For 2s orbital- 1 - 8 z - a 0 3 - 2 2- Zr - a 0 e zr - 2 a 0 then- hydrogen radial node will be at the distance of-

The correct option is B 2 a _ 0 Given 1√(8)(Za_0)^32(2 Zra_0)e^Zr2a_0 R=12√(2)(Za_0)^32(2 ρ)e^ρh=0 Here, ρ= 2, ∴ρ2Zrra_0=e For hydrogen ...

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Byju's Answer

Standard XII

Chemistry

Inductive Effect

For 2s orbita...

Question

For 2s orbital, $\frac{1}{\sqrt{8}} {(\frac{z}{a_{0}})}^{\frac{3}{2}} (2 - \frac{Z r}{a_{0}}) e^{\frac{z r}{2 a_{0}}}$ then, hydrogen radial node will be at the distance of?

a_{0}

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2 a_{0}

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a_{0} / 2

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a_{0} / 3

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Solution

The correct option is B $2 a_{0}$
Given $\frac{1}{\sqrt{8}} {(\frac{Z}{a_{0}})}^{\frac{3}{2}} (2 - \frac{Z r}{a_{0}}) e^{\frac{Z r}{2 a_{0}}}$
$R = \frac{1}{2 \sqrt{2}} {(\frac{Z}{a_{0}})}^{\frac{3}{2}} (2 - ρ) e^{\frac{ρ}{h}} = 0$
Here, $ρ = 2, ∴ ρ \frac{2 Z r}{r a_{0}} = e$
For hydrogen $Z = 1$
$\Rightarrow ρ = \frac{2 (1) r}{2 a_{0}} = 2$
$\Rightarrow r = 200$ .

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Similar questions

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

-orbitals given by:

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}

,radial node is formed.If

r_{0}

x a_{0}

, find the value of

x

Q. The Schrondinger equation for hydrogen atom is

ψ = \frac{1}{4 \sqrt{2 π}} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r_{0}}{a_{0}}) e^{-} \frac{r_{0}}{a_{0}}

, where

a_{0}

is Bohr's radius. If the radial node is

2 s

be at

r_{0}

.
Then,

r_{0}

in terms of

a_{0}

is given as

x a_{0} = r_{0}

, value of

x

Q. The Schrondinger equation for hydrogen atom is

ψ = \frac{1}{4 \sqrt{2 π}} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r_{0}}{a_{0}}) e^{-} \frac{r_{0}}{a_{0}}

, where

a_{0}

is Bohr's radius. If the radial node is

2 s

be at

r_{0}

.
Then,

r_{0}

in terms of

a_{0}

is given as

x a_{0} = r_{0}

, value of

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Q. The complete wave function of an orbital of hydrogen is given as

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

,
(

a_{0}

is first Bohr radius)
Which of the following option(s) is/are correct?

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For 2s orbital, 1√8(za0)32(2−Zra0)ezr2a0 then, hydrogen radial node will be at the distance of?

The correct option is B 2a0Given 1√8(Za0)32(2−Zra0)eZr2a0R=12√2(Za0)32(2−ρ)eρh=0Here, ρ=2,∴ρ2Zrra0=eFor hydrogen Z=1⇒ρ=2(1)r2a0=2⇒r=200 .

For 2s orbital, $\frac{1}{\sqrt{8}} {(\frac{z}{a_{0}})}^{\frac{3}{2}} (2 - \frac{Z r}{a_{0}}) e^{\frac{z r}{2 a_{0}}}$ then, hydrogen radial node will be at the distance of?