Question

A radio active nucleus (initial mass number A and atomic number Z emits 4$\alpha$- particles and 3 positrons).The ratio of numbers of neutrons to that of protons in the final nucleus will be

• A. $\frac{A-Z-4}{Z-8}$
• B. $\frac{A-Z-3}{Z-2}$
• C. $\frac{A-Z-2}{Z-3}$
• D. $\frac{A-Z-5}{Z-11}$ -

Answer 1

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

When a radio active nucleus emits an alpha particle, its mass number decreases by $4$ while the atomic number decreases by $2$.

When a radioactive nucleus, emits a ${\beta }^{+}$ particle (or position) $\left(e^{+}\right)$ its mass number remains unchanged while the atomic number decreased by $1$.

${}_Z^{A}X{\stackrel{3\alpha }{⟶}}_{Z-6}^{A-12}Y{\stackrel{{2e}^{+}}{⟶}}_{Z-8}^{A-P2}W$

In the final nucleus

Number of protons $N_P=Z-8$

Number of neutrons,

$N_n=A-12-(Z-8)=A-Z-4$

$\therefore$ $\frac{N_n}{N_P}=\frac{A-Z-4}{Z-8}$