Question

If a nuclues ${}_Z{X}^A$ emits $9\alpha$ and $5\beta$ particles, then the ratio of the total protons and neutrons in the final nucles is

• A. $\frac{(Z-13)}{(A-36)}$
• B. $\frac{(Z-13)}{(A-Z-13)}$
• C. $\frac{(Z-18)}{(A-36)}$
• D. $\frac{(Z-13)}{(A-Z-23)}$

Answer 1

The correct option is D $\frac{(Z-13)}{(A-Z-23)}$

The given situation can be shown as
${}_Z{X}^A{\stackrel{9\alpha }{\to }}_{Z-18}{X}^{A-36}{\stackrel{5\beta }{\to }}_{Z-13}{X}^{A-36}$
$\therefore$ Number of protons, $P=(Z-13)$
and number of neutrons,
$N=(A-36)-(Z-13)=(A-Z-23)$
$\therefore \frac{P}{N}=\frac{(Z-13)}{(A-Z-23)}$.