10^{-11} \, J$.Total energy liberated $E = N 	imes 200 \, MeV = 2.563 	imes 10^{24} 	imes 3.204 	imes 10^{-11} \, J$.$E \approx 8.21 	imes 10^{13} \, J \approx 8.2 	imes 10^{13} \, J$.

The energy liberated on complete fission of $1\, kg$ of $_{92}U^{235}$ is (Assume $200\, MeV$ energy is liberated on fission of $1$ nucleus).

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