Solutions to Quantum Mechanics Problems (2)

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

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.

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.

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 + ℓ 2 = 2 + 0 = 2

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: