In a reactor, 2 , kg of { }_{92} U ^{235} fue | vedclass

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(B) $1$. Calculate the number of uranium atoms in $2 \, kg$ of ${ }_{92} U ^{235}$:$n = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{2 \,

Vedclass · Accepted Answer

$60$

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(B) $1$. Calculate the number of uranium atoms in $2 \, kg$ of ${ }_{92} U ^{235}$:$n = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{2 \, kg}{235 \, kg/kmol} 	imes 6.023 	imes 10^{26} \, 	ext{atoms/kmol} \approx 5.126 	imes 10^{24} \, 	ext{atoms}$.$2$. Calculate the total energy released $(E)$:$E = n 	imes 200 \, MeV = 5.126 	imes 10^{24} 	imes 200 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J \approx 1.64 	imes 10^{14} \, J$.$3$. Calculate the power output $(P)$:$P = \frac{E}{t}$, where $t = 30 \, 	ext{days} = 30 	imes 24 	imes 3600 \, s = 2.592 	imes 10^6 \, s$.$P = \frac{1.64 	imes 10^{14}}{2.592 	imes 10^6} \approx 6.326 	imes 10^7 \, W = 63.26 \, MW$.The value is closest to $60 \, MW$.

$A. \text{ } _0^1 n + { }_{92}^{235} U \rightarrow { }_{54}^{140} Xe + { }_{38}^{94} Sr + 2_0^1 n$	$I. \text{ Chemical reaction}$
$B. \text{ } 2H_2 + O_2 \rightarrow 2H_2O$	$II. \text{ Fusion with } +ve \ Q \text{ value}$
$C. \text{ } _1^2 H + _1^2 H \rightarrow _2^3 He + _0^1 n$	$III. \text{ Fission}$
$D. \text{ } _1^1 H + _1^3 H \rightarrow _1^2 H + _1^2 H$	$IV. \text{ Fusion with } -ve \ Q \text{ value}$

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