(c) Since
L
orb,
z
=
m
A
h
/2
π
, the greatest allowed value of
L
orb,
z
is given by 
m
A

max
h
/2
π
=
3
h
/2
π
.
(d) Similar to part (c), since
µ
orb,
z
= –
m
A
B
, the greatest allowed value of
orb,
z
is given by

m
A

max
B
= 3
eh
/4
π
m
e
.
(e) From Eqs. 3223 and 3229 the
z
component of the net angular momentum of the
electron is given by
net,
orb,
,
.
22
s
zz
s
z
mh
LLL
=+
=
+
π
π
A
For the maximum value of
L
net,
z
let
m
A
= [
m
A
]
max
= 3 and
m
s
=
1
2
. Thus
L
hh
z
net,
max
.
.
F
H
G
I
K
J
=
3
1
35
2
ππ
(f) Since the maximum value of
L
net,
z
is given by [
m
J
]
max
h
/2
π
with [
m
J
]
max
= 3.5 (see the
last part above), the number of allowed values for the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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