The fission properties of _{94}^{239} Pu are | vedclass

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Question

(D) Average energy released per fission of $_{94}^{239} Pu$, $E_{avg} = 180 \; MeV$.Mass of pure $_{94}^{239} Pu$, $m = 1 \; kg = 1000 \; g$.Avogadro

Vedclass · Accepted Answer

$4.536 	imes 10^{26}$

Vedclass · Answer

(D) Average energy released per fission of $_{94}^{239} Pu$, $E_{avg} = 180 \; MeV$.Mass of pure $_{94}^{239} Pu$, $m = 1 \; kg = 1000 \; g$.Avogadro number, $N_A = 6.023 	imes 10^{23} \; 	ext{atoms/mol}$.Atomic mass of $_{94}^{239} Pu = 239 \; g/mol$.The number of atoms $N$ in $1 \; kg$ of $_{94}^{239} Pu$ is given by:$N = \frac{m}{M} 	imes N_A = \frac{1000}{239} 	imes 6.023 	imes 10^{23} \approx 2.52 	imes 10^{24} \; 	ext{atoms}$.Total energy released $E = N 	imes E_{avg}$.$E = (2.52 	imes 10^{24}) 	imes 180 \; MeV = 4.536 	imes 10^{26} \; MeV$.Therefore, the total energy released is $4.536 	imes 10^{26} \; MeV$.

The fission properties of $_{94}^{239} Pu$ are very similar to those of $_{92}^{235} U$. The average energy released per fission is $180 \; MeV$. How much energy, in $MeV$, is released if all the atoms in $1 \; kg$ of pure $_{94}^{239} Pu$ undergo fission?

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