A radioactive nucleus emits $4 \alpha$ particles and $7 \beta$ particles in succession. The ratio of number…
A radioactive nucleus emits $4 \alpha$ particles and $7 \beta$ particles in succession. The ratio of number of neutrons to that of protons is $[\mathrm{A}=$ mass number, $\mathrm{Z}=$ atomic number $]$
$\frac{\mathrm{A}-\mathrm{Z}-13}{\mathrm{Z}-2}$
$\frac{A-Z-13}{Z-1}$
$\frac{A-Z-15}{Z-1}$
$\frac{A-Z-11}{Z-2}$
Solution
$\frac{A-Z-15}{Z-1}$
Explanation:
Let us assume, a particle X having atomic number Z and mass number A .
When an $\alpha$-particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4 . So, for given case,
$\mathrm{z}^{\mathrm{A}} \xrightarrow{4 \alpha-\text { particle }} \mathrm{Z}-\mathrm{a} \mathrm{Y}^{\mathrm{A}-16}$
When a $\beta$-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case,
$\mathrm{Z}-8 \mathrm{Y}^{\mathrm{A}-16} \xrightarrow{7 \beta-\text { particle }} \mathrm{Z}-1 \mathrm{Z}^{\mathrm{A}-16}$
$\therefore \frac{\text { Number of neutrons }}{\text { Number of protons }}=\frac{(\mathrm{A}-16)-(\mathrm{Z}-1)}{(\mathrm{Z}-1)}$
$=\frac{A-Z-15}{(Z-1)}$