On bombarding U^{235} by slow neutrons,200 te | vedclass

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(B) The energy released per fission is $E = 200 	ext{ MeV} = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-11} 	ext{ J}$.Gi

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$5 	imes 10^{16} / s$

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(B) The energy released per fission is $E = 200 	ext{ MeV} = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-11} 	ext{ J}$.Given power output $P = 1.6 	ext{ MW} = 1.6 	imes 10^6 	ext{ W}$.The rate of fission $R$ is given by the formula $R = P / E$.Substituting the values: $R = (1.6 	imes 10^6) / (3.2 	imes 10^{-11})$.$R = 0.5 	imes 10^{17} = 5 	imes 10^{16} 	ext{ fissions/s}$.

On bombarding $U^{235}$ by slow neutrons,$200 \text{ MeV}$ of energy is released. If the power output of an atomic reactor is $1.6 \text{ MW}$,then the rate of fission will be:

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