1)
We see that from the diagram below that if L = 3 we get one values for ℓ. and three for j :

_{}

^{}

_{}

^{}

So
ℓ 1 = 1 and
ℓ 2 = 2
therefore: ℓ 1 + ℓ 2 = 1 + 2 = 3.

Meanwhile, S can be defined by only one value of s, or s = ½

The possible j-values are:

j = ℓ + s = ℓ 1 + s = 1 + ½ = 3/2

j = ℓ2 +s = 2 + ½ = 5/2

Meanwhile, S can be defined by only one value of s, or s = ½

The possible j-values are:

j = ℓ + s = ℓ 1 + s = 1 + ½ = 3/2

j = ℓ2 +s = 2 + ½ = 5/2

j
= ℓ - s = ℓ 1 - s = 1 - ½ = ½

j = ℓ 2 - s = 2 - ½ = 3/2

So in total: j = ½, 3/2 and 5/2

Note that, conforming to

To obtain any J (total angular momentum) we need an L-S coupling vector that yields J = 5/2. Two possible L-S couplings are available: [L + S] and [L + S – 1] and it is the last that yields the appropriate result: [3 + ½ - 1] = 5/2

This means we need values such that ℓ 1 = 1, ℓ 2 = 2 and s = ½ to make this work.

j = ℓ 2 - s = 2 - ½ = 3/2

So in total: j = ½, 3/2 and 5/2

Note that, conforming to

**j-selection rules**, all the j's differ by**, though they are half-integral (e.g. 3/2, 5/2) themselves.***an integral amount*To obtain any J (total angular momentum) we need an L-S coupling vector that yields J = 5/2. Two possible L-S couplings are available: [L + S] and [L + S – 1] and it is the last that yields the appropriate result: [3 + ½ - 1] = 5/2

This means we need values such that ℓ 1 = 1, ℓ 2 = 2 and s = ½ to make this work.

2)
The vector solutions (which the reader can do) following the same tack as the
previous probme, will show:

j = ℓ + s = 3 + ½ = 7/2

and:

m

meanwhile:

j = ℓ - s = 3 - ½ = 5/2

So:

m

Note that these last m

j = ℓ + s = 3 + ½ = 7/2

and:

m

**= -7/2, -5/2, -3/2, -½, ½, 3/2, 5/2 and 7/2**_{J}meanwhile:

j = ℓ - s = 3 - ½ = 5/2

So:

m

**= -5/2, -3/2, -½, ½, 3/2, 5/2**_{J}Note that these last m

**quantum numbers would also be the ones applicable to the original problem for which: L = 3 , S = ½, and J = 5/2.**_{J}
3) a) The numerical value of the total angular
momentum is given by:

L
= [ℓ (ℓ + 1)]

^{1/2}(**ħ**)
Where
ℓ = 3,
then:

L
= [3 (3 + 1)]

^{1/2}(**ħ**) = [3 (4)]^{1/2}(**ħ**) = 3Ö2 (**ħ**)
b)
The z-component of the orbital angular momentum is given by:

L(z)
= m

_{ℓ}**ħ**
For
this election, m

_{ℓ}= 3, so that:
L(z)
= 3

**ħ**
(4)
We have ℓ1 =3 and ℓ2 = 2, then:

Therefore,
the possible values of L will be found
from letting ℓ1 =3 and adding each next descending value of m

_{ℓ}from 2, to 1, to 0, to -1, to -2:
(3) + 1 = 4

(3) + 0 = 3

(3) + (-1) = 2

(3) + (-2) = 1

So the total angular momentum L can have the values:

5, 4, 3, 2 and 1.

The

**has ℓ =3 so that the***f electron**total angular momentum quantum number*possibilities are:
j = ℓ + ½,
ℓ - ½

Then: j = 7/2,
5/2

(5) For

ℓ 1 = 2, s1 = 1/2, ℓ 1 = 0 and s1 = 1/2. Then for the maximum value:

*4s 3d*we have:ℓ 1 = 2, s1 = 1/2, ℓ 1 = 0 and s1 = 1/2. Then for the maximum value:

ℓ
1 + ℓ 2 = 2
+ 0 = 2

and: s= s1 + s2 = 1/2 + 1/2 = 1

The

and: s= s1 + s2 = 1/2 + 1/2 = 1

The

*lowest*energy level is then:
4s
3d (

^{3}*)***D1**
Since
2s' + 1 = 3, leading to minimum:

j' = [s' - ℓ '] = 1.

Using the assorted combinations, for ℓ '= 0 and ℓ ' = 2, to get the respective j' values (in combination with s' = 0) and then further for s' = 1, we arrive at the energy configuration diagram shown below:

Using the assorted combinations, for ℓ '= 0 and ℓ ' = 2, to get the respective j' values (in combination with s' = 0) and then further for s' = 1, we arrive at the energy configuration diagram shown below:

_{}^{}
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