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 \[{{A}_{2}}=71\]
You need to login to perform this action.
You will be redirected in
3 sec