Question

When the radioactive isotope ${}_{88}R{a}^{226}$ decays in a series by emission of three alpha$\left(\alpha \right)$ and a beta $\left(\beta \right)$ particle, the isotope X which remains undecay is?

• A. ${}_{83}{X}^{214}$
• B. ${}_{84}{X}^{218}$
• C. ${}_{84}{X}^{220}$
• D. ${}_{87}{X}^{223}$

Answer 1

The correct option is A ${}_{83}{X}^{214}$

Due to emission of $3\alpha$ particles, the mass number is reduced by $12$ while atomic number is decreased by $6$. Due to emission of $\beta$ particle, the atomic number is increased by one while there is no change in mass number. Now, resulting mass will be decreased by $12$ and atomic number is decreased by $5$. So, isotope X is ${}_{83}{X}^{214}$.