What are the respective numbers of \( \alpha \) and \( \beta \) particles emitted respectively in the following radioactive decay? \( {}^{200}_{90}\mathrm{X} \rightarrow {}^{168}_{80}\mathrm{Y} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8 and 6
An alpha particle \( ({}^{4}_{2}\mathrm{He}) \) takes away 4 units of mass number and 2 units of charge. A beta-minus particle leaves the mass number unchanged and raises the atomic number by 1. So the mass number fixes the number of alpha particles first, and the atomic number then fixes the beta particles.
\( \displaystyle \text{number of } \alpha = \frac{200 - 168}{4} = \frac{32}{4} = 8 \)
After 8 alpha decays the atomic number would be
\( \displaystyle 90 - 8 \times 2 = 74 \)
The final atomic number is 80, so it has to rise by \( 80 - 74 = 6 \); each \( \beta^{-} \) decay raises it by 1:
\( \text{number of } \beta = 6 \)