Q. A radioactive nucleus (initial mass number A and atomic number $Z$) emits 3 $\alpha - particles$ and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be
A) $\frac{A - Z - 8}{Z - 4}$
B) $\frac{A - Z - 4}{Z - 8}$
C) $\frac{A - Z - 12}{Z - 4}$
D) $\frac{A - Z - 4}{Z - 2}$
Explanation
(b) In positive beta decay a proton is transformed into a neutron and a positron is emitted.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p^{+} \rightarrow n^{0}\ + \ e^{+}$
Number of neutrons initially was $A - Z.$
Number of neutrons after decay $(A - Z)\ - 3 \times 2$(due to alpha particles) + $2 \times 1$(due to positive beta decay).
The number of protons will reduce by 8 [as 3$\times$2 (due to alpha particles) + 2(due to positive beta decay)].
Hence, atomic number reduces by 8.
So, the ratio number of neutrons to that of protons
$= \frac{A - Z - 4}{Z - 8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.