41.1:
2
34
34
2
sJ10054.1
sJ104.716
)1()1(






















L

llllL
.40.20)1(





lll
41.2: a)
.2so,2
maxmax

zl
Lm
b)
.45.26)1(  ll
c)
The angle is arccos
,
6
arccos














l
z
m
L
L
and the angles are, for
,0.90,1.114,7.144,2to2 
ll
mm
.3.35,9.65


The angle corresponding to
lm
l

will always be larger for larger
.l
41.3:
.)1(  llL
The maximum orbital quantum number
:ifSo.1


nl




5.199)200(199199200
49.19)20(191920
41.12)1(12



Lln
Lln
llLln
The maximum angular momentum value gets closer to the Bohr model value the
larger the value of
.
n
41.4: The
),(
l
ml
combinations are (0, 0), (1, 0),
)1,1(

, (2, 0),
),1,2(

),2,2(

(3, 0),
4),(4,and),3,4(),2,4(),1,4(),0,4(),3,3(),2,3(),1,3(








a total
of 25.
b) Each state has the same energy (
n
is the same),
eV.544.0
25
eV60.13

41.5:
J103.2
m100.1
C)1060.1(
4
1
.
4
1
18
10
219
0
21
0








πεr
qq
πε
U
eV.4.14
eVJ1060.1
J103.2
19
18






41.6:
a) As in Example 41.3, the probability is


















2/
0
2
0
2
322
3
22
1
422
4
4||
a
a
ar
s
e
araar
a

dr
πrψP
.0803.0
2
5
1
1


e
b) The difference in the probabilities is
.243.0)2)(25())25(1()51(
2112


eeee
41.7: a)
))((|)(||)(|||
222

ll
imim
*
AeAeθrRψψψ


,|)(||)(|
222
θrRA 
which is independent of


b)


ππ
π
AπAdAd
2
0
2
2
0
22
.
2
1
12|)(|

41.8:
.)75.0(
22)4(
1
11
2
1
1212
22
4
2
0

EE
E
EEE
n
em
πε
E
r
n


a)
kg1011.9If
31
 mm
r
 
2
29
234
49131
22
0
4
CNm10988.8
s)J10055.1(2
C)10kg)(1.602109.109
)4(







πε
em
r

eV.59.13J10177.2
18


For
12

transition, the coefficient is (0.75)(13.59 eV)=10.19 eV.
b) If
,
2
m
m
r

using the result from part (a),
eV.795.6
2
eV59.132
eV)59.13(
)4(
22

0
4















m
m
πε
em
r

Similarly, the
12

transition,
eV.095.5
2
eV19.10









c) If
,8.185 mm
r

using the result from part (a),
eV,2525
185.8
eV)59.13(
)4(
22
0
4








m
m

πε
em
r

and the
12

transition gives

(10.19 eV)(185.8)=1893 eV.
41.9: a)
21931
234
0
2
2
0
1
C)10kg)(1.60210109.9(
)sJ10626.6(
:





π
ε
πme
hε

amm
r
m10293.5
11
1

 a
b)
.m10059.12
2
10
12

 aa
m
m
r
c)
.m10849.2
185.8
1
m8.185
13
13

 aam
r
41.10:
),sin()cos(



ll
im
mime
l

and to be periodic with period
πmπ
l
2,2
must be an
integer multiple of
l
mπ so,2
must be an integer.
41.11:
 


a
a
ar
s
drπre
πa
dVψaP
0
0
22
3

1
)4(
1
)(

.51)(
4422
4
422
44
)(
2
0
3
2
333
3
0
2
222
3
22
3










































eaP
e
a
e
aaa
a
e
araar
a
drer
a
aP
a
arar
a
o
41.12: a)
,2b),eV102.32T)400.0)(TVe1079.5(
55
B


l
mBμE
the
lowest possible value of
.

