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 He+ ion in mathringA .

Question

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 He+ ion in mathringA . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As at nodal surface probability of finding an electron is zero, therefor, ψ=0 So 2-Zr=0 r=(2/Z)=(2/2)=1 AA

Q. The wave function of atomic orbital of H like species is given as
$(ψ)_{2 s} = \frac{1}{4 2 π} Z^{\frac{3}{2}} (2 - Z r) e^{- \frac{Z r}{2}}$
The radius for nodal surface for $H e^{+}$ ion in $\overset{˚}{A}$ .

As at nodal surface probability of finding an electron is zero, therefor, $ψ = 0$
So $2 - Z r = 0$
$r = \frac{2}{Z} = \frac{2}{2} = 1 \overset{˚}{A}$

Q. The wave function of atomic orbital of H like species is given as (ψ)2s​=42​π1​Z23​(2−Zr)e−2Zr​ The radius for nodal surface for He+ ion in A˚ .

As at nodal surface probability of finding an electron is zero, therefor, ψ=0 So 2−Zr=0 r=Z2​=22​=1A˚

Q. The wave function of atomic orbital of H like species is given as
$(ψ)_{2 s} = \frac{1}{4 2 π} Z^{\frac{3}{2}} (2 - Z r) e^{- \frac{Z r}{2}}$
The radius for nodal surface for $H e^{+}$ ion in $\overset{˚}{A}$ .

As at nodal surface probability of finding an electron is zero, therefor, $ψ = 0$
So $2 - Z r = 0$
$r = \frac{2}{Z} = \frac{2}{2} = 1 \overset{˚}{A}$