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An element ${ }_{x} A^{y}$ emits $5 \alpha$ and $4 \beta$ particles to give ${ }_{82} B^{207}$. The number of protons and neutrons in $A$ are respectively
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The correct answer is:
88139
${ }_{x} A^{y} \longrightarrow{ }_{82} B^{207}+5_{2} \mathrm{He}^{4}+4{ }_{-1} \beta^{0}$
On comparing,
$$
\begin{aligned}
x &=82+5 \times 2+4(-1)=88 \\
y &=207+5 \times 4+4 \times 0 \\
&=227
\end{aligned}
$$
$\operatorname{In}_{88} A^{227}$
the number of protons $=88$ the number of neutrons $=227-88$
On comparing,
$$
\begin{aligned}
x &=82+5 \times 2+4(-1)=88 \\
y &=207+5 \times 4+4 \times 0 \\
&=227
\end{aligned}
$$
$\operatorname{In}_{88} A^{227}$
the number of protons $=88$ the number of neutrons $=227-88$
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