question_answer

A radioactive nucleus 'A' decays as given below : \[A\xrightarrow{_{-1}^{0}e}{{A}_{1}}\xrightarrow{\alpha }{{A}_{2}}\] If the mass number of atomic number of \[{{A}_{1}}\] are 180 and 73 respectively, find the mass number and atomic number of A and \[{{A}_{2}}\]                               

Answer:

                Using displacement laws of radioactive transformation, we can represent the successive decays as follows: \[_{72}^{180}A\xrightarrow{_{-1}^{0}e}_{73}^{180}{{A}_{1}}\xrightarrow{_{2}^{4}He}_{71}^{176}{{A}_{2}}\] Mass number of  \[\text{A }=\text{ 18}0\] Atomic number of \[\text{A }=\text{ 72}\] Mass number of \[{{A}_{2}}=176\] Atomic number of \[{{A}_{2}}=71\]