411
A
TOMIC
S
TRUCTURE
41.1.
I
DENTIFY
and
S
ET
U
P
:
(
1)
L
l l
=
+
=
.
z
l
L
m
=
=
.
0, 1, 2,...,
1
l
n
=
−
.
0,
1,
2,...,
l
m
l
=
±
±
±
.
cos
/
z
L
L
θ
=
.
E
XECUTE
:
(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
L
L
θ
=
.
0
L
=
:
θ
not defined.
2
L
=
=
:
45.0
, 90.0
, 135.0
°
°
°
.
6
L
=
=
:
35.3
, 65.9
, 90.0
, 114.1
, 144.7
°
°
°
°
°
.
E
VALUATE
:
There is no state where
G
L
is totally aligned along the
z
axis.
41.2.
I
DENTIFY
and
S
ET
U
P
:
(
1)
L
l l
=
+
=
.
z
l
L
m
=
=
.
0,1,2,...,
1
l
n
=
−
.
0, 1, 2,...,
l
m
l
=
±
±
±
. cos
/
z
L
L
θ
=
.
E
XECUTE
:
(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
=
:
2
3
L
=
=
,
3 ,2 , ,0,
, 2 , 3
z
L
=
−
−
−
=
= =
=
=
=
.
4
l
=
:
2
5
L
=
=
,
4 ,3 ,2 , ,0,
, 2 , 3 , 4
z
L
=
−
−
−
−
=
=
= =
=
=
=
=
.
(b)
0
L
=
:
θ
not defined.
2
L
=
=
:
45.0 ,90.0 ,135.0
°
°
°
.
6
L
=
=
: 35.3 ,65.9 ,90.0 ,114.1 ,144.7
°
°
°
°
°
.
2
3
L
=
=
:
54.7 ,73.2 ,90.0 ,106.8 ,125.3 ,150.0
°
°
°
°
°
°
.
2
5
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.
I
DENTIFY
and
S
ET
U
P
:
The magnitude of the orbital angular momentum
L
is related to the quantum number
l
by
Eq.(41.4):
(
1)
, 1
0, 1, 2,
L
l l
=
+
=
=
…
E
XECUTE
:
2
34
2
34
4.716
10
kg m /s
(
1)
20
1.055
10
J s
L
l l
−
−
⎛
⎞
×
⋅
⎛
⎞
+
=
=
=
⎜
⎟
⎜
⎟
×
⋅
⎝
⎠
⎝
⎠
=
And then (
1)
20
l l
+
=
gives that
4.
l
=
E
VALUATE
:
l
must be integer.
41.4.
(a)
max
max
(
)
2, so (
)
2 .
l
z
m
L
=
=
=
(b)
(
1)
6
2.45 .
l l
+
=
=
=
=
=
(c)
The angle is arccos
arccos
,
6
z
l
L
m
L
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
and the angles are, for
2 to
2, 144.7
,
l
l
m
m
= −
=
°
114.1 , 90.0 ,
°
°
65.9 , 35.3 .
°
°
The angle corresponding to
l
m
l
=
will always be larger for larger
.
l
41.5.
I
DENTIFY
and
S
ET
U
P
:
The angular momentum
L
is related to the quantum number
l
by Eq.(41.4),
(
1) .
L
l l
=
+
=
The maximum
l
,
max
,
l
for a given
n
is
max
1.
l
n
=
−
E
XECUTE
:
For
max
2,
1 and
2
1.414 .
n
l
L
=
=
=
=
=
=
For
max
20,
19 and
(19)(20)
19.49 .
n
l
L
=
=
=
=
=
=
For
max
200,
199 and
(199)(200)
199.5 .
n
l
L
=
=
=
=
=
=
E
VALUATE
:
As
n
increases, the maximum
L
gets closer to the value
n
=
postulated in the Bohr model.
41.6.
The ( ,
)
l
l m
combinations are (0,
0), (1,
0),
(1,
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
1
2
10
0
0
1
1
(1.60
10
C)
2.3
10
J
4
4
1.0
10
m
q q
U
π
r
π
−
−
−
−
×
=
=
= −
×
×
P
P
18
19
2.3
10
J
14.4 eV.
1.60
10
J eV
U
−
−
−
×
=
= −
×
41
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412
Chapter 41
41.8.
(a)
As in Example 41.3, the probability is
2
2
2
3
1
/ 2
2
2
2
1
3
0
0
4
5

 4
1
0.0803
2
2
4
2
a
a
r a
s
ar
a r
a
e
P
ψ
π
r dr
e
a
−
−
⎡
⎤
⎛
⎞
=
=
−
−
−
=
−
=
⎢
⎥
⎜
⎟
⎝
⎠
⎣
⎦
∫
.
(b)
The difference in the probabilities is
2
1
1
2
(1
5
)
(1
(5 2)
)
(5 2)(
2
)
0.243.
e
e
e
e
−
−
−
−
−
−
−
=
−
=
41.9.
(a)
2
2
2



( )  
( )  (
)(
)
l
l
im
im
*
ψ
ψ ψ
R r
θ
Ae
Ae
φ
φ
−
+
=
=
Θ
2
2
2

( )  
( )  ,
A
R r
θ
=
Θ
which is independent of
φ
.
(b)
2
2
2
2
2
0
0
1

( ) 
2
1
.
2
π
π
d
A
d
π
A
A
π
φ
φ
φ
Φ
=
=
=
⇒
=
∫
∫
41.10.
4
1
12
2
1
1
1
2
2
2
2
0
1
(0.75)
.
(4
)
2
2
r
n
m e
E
E
E
E
E
E
E
π
n
= −
Δ
=
−
=
−
= −
=
P
(a)
31
r
If
9.11 10
kg
m
m
−
=
=
×
(
)
4
31
19
4
2
9
2
18
r
2
2
34
2
0
(9.109
10
kg)(1.602
10
C)
8.988
10 N
m
C
2.177
10
J
13.59 eV
(4
)
2(1.055
10
J s)
m e
π
−
−
−
−
×
×
=
×
⋅
=
×
=
×
⋅
=
P
For 2
1
→
transition, the coefficient is (0.75)(13.59 eV)
=
10.19 eV.
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 Spring '11
 Shaefer
 Atom, Angular Momentum, Energy, Photon, Atomic orbital

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