The Schrodinger wave function for Hydrogen atom of 4s orbital is given by-4s - 116 -31a0 3-2- -1 -2 - 8- - 12- e-2where a0 - 1st Bohr radius and - - 2rao-The distance from the nucleus where there is no radial node will be-

We know, wave function ψ = 0 where a node is present. On putting equation of nbsp;ψ = 0, we will get nbsp; ( σ 1)(σ^2 8σ + 12) = 0 On solving, σ = 1 and ...

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Standard XII

Chemistry

Visualising Orbitals

The Schroding...

Question

The Schrodinger wave function for Hydrogen atom of 4s orbital is given by:
$Ψ_{4 s} = \frac{1}{16 \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [(σ - 1) (σ^{2} - 8 σ + 12)] e^{- σ / 2}$
where $a_{0}$ = $1^{s t}$ Bohr radius and $σ = \frac{2 r}{a_{o}}$ .

The distance from the nucleus where there is no radial node will be:

r = 3 a_{0}

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r = a_{0}

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r = 2 a_{0}

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r =

\frac{a_{0}}{2}

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Solution

The correct option is B $r = a_{0}$
We know, wave function $ψ$ = 0 where a node is present.
On putting equation of $ψ$ = 0, we will get
$(σ - 1$ )( $σ^{2} - 8 σ + 12$ ) = 0
On solving, $σ = 1$ and
$σ$ = $\frac{8 \pm \sqrt{8^{2} - 4 \times 12}}{2}$ = 6 and 2
Given, $σ = \frac{2 r}{a_{0}}$
So, $σ = \frac{2 r}{a_{0}}$ = 1,
$σ = \frac{2 r}{a_{0}}$ = 6,
$σ = \frac{2 r}{a_{0}}$ = 2
which gives, r = $\frac{a_{0}}{2}$ , r = 3 $a_{0}$ and r = $a_{0}$
Since, in 4s orbital total number of nodes is 3
$∴$ The distance from the nucleus where there is no radial node is present is r = $a_{0}$

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Similar questions

Q. The radial wave function for an orbital in a hydrogen atom is:

ψ = \frac{1}{16 \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} [(x - 1) (x^{2} - 8 x + 12)] e^{- \frac{x}{2}}

where,

x = \frac{2 r}{a_{0}}; a_{0} =

radius of first Bohr's orbit. The minimum and maximum position of radial nodes from the nucleus are:

Q. The radial part of the Schrodinger wave equation for hydrogen atom is given below :

γ (r) = \frac{1}{16 \sqrt{4} a_{0}^{3 / 2}} (σ - 1) (σ^{2} - 8 σ + 12) θ^{- σ / 2}

Here,

a_{0} =

constant,

σ = \frac{2 r}{n a_{0}}

and

n =

principal quantum number.

With respect to the above-given data, select the correct statements.

Q. The radial wave equation for hydrogen atom is

Ψ = \frac{1}{16 \sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} [(x - 1) (x^{2} - 8 x + 12)] e^{- x / 2}

where,

x = 2 r / a_{0}

;

a_{0} =

radius of first Bohr orbit.

The minimum and maximum position of radial nodes from nucleus are:

Q. For a

3 s -

orbital

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}}

What is the maxium radial distance of node from nucleus?

Q. 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

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Visualising Orbitals

Standard XII Chemistry

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The Schrodinger wave function for Hydrogen atom of 4s orbital is given by: Ψ4s=116√3(1a0)32[(σ−1)(σ2−8σ+12)]e−σ/2 where a0 = 1st Bohr radius and σ=2rao. The distance from the nucleus where there is no radial node will be: