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

Question

Examlex · 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 $\mathrm { n } + { } ^ { 235 } \mathrm { U } \rightarrow { } ^ { 137 } \mathrm { I } + { } ^ { 96 } \mathrm { Y } + 3 \mathrm { n }$

Definitions:

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

Definitions:

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