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Calculate the energy released by 1 g of ...

Calculate the energy released by `1 g` of natural uranium assuming `200 MeV` is released in eaech fission event and that the fissionable isotope `^235 U` has an abundance of 0.7% by weight in natural uranium.

Text Solution

Verified by Experts

The correct Answer is:
A
235 gm uranium contains `6.02 xx 10^(23)`
nuclei
So, `(6.02 xx 10^(23))/235 = 2.56 xx 10^(21) nuclei`
So total energy released
`= 2.56 xx 10^(21) xx 200 MeV`
`= 5.12 xx 10^(23) MeV`
According to energy released
`= 5.12 xx 10^(23) xx 1.6 xx 10^(-9)`
`xx 10^(6) xx 0.7/100 joule`
`=5.7 xx 10^8J`.
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Knowledge Check

  • The energy released by the fission of one uranium atom is 200 MeV. The number of fission per second required to prodice 6.4W power is

    A
    `2xx10^(11)`
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    D
    `2.5xx10^(2)` k Wh

  • The energy released by the fission of one uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 MW of power is :

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