Question

${}_{90}^{228}$Th $\to$${}_{83}^{212}$Bi. The number of $\alpha$ and $\beta$ particles given out during the process are:

• A. 4$\alpha$,7$\beta$
• B. 4$\alpha$,1$\beta$
• C. 4$\alpha$,
• D. 7$\beta$

Answer 1

The correct option is B 4$\alpha$,1$\beta$

In any $\alpha$ particle emission (i) mass number is decreased by 4 units.
and (ii) atomic number is decreased by 2 units.
In any $\beta$ particle emission (i) mass number remains same.
and (ii) atomic number is increased by 1 units.
So, in this question, mass number is decreased by 16 units
So, there are $\frac{16}{4}=4\alpha$-particles.
and atomic number is decreased by 7 units.
but by 4 $\alpha$-particles atomic number should be decreased by 8, so atomic number should be increased by 1. So, there is one $\beta$-particle.