The wave function of 2s electron is given by - - 2s - 1 4- 2- 1 a 0 3 -2 2- r a 0 e 2- r a 0 - If has a node at r- r 0- find relation between r 0 and a 0-

Given wave function: Ψ (2s)= 1 4√( 2π #160; ) #160; 1 a _ o #160; ^ 3 2 #160; ( 2 r a _ o #160; ) e ^( 2 r a _ o #160; ) #160; A ...

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Standard XII

Chemistry

Heisenberg's Uncertainty Principle

The wave func...

Question

The wave function of $2 s$ electron is given by :
$Ψ (2 s) = \frac{1}{4 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{3 / 2} (2 - \frac{r}{a_{0}}) e^{(2 - \frac{r}{a_{0}})}$ . If has a node at $r = r_{0}$ , find relation between $r_{0}$ and $a_{0}$ .

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Solution

Given wave function: $Ψ (2 s) = \frac{1}{4 \sqrt{2 π}} {\frac{1}{a_{o}}}^{\frac{3}{2}} (2 - \frac{r}{a_{o}}) e^{(2 - \frac{r}{a_{o}})}$
At node, any wave function becomes zero, because at node probability of finding an electron is zero.
Thus, $Ψ (2 s) = 0$ , which gives
$⟹ (2 - \frac{r}{a_{o}}) = 0$ as $e^{(2 - \frac{r}{a_{o}})}$ cannot be zero
$⟹ \frac{r}{a_{o}} = 2 ⟹ r = 2 a_{o}$
As $r = r_{o}$ at node, so $r_{o} = 2 a_{o}$

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Similar questions

Q. The wave function of 2s electrons is given by :

Ψ_{2 s} = \frac{1}{4 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{3 / 2} (2 - \frac{r}{a^{0}}) e^{(2 - \frac{r}{a_{0}})}

If it has a node at

r = r_{0}

, then the relation between

r_{0}

and

a_{0}

is:

The wavefunction of a 2s electron is given by $ψ_{2 s} = \frac{1}{4 \sqrt{2} π} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r}{a_{0}}) e^{\frac{- r}{a_{0}}}$
It has a node at $r = r_{0}$ . The value of $\frac{r_{0}}{a_{0}}$ is___

Q. The wave function

(Ψ)

2 s

is given by

Ψ_{2 s} = \frac{1}{2 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{1 / 2} {2 - \frac{r}{a_{0}}} e^{- r / 2 a_{0}}

r = r_{0}

, radial node is formed. Thus for

2 s

r_{0}

in terms of

a_{0}

is:

Q. The wave function

(Ψ)

2 s

is given by:

Ψ_{2 s} = \frac{1}{2 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{\frac{1}{2}} {2 - \frac{r}{a_{0}}} e^{(- r / a_{0})}

Where

a_{0}

is Bohr's radius. If the radial node in

2 s

is at

r_{0},

then find

r_{0}

in terms of

a_{0}

Q. The wave function

(Ψ)

2 s

is given by:

Ψ_{2 s} = \frac{1}{2 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{\frac{1}{2}} {2 - \frac{r}{a_{0}}} e^{- r / 2 a_{0}}

r = r_{0}

radial node is formed. Thus for

2 s

r_{0}

in terms of

a_{0}

is:

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The wave function of 2s electron is given by :Ψ(2s)=14√2π(1a0)3/2(2−ra0)e(2−ra0). If has a node at r=r0, find relation between r0 and a0.

Given wave function: Ψ(2s)=14√2π1ao32(2−rao)e(2−rao)At node, any wave function becomes zero, because at node probability of finding an electron is zero.Thus, Ψ(2s)=0, which gives⟹(2−rao)=0 as e(2−rao) cannot be zero⟹rao=2⟹r=2aoAs r=ro at node, so ro=2ao

The wave function of $2 s$ electron is given by :
$Ψ (2 s) = \frac{1}{4 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{3 / 2} (2 - \frac{r}{a_{0}}) e^{(2 - \frac{r}{a_{0}})}$ . If has a node at $r = r_{0}$ , find relation between $r_{0}$ and $a_{0}$ .