1)
We see that from the diagram below that if L = 3 we get one value 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 j-selection rules, all the j's differ by an integral amount, though they are half-integral (e.g. 3/2, 5/2) themselves.
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 an integral amount, though they are half-integral (e.g. 3/2, 5/2) themselves.
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 J = -7/2, -5/2, -3/2, -½, ½, 3/2, 5/2 and 7/2
meanwhile:
j = ℓ - s = 3 - ½ = 5/2
So:
m J = -5/2, -3/2, -½, ½, 3/2, 5/2
Note that these last m J quantum numbers would also be the ones applicable to the original problem for which: L = 3 , S = ½, and J = 5/2.
j = ℓ + s = 3 + ½ = 7/2
and:
m J = -7/2, -5/2, -3/2, -½, ½, 3/2, 5/2 and 7/2
meanwhile:
j = ℓ - s = 3 - ½ = 5/2
So:
m J = -5/2, -3/2, -½, ½, 3/2, 5/2
Note that these last m J quantum numbers would also be the ones applicable to the original problem for which: L = 3 , S = ½, and J = 5/2.
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
f
electron has ℓ =3 so that the total angular momentum quantum number
possibilities are:
j = ℓ + ½,
ℓ - ½
Then: j = 7/2,
5/2
(5) For
4s 3d
we have:
ℓ 1 = 2, s1 = 1/2, ℓ 1 = 0 and s1 = 1/2. Then for the maximum value:
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 lowest energy level is then:
and: s= s1 + s2 = 1/2 + 1/2 = 1
The lowest energy level is then:
4s
3d (3D1)
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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