For a 3s - orbital
$Ψ (3 s) = \frac{1}{a \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} (6 - 6 σ + σ^{2}) ϵ^{\frac{- σ}{2}},$
Where $σ = \frac{2 r Z}{3 a_{0}}$
What is the maximum radial distance of node from nucleus is

\frac{(3 + \sqrt{3}) a_{0}}{Z}

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\frac{a_{0}}{Z}

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\frac{3}{2} \frac{(3 + \sqrt{3}) a_{0}}{Z}

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\frac{2 a_{0}}{Z}

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Solution

The correct option is C $\frac{3}{2} \frac{(3 + \sqrt{3}) a_{0}}{Z}$
For a node,
$Ψ (3 s) = 0 6 - 6 σ + σ^{2} = 0 So, σ = \frac{(- 6) \pm \sqrt{36 - (4 (1) 6)}}{2 \times 1} σ = \frac{6 \pm \sqrt{12}}{2} σ = \frac{6 \pm 2 \sqrt{3}}{2} σ = 3 + \sqrt{3} o r 3 - \sqrt{3}$
Therefore, the maximum distance=
$= \frac{2 r Z}{3 a_{0}} = 3 + \sqrt{3} r = \frac{3}{2} (3 + \sqrt{3}) \frac{a_{0}}{Z}$

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Similar questions

3 s -

Ψ (3 s) = \frac{1}{9 \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} (6 - 6 σ + σ^{2}) e^{\frac{- σ}{2}}; where σ = \frac{2 r . Z}{3 a_{0}}

3 s -

ψ (3 s) = \frac{1}{a \sqrt{3}} (\frac{1}{a_{0}})^{\frac{3}{2}} (6 - 6 σ + σ^{2}) e^{\frac{- σ}{2}}

σ = \frac{2 r Z}{3 a_{0}}

Ψ (3 s) = \frac{1}{a \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} (6 - 6 σ + σ^{2}) ϵ^{\frac{- σ}{2}},

σ = \frac{2 r Z}{3 a_{0}}

Ψ (3 s) = \frac{1}{9 \sqrt{3}} {(\frac{1}{a_{0}})}^{3 / 2} (6 - 6 σ + σ^{2}) e^{- σ / 2}; where σ = \frac{2 r . Z}{3 a_{0}}

3 s -

Ψ (3 s) = \frac{1}{9 \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} (6 - 6 σ + σ^{2}) e^{\frac{- σ}{2}}; where σ = \frac{2 r . Z}{3 a_{0}}

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