_{92}U^{235} + n to text{fission product} + t | vedclass

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(C) The number of atoms in $1 \ g$ of $_{92}U^{235}$ is given by $\frac{1 \ g}{235 \ g/mol} 	imes 6.023 	imes 10^{23} \ 	ext{atoms/mol} \approx 2.5

Vedclass · Accepted Answer

$8.21 	imes 10^7 \ kJ$

Vedclass · Answer

(C) The number of atoms in $1 \ g$ of $_{92}U^{235}$ is given by $\frac{1 \ g}{235 \ g/mol} 	imes 6.023 	imes 10^{23} \ 	ext{atoms/mol} \approx 2.563 	imes 10^{21} \ 	ext{atoms}$.Energy released per atom is $3.20 	imes 10^{-11} \ J$.Total energy released $= (2.563 	imes 10^{21} \ 	ext{atoms}) 	imes (3.20 	imes 10^{-11} \ J/	ext{atom}) = 8.2016 	imes 10^{10} \ J$.Converting to $kJ$: $8.2016 	imes 10^{10} \ J = 8.2016 	imes 10^7 \ kJ \approx 8.21 	imes 10^7 \ kJ$.

$_{92}U^{235} + n \to \text{fission product} + \text{neutron} + 3.20 \times 10^{-11} \ J$. The energy released when $1 \ g$ of $_{92}U^{235}$ undergoes fission is

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