**a)**As a result of the stronger spin-orbit interaction, the orbital and the spin angular

momentum vectors of each individual electron are strongly coupled together to form a resultant angular monnentuin vector of magnitude :

*√ j ( j + 1).h/2π*

where,

**=***j**l – 1/2*and*l + 1/2*that is, j takes half-integral values only.This means that due to spin-orbit interaction, the unperturbed energy level is splitted into a number of well-spaced levels, each corresponding to a different combination of the possible j-values for the individual optical electrons; the level corresponding to all the electrons having their smaller j-value (j =

*l – 1/2*)being lowest.**b)**As a result of the residual electrostatic interaction and spin-spin corelation, the

resultant angular momentum vectors

**j**of the individual electrons are less stronglycoupled with one another to form the total angular momentum vector

**J**of the atom, ofmagnitude = √

*J(J + 1).h/2π*

The total angular momentum quantum number takes the values:

Let us know explain electronic configuration of 4p 4d under J – J coupling.

This gives four possible combinations :

{ 1/2 , 3/2 }, {1/2 ,5/2 } ,{3/2 ,3/2 } ,{3/2 ,5/2}

The four combinations of J values gives :

The complete splitting is shown in fig. below as :

**Selection Rule in j -j Coupling :**

**(i)**The parity of the configuration must change in an electric-dipole transition

(Laporte rule). This means that if only one electron jumps in the transition (as is usually

the case) then for this election we must have Δ

*l*= ± 1. If two electrons jump then,Δ

*l1*= ±1 and Δl2 = 0,± 2. This rule is exactly same as in L-S coupling.**(ii) Δ**j = 0, ±1 for the jumping electron, and Δj = 0 for all the other electrons.

**(iii)**For the atom as a whole,

ΔJ=0, ± 1 but J = 0 <—/—> J = 0.

The selection rules ΔS = 0 and Δ L = 0, ±1 no

longer good quantum numbers.

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