question_answer

A nucleus of mass number 220, initially at rest, emits an \[\alpha \]-particle. If the Q value of the reaction is 5.5 MeV, the energy of the emitted \[\alpha \]-particle will be

A)  4.8 MeV             

B)  5.4 MeV

C)  5.5 MeV             

D)  6.8 MeV

Correct Answer: B

Solution :

                Q-value of the reaction is 5.5 MeV i.e.,        \[{{K}_{1}}+{{K}_{2}}=5.5.\] MeV                              ... (i) By conservation of linear momentum                                 \[{{p}_{1}}={{p}_{2}}\] \[\Rightarrow \]               \[\sqrt{2(216){{K}_{1}}}=\sqrt{2(4){{K}_{2}}}\]                 \[{{K}_{2}}=54\,{{K}_{1}}\]                                           ... (ii) On solving Eqs. (i) and (ii), we get The energy of emitted a-particle                 \[{{K}_{2}}=5.4\,MeV\]