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 $\frac{A-Z-4}{Z-8}$

$\therefore$ Mass number of final nucleus = A - 12
Atomic number of final nucleus = Z - 8
$\therefore$ Number of neutrons = (A - 12) - (Z - 8)
= A - Z - 4
Number of protons = Z - 8
$\therefore$ Required ratio = $\frac{A-Z-4}{Z-8}$