A 1000 ; MW fission reactor consumes half of | vedclass

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Question

(A) The energy released per fission of one $_{92}^{235} U$ nucleus is $200 \; MeV = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} \; J = 3.2 	imes 10^{-

Vedclass · Accepted Answer

$3076$

Vedclass · Answer

(A) The energy released per fission of one $_{92}^{235} U$ nucleus is $200 \; MeV = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} \; J = 3.2 	imes 10^{-11} \; J$.The number of atoms in $1 \; g$ of $_{92}^{235} U$ is $N = \frac{6.023 	imes 10^{23}}{235} \approx 2.563 	imes 10^{21} \; 	ext{atoms/g}$.Energy released per gram of $_{92}^{235} U$ is $E_g = N 	imes 3.2 	imes 10^{-11} \; J/g \approx 8.20 	imes 10^{10} \; J/g$.The reactor power is $P = 1000 \; MW = 10^9 \; W$. It operates $80 \%$ of the time over $5 \; y$.Total time in seconds $T = 5 	imes 365.25 	imes 24 	imes 3600 \approx 1.578 	imes 10^8 \; s$.Total energy produced $E_{total} = P 	imes T 	imes 0.80 = 10^9 	imes 1.578 	imes 10^8 	imes 0.80 = 1.2624 	imes 10^{17} \; J$.Mass of fuel consumed $m = \frac{E_{total}}{E_g} = \frac{1.2624 	imes 10^{17}}{8.20 	imes 10^{10}} \approx 1.539 	imes 10^6 \; g = 1539 \; kg$.Since half the fuel is consumed in $5 \; y$, the initial mass $M_0 = 2 	imes 1539 \approx 3078 \; kg$. The closest option is $3076 \; kg$.

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