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If the radius of the hydrogen atom is $53 \mathrm{pm}$, the radius of the $\mathrm{He}^{+}$ion is closest to
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$27 \mathrm{pm}$
$r_{n}=\frac{R_{H} n^{2}}{Z}$
$r_{H_{e+}}=\frac{53 n^{2}}{Z}$ $=\frac{53 \times 1^{2}}{2}=27$ approx.
$r_{H_{e+}}=\frac{53 n^{2}}{Z}$ $=\frac{53 \times 1^{2}}{2}=27$ approx.
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