The Bohr radius of second energy level of He $^{+}$ ion is _____ nm.

0.106

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0.052

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0.424

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None of these

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Solution

The correct option is A 0.106
As we know,
r $_{n}$ = 0.053 $\frac{n^{2}}{m z}$ nm
$\Rightarrow$ r $_{n}$ = 0.106 nm [z = 2 n = 2]

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