Question

A radioactive element $Y$ emits an alpha particle and a beta particle successively and changes into an element ${}_{82}{X}^{206}$. Find the atomic mass and atomic number of radioactive element $Y$.

Answer 1

The element Z looses a beta particle to form element $X$.
The element Z should have atomic number one less than the atomic number of element $X$$.
${}_{81}{Z}^{206}{\underrightarrow{-\beta }}_{82}{X}^{206}{+}_{-1}{e}^0$
The element Y loses an alpha particle to form element Z.
The element Y should have atomic number 2 greater than that of $Z$ and mass number 4 greater than that of $Z$.
${}_{83}{Y}^{210}{\underrightarrow{-\alpha }}_{81}{Z}^{206}{+}_2{He}^4$
Hence, the atomic mass of $Y$ is 210 and atomic number is $83$.