The wave function for 2s orbital is given as :Ψ=1421a03/22−ra0⋅e−r/2a0Where a0=First Bohr's radius in H-atom =0.529Å Read the given statement and pick out the correct statement(s).

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The wave function for 2s orbital is given as : $Ψ = (\frac{1}{4 \sqrt{2}}) {(\frac{1}{a_{0}})}^{3 / 2} (2 - \frac{r}{a_{0}}) \cdot e^{- r / 2 a_{0}}$

Where $a_{0} =$ First Bohr's radius in H-atom $= 0.529 Å$ Read the given statement and pick out the correct statement(s).

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The probability density of finding electron at nucleus is non-zero.

The number of radial nodes is equal to three

The radial node occur at a distance $2 a_{0}$ from nucleus.

The probability density is independent of direction.

answer is B, C, D.

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Detailed Solution

No. of radial nodes $= n - l - 1 = 2 - 0 - 1 = 1$

Probability density $ψ^{2}$

Since $ψ$ contains no $θ$ or $ϕ$ terms - probability of finding electron does not depend upon direction.

Also, since at $r = 0$ ; $ψ = (\frac{1}{4 \sqrt{2}}) {(\frac{1}{a_{0}})}^{3 / 2}$

$ψ^{2}$ is non-zero at $r = 0$

$ψ = 0$ when the term $(2 - \frac{r}{a_{0}})$ vanishes putting $2 - \frac{r}{a_{0}} = 0$

$\Rightarrow r = 2 a_{0}$

at $r = 2 a_{0} ψ = 0$ : Corresponds to radial node.

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