The wave function for an orbital is H-atom is given as- - - - 2 81- 1 a 0 2 3 6- r a 0 r a 0 - r 3 a 0 -sin-sin-The orbital is-

The correct option is A 2s In the given equation ψ=0 then 6 ra_0 =0 r=6a_0 x=n+l=r3a_0 x=6a_03a_0 x=2 2s 2+0=2 ...

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Schrodinger Equation

The wave func...

Question

The wave function for an orbital is $H -$ atom is given as:
$Ψ = (\frac{\sqrt{2}}{81 \sqrt{π}}) {(\frac{1}{a_{0}})}^{\frac{2}{3}} (6 - \frac{r}{a_{0}}) {(\frac{r}{a_{0}})}^{- \frac{r}{3 a_{0}}} . sin θ sin ϕ$
The orbital is:

2 s

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3 p

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2 p

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3 d

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Solution

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Similar questions

Ψ_{3 s} = \frac{1}{(81) \sqrt{3 π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}}] e^{\frac{- r}{3 a_{0}}}

\frac{d}{1.5 a_{0}}

\sqrt{108} = 10

- Ψ = \frac{1}{81 \sqrt{6 π}} (\frac{1}{a_{0}}) \frac{3}{2} σ e^{\frac{σ}{3}} (3 c o s^{2} θ - 1); H e r e σ = r / a_{0} a n d a_{0} = \frac{h^{2} ϵ_{0}}{π m e^{2}}

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

a_{0}

(ψ)

2 s

ψ_{2 s} = \frac{1}{\sqrt{32 π}} {[\frac{1}{a_{0}}]}^{3 / 2} [2 - \frac{r}{a_{0}}] e^{- r / 2 a_{0}}

r = r_{0}

r_{0}

x a_{0}

x

Ψ_{3 s} = \frac{1}{(81) \sqrt{3 π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}}] e^{\frac{- r}{3 a_{0}}}

\frac{d}{1.5 a_{0}}

\sqrt{108} = 10

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