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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 $]$
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The correct answer is:
$\frac{A-Z-15}{Z-1}$
Let us assume, a particle $X$ having atomic number $Z$ and mass number $A$.
When an a-particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4 . So, for given case, $\mathrm{Z} \mathrm{X}^{\mathrm{A}} \stackrel{4 \text { a-particle }}{\longrightarrow} \mathrm{Z}_{-\mathrm{a}} \mathrm{Y}^{\mathrm{A}-16}$
When a $\beta$-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case,
$\mathrm{Z}-8 \mathrm{Y}^{\mathrm{A}-16} \stackrel{7 \beta-p a r t i c l e}{\longrightarrow} \mathrm{Z}-1 \mathrm{Z}^{\mathrm{A}-16}$
When an a-particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4 . So, for given case, $\mathrm{Z} \mathrm{X}^{\mathrm{A}} \stackrel{4 \text { a-particle }}{\longrightarrow} \mathrm{Z}_{-\mathrm{a}} \mathrm{Y}^{\mathrm{A}-16}$
When a $\beta$-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case,
$\mathrm{Z}-8 \mathrm{Y}^{\mathrm{A}-16} \stackrel{7 \beta-p a r t i c l e}{\longrightarrow} \mathrm{Z}-1 \mathrm{Z}^{\mathrm{A}-16}$
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