l
m
c)
41.13: a)
4).3,2,1,,0(91)(2isstatesof#4state-










l
mllg
b)
.J1056.5eV1047.3)T600.0)(TVe1079.5(
2455
B

 BμU
c)
eV1078.2)T600.0)(TVe1079.5(88
45
B44


 BμU


.J1045.4
23

41.14:
a) According to Fig. 41.8 there are three different transitions that are consistent
with the selection rules. The initial
l
m
values are 0,
;1

and the final
l
m
value is 0.
b) The transition from
0to0 
ll
mm
produces the same wavelength (122 nm)
that was seen without the magnetic field.
c) The larger wavelength (smaller energy) is produced from the
0to1 
ll
mm
transition.
d) The shorter wavelength (greater energy) is produced from the
0to1 
ll

mm
transition.
41.15: a)
B
B
,1,33
μ
U
BB
μUlnp 

.T468.0
T)Ve10(5.79
eV)1071.2(
5
5






B
b) Three
.1,0 
l
m
41.16: a)
B
m

e
U














22
)00232.2(


Bμ
B
2
)00232.2(


.eV1078.2
)T480.0)(TVe10788.5(
2
)00232.2(

5
5




b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if
0

n
there could be since l < n allows for
.0

l
41.17:
BS
m
e
BU
zz
2
)00232.2(


.
2
where)00232.2(
)(
2
)00232.2(

BB
m
e
μBmμU
Bm
m
e
U
s
s




So the energy difference
















2
1
2
1
)00232.2(
B
BμU

)T45.1)(TVe10788.5)(00232.2(
5
 U

.eV1068.1
4

And the lower energy level is
).direction
ˆ
theinpointssince(
2
1
zBm
s

41.18: The allowed
),( jl
combinations are
.
2
5

,2and
2
3
,2,
2
3
,1,
2
1
,1,
2
1
,0































41.19:
j
quantum numbers are either
.4then,27and29ifSo,
2
1
or
2
1
 ljll
The letter used to describe
is4

l
“g”
41.20: a)









λ
,cm21
eV)10(5.9
)sm10s)(300eV10136.4(
λ
6
815
c
f
E
hc
9
8
104.1
0.21m
)sm1000.3(


Hz, a short radio wave.
b) As in Example 41.6, the effective field is
for,T101.52
2
B


 μEB
smaller
than that found in the example.
41.21: a) Classically
2
5
2
and,
mRIIωL 
for a uniform sphere.
s.rad102.5
4
3
m)10(1.0kg)102(9.11
s)J105(1.054
4
3
2
5
4
3
5
2
30
21731
34
2
2








ω
mR
ω
RmωL


b)
.sm102.5)srad10(2.5m)10(1.0
133017


rωv
Since this is faster
than the speed of light this model is invalid.
41.22:
For the outer electrons, there are more inner electrons to screen the nucleus.
41.23: Using Eq. (41.27) for the ionization energy:
eV).(13.6
2
2
eff
n
Z
E

n


The
s5
electron sees
5and771.2
eff
 nZ
eV.4.18eV)(13.6
5
(2.771)
2
2
5



E
41.24:
However the number of electrons is obtained, the results must be consistent with
Table (43-3); adding two more electrons to the zinc configuration gives
622
221 pss
.43433
210262
pdsps
41.25:
The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells.
states.6:

2
1
,1,0,1,2
states.2:
2
1
,0,0,2
states.2:
2
1
,0,0,1



sl
sl
sl
mmln
mmln
mmln
41.26: For the
s4
state,
.26.2)6.13()339.4(4andeV339.4
eff
 ZE
Similarly,
79.1
eff
Z

for the 4p state and 1.05 for the 4d state. The electrons in the states
with higher l tend to be further away from the filled subshells and the screening is more
complete.
41.27: a) Nitrogen is the seventh element (Z = 7).
2
N
has two electrons removed, so
there are 5 remaining electrons

electron configuration is
.221
22
pss
b)
eV30.6eV)(13.6
2
4)(7
eV)(13.6
2
2
2
2
eff





n
Z

E
c) Phosphorous is the fifteenth element (Z = 15).
2
P
has 13 electrons, so the
electron configuration is
.33221
2622
pspss
d) The least tightly held electron:
eV.13.6eV)(13.6
3
12)(15
2
2



