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The radius of Bohr's first orbit in H-atom is 0.053 nm. The radius of second orbit in He⁺ would be :

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0.1060 nm

0.2120 nm

0.0265 nm

0.0530 nm

answer is C.

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Detailed Solution

$r_{n} for {He}^{+} = \frac{r_{n} for H}{Z} ∴ r_{2} for {He}^{+} = \frac{r_{2} for H}{2} = \frac{r_{1} for H \times 2^{2}}{2} (∵ r_{n} = r_{1} \times n^{2}) ∴ r_{2} for {He}^{+} = 0 .053 \times 2 = 0 .106 nm$

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