Calculate the energy of the gamma ray liberated during thermal neutron $$\text { capture by } { } ^ { 232 } \mathrm { Th } \text {. Note: } m \left( { } ^ { 232 } \mathrm { Th } \right) = 232.0380504 \mathrm { u } ; m \left( { } ^ { 233 } \mathrm { Th } \right) = 233.0415769 \mathrm { u } .$$

Question

Quizplus · Accepted Answer

The Answer is The reaction is

\mathrm { n } + { } ^ { 232 } \mathrm { Th } ightarrow { } ^ { 233 } \mathrm { Th } + \gamma

So the excess energy will be given by

E = \left[ m _ { \mathrm { n } } + m \left( { } ^ { 232 } \mathrm { Th } ight) - m \left( { } ^ { 233 } \mathrm { Th } ight) ight] c ^ { 2 }

or

E = [ 1.008664904 \mathrm { u } + 232.0380504 \mathrm { u } - 233.0415769 \mathrm { u } ] imes 931.494 \mathrm { MeV } / \mathrm { u } = 4.79 \mathrm { MeV }

Calculate the Energy of the Gamma Ray Liberated During Thermal