question_answer

In the \[\beta \] decay of\[N{{a}^{24}}\], the combined electron neutrino momentum has a magnitude equal to 4 MeV/c. What is the recoil energy of the daughter nucleus, given that its mass is 23.99 u? Take.

Answer:

                Here, momentum of electron and neutrino pair \[p=\frac{4MeV}{c}=\frac{4\times 1.6\times {{10}^{-13}}}{3\times {{10}^{8}}}=2.13\times {{10}^{-21}}kg\,m{{s}^{-1}}\] Recoil energy of daughter nucleus, E = ? Mass of this nucleus, \[m=23.99\,u=23.99\,\times 1.66\times {{10}^{-27}}kg\] As parent nucleus decays at rest, according to the law of conservation of linear momentum, [Fig. 4(HT).6], momentum of daughter nucleus =momentum of electron and neutrino pair, \[p=2.13\times {{10}^{-21}}kg\,m{{s}^{-1}}\] From \[E=\frac{{{p}^{2}}}{2m}\] \[E=\frac{{{\left( 2.13\times {{10}^{-21}} \right)}^{2}}}{2\times 23.99\times 1.66\times {{10}^{-27}}}J\] \[=\frac{2.13\times 2.13\times {{10}^{-15}}}{2\times 23.99\times 1.66\times {{10}^{-13}}}MeV=3.56\times {{10}^{-4}}MeV\]