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If the radius of $3^{rd}$ Bohr's orbit in $H$ is $x$ , then radius of $4^{th}$ orbit of ${Li}^{2 +}$ ion would be

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$\frac{27}{16} x$

$\frac{16}{27} x$

$\frac{9}{16} x$

None of these

answer is B.

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Detailed Solution

$r_{n} = \frac{n^{2} h^{2}}{z 4 π^{2} {me}^{2}} = \frac{n^{2}}{z}$ (constant)

$x = \frac{3^{2}}{1}$ (constant)

and $r_{4} = \frac{4^{2}}{3}$ (constant)

$\Rightarrow \frac{x}{r_{4}} = \frac{3^{2}}{4^{2}} \times 3 = \frac{27}{16} \Rightarrow r_{4} = \frac{16}{27} x$

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