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Question

(C) The number of nuclei in $1\, kg$ of $_{92}{U^{235}}$ is calculated using Avogadro's number $(N_A = 6.023 	imes 10^{23}\, 	ext{atoms/mol})$.Numbe

Vedclass · Accepted Answer

$8.2 	imes 10^{13}\, J$

Vedclass · Answer

(C) The number of nuclei in $1\, kg$ of $_{92}{U^{235}}$ is calculated using Avogadro's number $(N_A = 6.023 	imes 10^{23}\, 	ext{atoms/mol})$.Number of nuclei $N = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{1000\, g}{235\, g/mol} 	imes 6.023 	imes 10^{23}\, 	ext{nuclei/mol}$.$N \approx 2.563 	imes 10^{24}\, 	ext{nuclei}$.Energy released per nucleus is $200\, MeV$.Total energy $E = N 	imes 200\, MeV = 2.563 	imes 10^{24} 	imes 200 	imes 1.602 	imes 10^{-13}\, J$.$E \approx 8.21 	imes 10^{13}\, J$.Thus, the correct option is $C$.

The energy liberated on complete fission of $1\, kg$ of $_{92}{U^{235}}$ is (Assume $200\, MeV$ energy is liberated on fission of $1$ nucleus).

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