A nuclear fission is represented by the following reaction: `U^(236) = X^(111) +Y^(122) +3n` If the binding energies per nucleon of `X^(111), Y^(122)`
A nuclear fission is represented by the following reaction: `U^(236) = X^(111) +Y^(122) +3n`
If the binding energies per nucleon of `X^(111), Y^(122)` and `U^(236)` are `8.6MeV, 8.5 MeV` and `7.6 MeV` respectively, then the energy released in the reaction will be-
A. `200 MeV`
B. `202 MeV`
C. `195 MeV`
D. `198 MeV`
6 views
1 Answers
Correct Answer - D
Energy released
`= BE` of products - BE of reactants
`= 111 xx 8.6 +122 xx 8.5 - 236 xx 7.6`
`= 954.6 + 1037.0 - 173.6 =198 MeV`
6 views
Answered