The fission of a _{92}U^{235} nucleus release | vedclass

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(B) Power $P$ is defined as the total energy $E$ released per unit time $t$,given by $P = \frac{N 	imes E'}{t}$,where $N$ is the number of nuclei and

Vedclass · Accepted Answer

$1.56 	imes 10^{11} \, s^{-1}$

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(B) Power $P$ is defined as the total energy $E$ released per unit time $t$,given by $P = \frac{N 	imes E'}{t}$,where $N$ is the number of nuclei and $E'$ is the energy per fission.The rate of fission is given by $\frac{N}{t} = \frac{P}{E'}$.Given: $P = 5 \, W$ (Joules/second) and $E' = 200 \, MeV = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-11} \, J$.Substituting the values:$\frac{N}{t} = \frac{5}{3.2 	imes 10^{-11}} = 1.5625 	imes 10^{11} \, s^{-1}$.Thus,the rate of fission is $1.56 	imes 10^{11} \, s^{-1}$.

The fission of a $_{92}U^{235}$ nucleus releases $200 \, MeV$ of energy. Find the rate of fission of $_{92}U^{235}$ required to operate a reactor at a constant power of $5 \, W$.

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