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Q.
The radius of the hydrogen atom in its ground state is $0.53 \mathring{A}$. The radius of $He^+$ ion in second excited state is
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Solution:
The radius of $n^{th}$ orbit of hydrogen like atoms is
$r_{n} =\frac{ n^{2} a_{0}}{Z}$ where $a_{0} = 0.53 \mathring{A}$
For $He^{+}$ ion, $Z = 2 $
For second excited state $n = 3$
$ \therefore \left(r_{3}\right)_{He^{+}} =\frac{ \left(3\right)^{2}\times0.53 \mathring{A}}{\left(2\right)} $
$= 2.4\mathring{A}$