A Common Induced Fission Process in a Thermal Nuclear Reactor$$\mathrm { n } + { } ^ { 235 } \mathrm { U } \rightarrow { } ^ { 236 } \mathrm { U } \rightarrow { } ^ { 137 } \mathrm { Xe } + { } ^ { 97 } \mathrm { Sr } + x \mathrm { n }$$

Question

Examlex · Accepted Answer

The Answer is (a) On the left hand side of the equation the number of neutrons is

1 + ( 235 - 92 ) = 144

On the right hand side the nuclei contain

( 137 - 54 ) + ( 97 - 38 ) = 142

So the number of excess neutrons is

144 - 142 = 2

(b) We need to consider the mass of the neutron plus the

{ } ^ { 235 } \mathrm { U }

on the left hand side. We should not use the mass of the

{ } ^ { 236 } \mathrm { U }

because it is not in the ground state when it is formed. So we have

E = \left[ m _ { \mathrm { n } } + m \left( { } ^ { 235 } \mathrm { U } ight) - m \left( { } ^ { 137 } \mathrm { Xe } ight) - m \left( { } ^ { 97 } \mathrm { Sr } ight) - 2 imes m _ { \mathrm { n } } ight] c ^ { 2 }

Using atomic masses (since the number of protons remains the same in the atoms) we obtain

\begin{aligned}E & = [ 235.0439231 \mathrm { u } - 136.9115629 \mathrm { u } - 96.926153 \mathrm { u } - 1.008664904 \mathrm { u } ] imes 931.494 \mathrm { MeV } / \mathrm { u } \& = 184 \mathrm { MeV }\end{aligned}

A Common Induced Fission Process in a Thermal Nuclear Reactor

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