Question

15. 92U235 nucleus releases energy in fission:
(A) 3.4 MeV
(B) 26 Mev
(C) 200 MeV
(D) 931 MeV​

Answer 1

Answer:

200MeV

Explanation:

The fission process represented by the equation,

92

U

235

+

0

n

1

→

56

Ba

144

+

36

Kr

89

+3

0

n

1

Masses of reactants =234.39+1.01=235.4amu

Masses of products =143.28+88.89+3(1.01)=235.2amu

Energy released = mass difference =235.4−235.2=0.2amu=0.2×931∼200MeV