Question

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. $\frac{A-Z-4}{Z-2}$

Answer 1

The correct option is B $\frac{A-Z-4}{Z-8}$

When a radioactive nucleus emits 1 α-particle mass number decreases by 4 and the atomic number decreases by 2.

So after the emission of $3-\alpha$ particles

New atomic mass, $A^{\prime }=A-4\times 3=A-12$

New atomic number $Z^{\prime }=Z-6$

When this nucleus emits $1\beta$-particle (positron), the atomic mass remains unchanged but the atomic number decreases by 1.

So after emission of 2 positrons

$A"=A^{\prime }=A-12$

$Z"=Z^{\prime }-2\times 1=Z-8$

So, the ratio of neutrons to protons-

$\frac{n}{p}=\frac{A-Z-4}{Z-8}$