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?

Solution

Given,
Average amount of energy released per fission, ${}_{94}^{239}{Pu} = 180 MeV$

Quantity of fissionable material = 1 kg

In 239 gm Pu, number of fissionable atom or nuclei $= 6.023 \times 10^{23}$

In 1 g of Pu, number of fissionable atom or nuclei $= \frac{6.023 \times 10^{23}}{239}$

In 1000 gm of Pu, number of fissionable atom or nuclei,
$= \frac{6.023 \times 10^{23}}{239} \times 1000 = 25.2 \times 10^{23}$

Therefore,
Total energy released in fission of 25.2 x 10²³Pu nucleus or in fission of 1 kg pure Pu is,
= 180 x 25.2 x 10²³= 4536 x 10²³MeV
= 4.5 x 10²⁶MeV.

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Wave Optics

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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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 Pu94239 undergo fission?

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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?