For a 3s - orbital- 3s - 19 -31a0 32 6-6 - - -2 e-2- where - - 2r-Z3a0What is the maxium radial distance of node from nucleus-

Radial node occurs where probability of finding e^ is zero. ∴Ψ^2 = 0 or Ψ = 0 ∴ 6 6 σ + σ^2 = 0 ∴ nbsp;σ = 3 ±√(3) For maximum distance r = 32×(3 + √(3)) ...

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Byju's Answer

Standard XIII

Chemistry

Aufbau Principle

For a 3s - or...

Question

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?

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

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\frac{a_{0}}{Z}

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\frac{3}{2} \times \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 $\frac{3}{2} \times \frac{(3 + \sqrt{3}) a_{0}}{Z}$
Radial node occurs where probability of finding $e^{-}$ is zero.
$∴ Ψ^{2} = 0 or Ψ = 0 ∴ 6 - 6 σ + σ^{2} = 0 ∴ σ = 3 \pm \sqrt{3}$
For maximum distance $r = \frac{3}{2} \times \frac{(3 + \sqrt{3}) a_{0}}{Z}$

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Similar questions

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

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-orbitals

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}}

What is the maximum 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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For a 3s− orbital Ψ(3s)=19√3(1a0)32(6−6σ+σ2)e−σ2;where σ=2r.Z3a0 What is the maxium radial distance of node from nucleus?

The correct option is C 32×(3+√3)a0ZRadial node occurs where probability of finding e− is zero. ∴Ψ2=0 or Ψ=0∴6−6σ+σ2=0∴σ=3±√3 For maximum distance r=32×(3+√3)a0Z

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?

The correct option is C $\frac{3}{2} \times \frac{(3 + \sqrt{3}) a_{0}}{Z}$
Radial node occurs where probability of finding $e^{-}$ is zero.
$∴ Ψ^{2} = 0 or Ψ = 0 ∴ 6 - 6 σ + σ^{2} = 0 ∴ σ = 3 \pm \sqrt{3}$
For maximum distance $r = \frac{3}{2} \times \frac{(3 + \sqrt{3}) a_{0}}{Z}$