The energy released per fission of uranium-23 | vedclass

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Question

(C) The power produced by the reactor is $P = 1000 \, kW = 10^6 \, J/s$.The energy released per fission is $E_f = 200 \, MeV = 200 	imes 10^6 \, eV$.

Vedclass · Accepted Answer

$3 	imes 10^{16}$

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(C) The power produced by the reactor is $P = 1000 \, kW = 10^6 \, J/s$.The energy released per fission is $E_f = 200 \, MeV = 200 	imes 10^6 \, eV$.Converting this energy into Joules:$E_f = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-11} \, J$.The number of fissions per second $(n)$ is given by the ratio of total power to energy per fission:$n = \frac{P}{E_f} = \frac{10^6 \, J/s}{3.2 	imes 10^{-11} \, J/fission}$.$n = \frac{1}{3.2} 	imes 10^{17} \approx 0.3125 	imes 10^{17} = 3.125 	imes 10^{16} \, 	ext{fissions/s}$.Thus, the approximate number of nuclei undergoing fission per second is $3 	imes 10^{16}$.

The energy released per fission of uranium-$235$ is about $200 \, MeV$. $A$ reactor using $U-235$ as fuel is producing $1000 \, kW$ power. The number of $U-235$ nuclei undergoing fission per second is approximately:

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