question_answer

The sequence of stepwise decays of a radioactive nucleus is \[D\xrightarrow{\alpha }{{D}_{1}}\xrightarrow{{{\beta }^{-}}}{{D}_{2}}\xrightarrow{\alpha }{{D}_{3}}\] If the nucleon number and atomic number of \[{{D}_{2}}\] are 176 and 71 respectively, what are the corresponding values of D and \[{{D}_{3}}\]? Justify your answer in each case.     

Answer:

                The decays can be represented as follows : \[_{72}^{180}D\xrightarrow{_{2}^{4}He}_{70}^{176}\xrightarrow{_{-1}^{0}e}_{71}^{176}{{D}_{2}}\xrightarrow{_{2}^{4}He}_{69}^{172}{{D}_{3}}\] Nucleon number of \[\text{D }=\text{ 18}0\] Atomic number of \[\text{D }=\text{ 72}\] Nucleon number of \[{{\text{D}}_{3}}\text{ }=\text{172}\] Atomic number of \[{{\text{D}}_{3}}\text{ }=\text{ 69}\].