13.17 The fission properties of ${}_{94}^{239}{P u}$ 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 ${}_{94}^{239}{P u}$ undergo fission?

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Payal Gupta | Contributor-Level 10

8 months ago

13.17 The average energy released per fission of ${}_{94}^{239}{P u}$ $E_{a v g}$ = 180 MeV

Amount of pure ${}_{94}^{239}u$ , m = 1 kg = 1000 g

Avogadro's number,
$N_{A}$ = 6.023 $* 10^{23}$

Mass number of ${}_{94}^{239}{P u}$ = 239 gm

Hence, number of atoms in 1000 g ${}_{94}^{239}{P u}$ , N = $\frac{6.023 * 10^{23}}{239}$ $* 1000$ = 2.52 $* 10^{24}$

Total energy released during the fission of 1 kg of ${}_{94}^{239}{P u}$ , E = $E_{a v g}$ $*$ N

= 180 $*$ 2.52 $* 10^{24}$ MeV = 4.536 $* 10^{26}$ MeV

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13.17 The fission properties of Pu94239 are very similar to those of U92235 . 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 Pu94239undergo fission?

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13.17 The fission properties of ${}_{94}^{239}{P u}$ 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 ${}_{94}^{239}{P u}$ undergo fission?