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Question

The radius of the $\mathrm{2^{nd}}$ orbit of $\mathrm{Li^{2+}}$ is $x$. The expected radius of the $\mathrm{3^{rd}}$ orbit of $\mathrm{Be^{3+}}$ is

$\frac{16}{27}x$
$\frac{4}{9}x$
$\frac{9}{4}x$
$\frac{27}{16}x$

Solution

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