The wave function of atomic orbital of H-like atoms is given as-2 s -1-4 -2- Z 3 - 22- Zr e -Z r -2Given that the radius is in A- Then- which of the following will be the radius for nodal surface of He -ion- A- 1 A 0B- 2 0 AC- 2-5 0 AD- 4 A 0

The correct option is A 1 oAThe wave function is given as, ψ _2s=14√(2)πZ^3/2(2 Zr)e^ Zr2 At nodal surface, ψ =0 ⇒ 2 Zr=0 [for He^+, Z=2] ⇒ r=2Z=2Z oA=1 oA ...

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Standard XII

Chemistry

Shapes of Orbitals and Nodes

The wave func...

Question

The wave function of atomic orbital of H-like atoms is given as:
$ψ_{2 s} = \frac{1}{4 \sqrt{2} π} Z^{3 / 2} (2 - Z r) e^{- \frac{Z r}{2}}$
Given that the radius is in $o A$ . Then, which of the following will be the radius for nodal surface of $H e^{+}$ ion?

1 o A

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2 o A

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2.5 o A

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4 o A

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Solution

The correct option is A $1 o A$
The wave function is given as,
$ψ_{2 s} = \frac{1}{4 \sqrt{2} π} Z^{3 / 2} (2 - Z r) e^{- \frac{Z r}{2}}$
At nodal surface, $ψ = 0$
$\Rightarrow 2 - Z r = 0 [for H e^{+}, Z = 2]$
$\Rightarrow r = \frac{2}{Z} = \frac{2}{Z} o A = 1 o A$

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The wave function of atomic orbital of H-like atoms is given as: ψ2s=14√2πZ3/2(2−Zr)e−Zr2 Given that the radius is in oA. Then, which of the following will be the radius for nodal surface of He+ ion?

The correct option is A 1 oAThe wave function is given as, ψ2s=14√2πZ3/2(2−Zr)e−Zr2 At nodal surface, ψ=0 ⇒2−Zr=0 [for He+,Z=2] ⇒r=2Z=2Z oA=1 oA

The wave function of atomic orbital of H-like atoms is given as:
$ψ_{2 s} = \frac{1}{4 \sqrt{2} π} Z^{3 / 2} (2 - Z r) e^{- \frac{Z r}{2}}$
Given that the radius is in $o A$ . Then, which of the following will be the radius for nodal surface of $H e^{+}$ ion?

The correct option is A $1 o A$
The wave function is given as,
$ψ_{2 s} = \frac{1}{4 \sqrt{2} π} Z^{3 / 2} (2 - Z r) e^{- \frac{Z r}{2}}$
At nodal surface, $ψ = 0$
$\Rightarrow 2 - Z r = 0 [for H e^{+}, Z = 2]$
$\Rightarrow r = \frac{2}{Z} = \frac{2}{Z} o A = 1 o A$