Q. A certain radioactive material $\ _{Z}X^{A}$ starts emitting $\alpha\ and\ \beta$ particles successively such that the end product is $\ _{z - 3}Y^{A - b}$. The number of $\alpha and\beta$ particles emitted are
A) 4 and 3 respectively
B) 2 and 1 respectively
C) 3 and 4 respectively
D) 3 and 8 respectively
Explanation
(b) Let there be $x\alpha$-particles and $\gamma\beta$- particles
$\ _{z}X^{4} \rightarrow {xHe}_{2}^{4}{+ y\beta}_{- 1}^{0} + Y_{Z - 3}^{A - 8}$
Then equating the mass numbers
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A = 4x + A - 8$ …(i)
and Equating atomic number
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Z = 2x - y + Z - 3$ …(ii)
Solving Eqs.(i) and (ii),we get
$x$=2 and y=1
∴The number of α and β particles emitted are 2 and 1 respectively.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.