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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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    13.17 The average energy released per fission of Pu94239  Eavg= 180 MeV

    Amount of pureu94239 , m = 1 kg = 1000 g

    Avogadro's number,
    NA = 6.023 *1023

    Mass number of P u 94 239  = 239 gm

    Hence, number of atoms in 1000 g Pu94239, N =6.023*1023239 *1000= 2.52 *1024

    Total energy released during the fission of 1 kg of  Pu94239, E =Eavg *N

    = 180 * 2.52*1024 MeV = 4.536 *1026 MeV

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