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(B) Given: Power $P = 300 \, MW = 300 	imes 10^6 \, J/s$. Energy per fission $E = 170 \, MeV = 170 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J$. Powe

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$4 	imes 10^{22}$

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(B) Given: Power $P = 300 \, MW = 300 	imes 10^6 \, J/s$. Energy per fission $E = 170 \, MeV = 170 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J$. Power is defined as the total energy released per unit time: $P = \frac{n \cdot E}{t}$, where $n$ is the number of atoms fissioned in time $t$. Rearranging for the rate of fission per second: $\frac{n}{t} = \frac{P}{E} = \frac{300 	imes 10^6}{170 	imes 10^6 	imes 1.6 	imes 10^{-19}}$. $\frac{n}{t} = \frac{300}{170 	imes 1.6 	imes 10^{-19}} \approx 1.103 	imes 10^{19} \, 	ext{atoms/second}$. To find the number of atoms fissioned per hour, multiply by $3600 \, s$: $n_{	ext{hour}} = 1.103 	imes 10^{19} 	imes 3600 \approx 3.97 	imes 10^{22} \approx 4 	imes 10^{22} \, 	ext{atoms}$. Thus, the correct option is $B$.

An atomic power nuclear reactor can deliver $300 \, MW$. The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 \, MeV$. The number of uranium atoms fissioned per hour will be (approximately):

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