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The energy liberated on complete fission...

The energy liberated on complete fission of `1 kg` of `._92 U^235` is (Assume `200 MeV` energy is liberated on fission of `1` nucleus).

A

`8.2 xx 10^(10)J`

B

`8.2 xx 10^(9)J`

C

`8.2 xx 10^(13)J`

D

`8.2 xx 10^(16)J`

Text Solution

Verified by Experts

The correct Answer is:
C
Mass of uranium nucleus
`=92 xx 1.6725 xx 10^(-27) + (235- 92) xx 1.6747 xx 10^(-27)`
`= 393.35 xx 10^(-27) kg`
No. of nuclei `= (1)/(393.35 xx 10^(-27)) = 2.542 xx 10^(24)`
For given mass,
Energy released `= 200 xx 2.542 xx 10^(24) xx 10^(6) xx 1.6 xx 10^(-19) = 8.2 xx 10^(13)J`
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