Radius of $2 n d$ shell of $H e^{+}$ (where $a_{0}$ - Bohr radius)

3 a_{0}

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a_{0}

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\frac{3}{2} a_{0}

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2 a_{0}

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Solution

The correct option is D $2 a_{0}$
Radius of the Bohr's orbit is given by,
$r = \frac{h^{2} n^{2}}{4 π^{2} m Z e^{2}}$
Substituting values of h, m for helium ion and the electronic charge r may be calculated as
$r = \frac{0.529 n^{2}}{Z} A^{0}$
Where $0.529 A^{0}$ is is the Bohr's radius $a_{0} = 0.529$
Substituting $Z = 2$ , $n = 2$
$r = 2 a_{0} A^{0}$

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L i^{2 +} i s :

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ψ = \frac{1}{4 a_{0} \sqrt{2 π a_{0}}} (2 - \frac{r}{a_{0}}) e^{\frac{- r}{2 a_{0}}}

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a_{0}, in L i^{2 +}

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