Question

Actinium $231,{}^{231}{\text{Ac}}_{89}$, emit in succession two $\beta -$particles, four alphas, one $\beta$ and one alpha plus several $\gamma$ rays. What is the resultant isotope?

• A. ${}^{221}A{u}_{79}$
• B. ${}^{211}A{u}_{79}$
• C. ${}^{211}P{b}_{82}$
• D. ${}^{221}P{b}_{82}$

Answer 1

The correct option is C ${}^{211}P{b}_{82}$

Apart from several gamma particles, total five alpha and three beta particles are emitted. This converts
$231{,}^{231}A{c}_{89}$ into ${}^{221}P{b}_{82}$.
${}^{231}A{c}_{89}{\to }^{82}P{b}_{82}$
Note: When an alpha particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4. When a beta particle is emitted, the atomic number increases by 1 and the mass number remains unchanged.
The atomic number of daughter nucleus $=89-5\times 2+3\times 1=82$
The mass number of daughter nucleus $=231-5\times 4=211$

Hence, the correct option is $\text{C}$