question_answer

A heavy nucleus \[\mathbf{X}\] of mass number \[\mathbf{A}=\mathbf{240}\]and binding energy per nucleon \[\mathbf{7}.\mathbf{6}\text{ }\mathbf{MeV}\] is split into two nearly equal fragments \[\Upsilon \] and \[\mathbf{Z}\] of mass numbers \[{{A}_{1}}=110\] and \[{{A}_{2}}=130\]. The binding energy of each one of these nuclei is \[\mathbf{8}.\mathbf{5}\text{MeV}\] per nucleon. Calculate the total binding energy of each of the nuclei \[\mathbf{X},\text{ }\mathbf{Y}\] and \[\mathbf{Z}\] and hence the energy Q released per fission in \[\mathbf{MeV}\].                      

Answer:

                For nucleus \[~X:\text{A }=\text{ 24}0\] B.E. per nucleon \[=\text{ 7}.\text{6 MeV}\] Total B.E. of \[\text{X}\] \[=\text{ 24}0\text{ }\times \text{ 7}.\text{6 }=\text{ 1824 MeV}\]For nucleus \[\Upsilon \]: \[{{A}_{1}}=110\] B.E. per nucleon \[=\text{ 8}.\text{5 MeV}\] Total B.E. of \[\Upsilon \]\[110\times 8.5=935\text{MeV}\] For nucleus Z : \[{{A}_{2}}=130\]  B.E. per nucleon\[~=\text{ 8}.\text{5 MeV}\] Total B.E, of Z \[=\text{ 13}0\text{ }\times \text{ 8}.\text{5 }=\text{ 11}0\text{5 MeV}\] Energy released per fission, \[Q=B.E\] of \[\Upsilon +B.E\]of \[Z-B.E\]of \[X\] \[=935+1105-1824=2040-1824=\mathbf{216MeV}\]