How Much Energy (In MeV)is Released When A ${ } _ { 92 } ^ { 235 } \mathrm { U }$

How much energy (in MeV) is released when a ${ } _ { 92 } ^ { 235 } \mathrm { U }$ fissions to ${ } _ { 56 } ^ { 141 } \mathrm { Ba }$ and $_{36}^{92} \mathrm{Kr}$ in the reaction $\mathrm { n } + { } _ { 92 } ^ { 235 } \mathrm { U } \rightarrow { } _ { 56 } ^ { 141 } \mathrm { Ba } + { } _ { 36 } ^ { 92 } \mathrm { Kr } + 3 \mathrm { n }$ m(n) = 1.008665 u m(U) = 235.043915 u
M(Ba) = 140.9139 u
M(Kr) = 91.8973 u
U = 1.66 *10 ^{- 27} kg

A) 100
B) 50
C) 200
D) 150
E) 250

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Access For Free