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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
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The correct answer is:
$\frac{A-Z-4}{Z-8}$
$\frac{A-Z-4}{Z-8}$
In positive beta decay a proton is transformed into a neutron and a positron is emitted.
$$
\mathrm{p}^{+} \longrightarrow \mathrm{n}^0+\mathrm{e}^{+}
$$
no. of neutrons initially was $A-Z$
no. of neutrons after decay $(A-Z)-3 \times 2$ (due to alpha particles) $+2 \times 1$ (due to positive beta decay)
The no. of proton will reduce by 8 . [as $3 \times 2$ (due to alpha particles) $+2$ (due to positive beta decay)] Hence atomic number reduces by 8 .
$$
\mathrm{p}^{+} \longrightarrow \mathrm{n}^0+\mathrm{e}^{+}
$$
no. of neutrons initially was $A-Z$
no. of neutrons after decay $(A-Z)-3 \times 2$ (due to alpha particles) $+2 \times 1$ (due to positive beta decay)
The no. of proton will reduce by 8 . [as $3 \times 2$ (due to alpha particles) $+2$ (due to positive beta decay)] Hence atomic number reduces by 8 .
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