Problem 6.11
The magnetic quantum number, ml, is limited to the values 0, ñ 1, ñ 2,
.... ñ l. There are a total of 2l + 1 values.
For an orbital quantum number of l = 4, there are nine possible
values for m.l:
m
l
4
3
,
2
,
1
,
0
,
1
,
2
,
3
,
4
,
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Problem 6.12
The orbital quantum number l may take the values l = 0, 1, 2,
......
n-1, where n is the principal quantum number.
Each orbital
quantum number has an associated set of magnetic quantum
numbers, m.l.
The magnetic quantum number m.l is limited to the values 0, ñ 1, ñ 2,
.... ñ l. There are a total of 2l + 1 values for each value of the orbital
quantum number l.
For a principal quantum number n = 4, l takes the values l = 0, 1, 2, 3.
l = 0
m.l = 0
l = 1
m.l = -1, 0-, 1
l = 2
m.l = -2, -1, 0, 1, 2
l = 3
m.l = -3, -2, -1, 0, 1, 2, 3
The total number of orbitals with n = 4 is therefore N = 1 + 3 + 5 + 7 = 16.
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Problem 6.13
s p d f
l=
0 1 2 3
Ll
l
1
()
.
hbar
.
L
z
m
l
hbar
.
because m
l
ranges from -l to l, the maximum value of m
l
is l and the
maximum value of L
z
is l*hbar
L
zmax
l hbar
.
hbar cancels out in the equation for percentage difference, so for
convenience, I will just set it equal to 1
hbar
1
percent l
l
() l 1
.
hbar
.
l hbar
.
l
.
hbar
.
100
.
2