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Consider the nuclear reaction X^(200)toA...

Consider the nuclear reaction `X^(200)toA^(110)+B^(80)+10n^(1)`. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction:

A

70 MeV

B

200 MeV

C

190 MeV

D

1480 MeV

Text Solution

Verified by Experts

The correct Answer is:
A
For X, energy `=200xx7.4=1480 MeV`
For A, energy `=110xx8.2=902MeV`
For B, energy `=80xx8.1="648 MeV"`
`therefore" Energy released"=(902+648)-1480`
`=1550-1480="70 MeV"`
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