The wave function of atomic orbital of H-like species is given as ψ 2 s = 1 4 2 π Z 3 / 2 ( 2 - Z r ) e - Z r / 2 The radius for nodal surface for H e + ion in Å is

At nodal surface ψ 2 = 0 So 2 - Z r = 0 r = 2 Z = 2 2 = 1 Å .

Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

The wave function of atomic orbital of H-like species is given as

$ψ_{2 s} = \frac{1}{4 \sqrt{2} π} Z^{3 / 2} (2 - Z r) e^{- Z r / 2}$

The radius for nodal surface for $H e^{+}$ ion in $Å$ is

Answer:

1.00

Solution:

At nodal surface $ψ^{2} = 0$
So $2 - Z r = 0$
$r = \frac{2}{Z} = \frac{2}{2}$
$= 1 Å$ .

Stream:NTA_ABHYASSubject:ChemistryTopic:Atomic StructureSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...