question_answer

If the nucleons of a nucleus are separated far apart from each other, the sum of masses of all these nucleons is larger than the mass of the nucleus. Where does this mass difference come from? Calculate the energy released, if \[{}^{238}U\] nucleus emits an \[\alpha \,\,\text{-}\,\text{particle}.\]

Given, atomic mass of \[{}^{238}U=\text{ }238.05084\] amu atomic mass of \[{}^{234}Th=\text{ }234.04363\] amu atomic mass of \[\alpha -particle=4.00260\] amu and 1 amu\[=931\text{ }MeV/{{c}^{2}}\]

Answer:

If we separate a nucleus into its nucleons, we would have supply a total energy \[{{E}_{b}}\] called binding energy of that nucleus. The mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons in their free state. This mass difference is called mass defect. If \[\Delta \,m\] is mass defect of a nucleus, its binding energy, \[{{E}_{b}}=\Delta \,m{{c}^{2}}\] The reaction for \[\alpha \,-\]emission from \[{{U}^{238}},\] \[_{92}{{U}^{238}}\] \[\to \] \[{}_{90}^{234}\,\text{Th+}{}_{2}^{4}\text{He}\] Mass defect, \[\Delta \,m=238.05076-(4.00260+234.04357)\] \[=0.00467\] \[\therefore \] Energy released, \[Q=931\times 0.00467\]  \[=4.35\,\,\text{MeV}\]