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IN a nuclear reactor, fission is produce...

IN a nuclear reactor, fission is produced in 1 g of `.^(235)U`
`(235.0349 am u)`. In assuming that `._(53)^(92)Kr (91.8673 am u)`
and `._(36)^(141)Ba (140.9139 am u)` are produced in all reactions and no energy is lost, calculate the total energy produced in killowatt. Given: `1 am u =931 MeV`.

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To solve the problem step by step, we will follow the outlined process of calculating the energy produced during the fission of Uranium-235. ### Step 1: Determine the mass of reactants The reaction involves the fission of Uranium-235 and the production of Krypton-92, Barium-141, and neutrons. The mass of Uranium-235 is given as: - Mass of \( ^{235}U = 235.0349 \, \text{amu} \) - Mass of one neutron \( = 1.0087 \, \text{amu} \) ...
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Find the amount of energy released when 1 atom of Uranium ._(92)U^(235)(235.0439 "amu") undergoes fission by slow neturon (1.0087 "amu") and is splitted into Krypton ._(36)Kr^(92)(91.8973 "amu") and Barium ._(56)Br^(141)(140.9139 "amu") assuming no energy is lost. Hence find the enrgy in kWh , when 1g of it undergoes fission.

A neutron strikes a ""_(92)U^(235) nucleus and as a result ""_(36)Kr^(93) and ""_(56)Ba^(140) are produced with

Knowledge Check

  • The fission properties of ._94^239Pu are very similar to those of ._92^235 U. The average energy released per fission is 180 MeV. If all the atoms in 1 kg of pure ._94^239Pu undergo fission, then the total energy released in MeV is

    A
    `4.53xx10^26` MeV
    B
    `2.21xx10^14` MeV
    C
    `1xx10^13` MeV
    D
    `6.33xx10^24` MeV

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