Consider the fission process in equation (6.3); $$\mathrm { n } + { } ^ { 235 } \mathrm { U } \rightarrow { } ^ { 137 } \mathrm { I } + { } ^ { 96 } \mathrm { Y } + 3 \mathrm { n }$$ (a) Calculate the total binding energy for each of the components in the reaction.
(b) Show that the fission energy is the total binding energy on the right hand side of the
equation minus the total binding energy on the left hand side.

Question

Quizplus · Accepted Answer

The Answer is (a) The total binding energy is given by

B = \left[ N m _ { \mathrm { n } } + Z \left( m _ { \mathrm { p } } + m _ { \mathrm { e } } ight) - m _ { ext {atom } } ight] c ^ { 2 }

For

{ } ^ { 235 } \mathrm { U }

this gives

\begin{aligned}B \left( { } ^ { 235 } \mathrm { U } ight) & = [ 143 imes 1.008664904 \mathrm { u } + 92 ( 1.00727647 \mathrm { u } + 0.0005485799 \mathrm { u } ) \& - 235.0439231 \mathrm { u } ] imes 931.494 \mathrm { MeV } / \mathrm { u } \& = 1783.0 \mathrm { MeV }\end{aligned}

For

{ } ^ { 137 }

I this gives

\begin{aligned}B \left( { } ^ { 137 } \mathrm { I } ight) & = [ 84 imes 1.008664904 \mathrm { u } + 53 ( 1.00727647 \mathrm { u } + 0.0005485799 \mathrm { u } ) \& - 136.91787084 \mathrm { u } ] imes 931.494 \mathrm { MeV } / \mathrm { u } \& = 1140.8 \mathrm { MeV }\end{aligned}

For

^ { 96 } \mathrm { Y }

this gives

\begin{aligned}B \left( { } ^ { 96 } \mathrm { Y } ight) & = [ 57 imes 1.008664904 \mathrm { u } + 39 ( 1.00727647 \mathrm { u } + 0.0005485799 \mathrm { u } ) \& - 95.91589779 \mathrm { u } ] imes 931.494 \mathrm { MeV } / \mathrm { u } \& = 822.7 \mathrm { MeV }\end{aligned}

(b) The fission energy will be given by the difference of the total binding energies on the right hand side and the left hand side as

E = B \left( { } ^ { 137 } \mathrm { I } ight) + \mathrm { B } \left( { } ^ { 96 } \mathrm { Y } ight) - \mathrm { B } \left( { } ^ { 235 } \mathrm { U } ight) = ( 1140.8 + 822.7 - 1783.9 ) \mathrm { MeV } = 179.6 \mathrm { MeV }

Compare this to the result shown in Example 6.2.

Consider the Fission Process in Equation (6 n+235U→137I+96Y+3n\mathrm { n } + { } ^ { 235 } \mathrm { U } \rightarrow { } ^ { 137 } \mathrm { I } + { } ^ { 96 } \mathrm { Y } + 3 \mathrm { n }n+235U→137I+96Y+3n

Consider the Fission Process in Equation (6 $\mathrm { n } + { } ^ { 235 } \mathrm { U } \rightarrow { } ^ { 137 } \mathrm { I } + { } ^ { 96 } \mathrm { Y } + 3 \mathrm { n }$