The radius of the first orbit of hydrogen is $0.53 A^{0}$ . The radius of second orbit of $H e^{+}$ would be:

0.265 A^{0}

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1.06 A^{0}

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2.12 A^{0}

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0.53 A^{0}

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Solution

The correct option is A $1.06 A^{0}$
$r = r_{0} \frac{n^{2}}{z}$ In given question $r_{0} = 0.53 A^{0}$
Thus $r_{{H e}^{+}, 2} = r_{0} \frac{{(2)}^{2}}{2} = 2 r_{0} = 1.06 A^{0}$
Hence answer is option B

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