Question

A radioactive nucleus (initial mass number $A_o$ 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. $\frac{A-Z-4}{Z-2}$

Answer 1

Answer: (B)

In positive beta decay a proton is transformed into a neutron and a positron is emitted.
$P^+ \longrightarrow n^0 + e^+$
no. of neutrons initially was A - Z
no. of neutrons after decay (A - Z) - 3 x 2 (due to alpha particles) + 2 x 1 (due to positive beta decay)
The no. of proton will reduce by 8. [as 3 x 2 (due to alpha particles) + 2(due to positive beta decay)] Hence atomic number reduces by 8.