What is the radius of second Bohr orbit of single ionized helium ion? (Assume $a_{0} = 53 p m$ )

53 pm

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106 pm

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159 pm

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212 pm

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Solution

The correct option is B 106 pm
The general formula for radius of Bohr's orbit is
$r = \frac{n^{2} a_{0}}{Z},$
Now, second Bohr's orbit implies, n = 2
And we are talking about He+. Hence Z=2.
Now substituting all the values,
$r = \frac{n^{2} a_{0}}{Z} = 2 a_{0}$
Now, $a_{0} = 53 p m$ Hence, r=106 pm

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