41-1
A
TOMIC
S
TRUCTURE
41.1.
IDENTIFY
and
SET UP:
(1
)
Ll
l
=+
=
.
zl
Lm
=
=
.
0, 1, 2,.
..,
1
ln
=−
.
0, 1,
2,.
l
ml
=±±
±
.
cos
/
z
LL
θ
=
.
EXECUTE: (a)
0
l
=
:
0
L
=
,
0
z
L
=
.
1
l
=
:
2
L
=
=
,
,0,
z
L
==
.
2
l
=
:
6
L
=
=
,
2 , ,0,
, 2
z
L
−
=
=
.
(b)
In each case cos
/
z
=
.
0
L
=
:
not defined.
2
L
=
=
:
45.0 , 90.0 , 135.0
°°
°
.
6
L
=
=
:
35.3 , 65.9 , 90.0 , 114.1 , 144.7
°
°
°°°
.
EVALUATE:
There is no state where
G
L
is totally aligned along the
z
axis.
41.2.
IDENTIFY
and
SET UP:
)
l
=
.
=
=
.
0,1,2,.
..,
1
.
0, 1, 2,.
..,
l
±
. cos
/
z
=
.
EXECUTE: (a)
0
l
=
:
0
L
=
,
0
z
L
=
.
1
l
=
:
2
L
=
=
,
,0,
z
L
.
2
l
=
:
6
L
=
=
,
2 , ,0,
, 2
z
L
−
=
=
.
3
l
=
:
23
L
=
=
,
3,2,,0
, ,2,3
z
L
−
−
= = = =
.
4
l
=
:
25
L
=
=
,
4,3,2,,0
, ,2,3,4
z
L
−
−
−
=
=
=
= ====
.
(b)
0
L
=
:
not defined.
2
L
=
=
:
45.0 ,90.0 ,135.0
°
.
6
L
=
=
: 35.3 ,65.9 ,90.0 ,114.1 ,144.7
°
°
.
L
=
=
:
54.7 ,73.2 ,90.0 ,106.8 ,125.3 ,150.0
°
°
°°°°
.
L
=
=
: 26.6 ,47.9 ,63.4 ,77.1 ,90.0 ,102.9 ,116.6 ,132.1 ,153.4
°
°
°
°
°°°°°
.
(c)
The minimum angle is 26.6
°
and occurs for
4
l
=
,
4
l
m
. The maximum angle is 153.4
°
and occurs for
4
l
=
,
4
l
m
.
41.3.
IDENTIFY
and
SET UP:
The magnitude of the orbital angular momentum
L
is related to the quantum number
l
by
Eq.(41.4):
)
,
10
,
1
,
2
,
l
=
=
…
EXECUTE:
2
34
2
34
4.716 10
kg m /s
)
2
0
1.055 10
J s
L
ll
−
−
⎛⎞
×⋅
+=
=
=
⎜⎟
⎝⎠
=
And then (
1)
20
gives that
4.
l
=
EVALUATE:
l
must be integer.
41.4.
(a)
max
max
()
2
,
s
o
2
.
lz
mL
=
(b)
(
6
2.45 .
=
=
(c)
The angle is arccos
arccos
,
6
L
=
and the angles are, for
2 to
2, 144.7 ,
mm
=
°
114.1 , 90.0 ,
65.9 , 35.3 .
The angle corresponding to
l
=
will always be larger for larger .
l
41.5.
IDENTIFY
and
SET UP:
The angular momentum
L
is related to the quantum number
l
by Eq.(41.4),
)
.
l
=
The maximum
l
,
max
,
l
for a given
n
is
max
1.
EXECUTE:
For
max
2,
1 and
2
1.414 .
nl
L
=
=
For
max
20,
19 and
(19)(20)
19.49 .
L
=
=
For
max
200,
199 and
(199)(200)
199.5 .
L
=
=
EVALUATE:
As
n
increases, the maximum
L
gets closer to the value
n
=
postulated in the Bohr model.
41.6.
The ( ,
)
l
lm
combinations are (0,
0), (1,
0), (1,
±
, (2,
0), (2,
1),
±
(2,
2),
±
(3,
0),
(3,
1), (3,
2), (3,
3), (4, 0), (4,
1), (4,
2), (4,
3), and (4,
4),
±±
±
±
±
a total of 25.
(b)
Each state has the same energy (
n
is the same),
13.60 eV
0.544 eV.
25
−=
−
41.7.
19
2
18
12
10
00
11
(
1
.
6
0
1
0
C
)
2.3 10
J
44
1
.
0
1
0
m
qq
U
π
r
π
−
−
−
−×
=
−
×
×
PP
18
19
2.3 10
J
14.4 eV.
1.60 10
J eV
U
−
−
−
×
41