Q. A radioactive element $\ _{90}X^{238}$ decays into $\ _{83}Y^{222}$. The number of $\beta -$particles emitted are
A) 4
B) 6
C) 2
D) 1
Explanation
(d) Number of $\alpha -$particles emitted $= \frac{238 - 222}{4} = 4$
This decreases atomic number to $90 - 4 \times 2 = 82$
Since atomic number of $\ _{83}Y^{222}$ is 83, this is possible of one $\beta -$particle is emitted
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.