E
41.28: a)
.26.1so,
4
e6.13
eff
2
eff2
 ZZ
V
E
b) Similarly,

2.26.
eff
Z
c)
eff
Z
becomes larger going down the columns in the periodic table.
41.29: a) Again using
eV),(13.6
2
2
eff
n
Z
E
n


the outermost electron of the
L

Be
shell
(n = 2) sees the inner two electrons shield two protons so
2.
eff

Z
eV.13.6eV)(13.6
2

2
2
2



2
E
b) For
,Ca

outer shell has n = 4, so
eV.3.4eV)(13.6
4
2
2
2
4



E
41.30:
eV)(10.2)1(
2
 ZE
kx
,0.28
eV10.2
eV107.46

1
3



Z
which corresponds to the element Nickel (Ni).
41.31: a)
Hz108.951)Hz)(2010(2.48:20
17215
 fZ
m.103.35
Hz108.95
sm103.00
λ
keV3.71Hz)10(8.95s)eV10(4.14
10
17
8
1715







f
c
hfE

b) Z = 27:
m.101.79λ
keV6.96
Hz101.68
10
18




E
f
c)
m.105.47λkeV,22.7Hz,105.48:48
1118 
 EfZ
41.32: See Example 41.3;
)),2(2(,
22
2
2
/22
2
2
arrCe
dr
ψdr
eCr
ψr
arar



and for a
maximum, r = a, the distance of the electron from the nucleus in the Bohr model.
41.33: a)



2
4
2
0
1
2
0
2
4
2
0
1
2)4(
1
then),(.
4
1
)(and
2)4(
1

me

πε
rUEIf
r
e
πε
rU
me
πε
E
ss

a
me
πε
r
r
e
πε
2
24
)4(
1
2
2
0
2
0




b)
.
1
isand4)2(
/
3
22
2
1
2
2
1
ar
a
s
a
s
e
a
π
ψ
drrψπdVψarP




.
4
2
4

)2(
422
4
)2(
4
)2(
3
334
3
2
322
2
3
2
22
3
































a
aae
a
arP
araar
e
a
arP
drer
a
arP
a
ar

a
ar

.238.013
4


e
41.34:
a) For large values of n
, the inner electrons will completely shield the nucleus, so
1
ff

e
Z
and the ionization energy would be
2
eV60.13
n
.
b)
m106.48m)10(0.529(350))350(,eV101.11
350
eV13.60
6102
0
2
350
4

2

 ar
.
c) Similarly for n = 650,
650
5
2
,eV103.22
(650)
eV13.60
r

  (0.529(650)
2
m.102.24m)10
510 

41.35:
a) If normalized, then
.
4
4
8
1
2
32
4
14
0

2
43
2
3
2
2
0
3
2
0
2
2
0
2
2
dre
a
r
a
r
r
a
dre
a
r
r
πa
π
I
drr

ψπdVψ
ar
ar
ss

























But recall






0
1
.
!
n
axn
n
dxex

So







5
2
4
3
)24(
1
)6(
4

)2(4
8
1
3
a
a
a
a
a
a
I



.normalizedisand124248
8
1
2
333
3
s
ψaaa
a
I 
b) We carry out the same calculation as part (a) except now the upper limit on the
integral is 4a, not infinity.
So
.44
8
1

4
0
2
43
2
3
dre
a
r
a
r
r
a
I
ar
a











Now the necessary integral formulas are:










)2424124(
)663(
)22(
54322344
332233
3222
araarararedrer
araararedrer
araaredrer
arar
arar
arar
All the integrals are evaluated at the limits
0

r
and
.4a
After carefully plugging in the
limits and collecting like terms we have:
)]824568104()24248[(
8
1

43
3


ea
a
I

).4(Prob176.0)3608(
8
1
4
areI 

41.36: a) Since the given
.||real,is)(
2222
ψrψrrψ 
The probability density will be an
extreme when
.022)(
2222
















dr
d
ψ
rψrψ
dr
ψ
ψ
rrψψr
dr
d
This occurs at
,0

r
a minimum, and when
,0

ψ
also a minimum. A maximum must
correspond to
.0
dr
d

ψ
rψ
Within a multiplicative constant,
,)22(
1
,)2()(
22 arar
ear
a
dr
dψ
earrψ


and the condition for a maximum is
.046or),22()()2(
22
 ararararar
The solutions to the quadratic are
).53(  ar
The ratio of the probability densities at
these radii is 3.68, with the larger density at
.2at0)).53( arψbar 
Parts (a)
and (b) are consistent with Fig.(41.4); note the two relative maxima, one on each side of
the minimum of zero at
.2ar

41.37: a)
.arccos








L
L
θ
z
L
This is smaller for
LL
z
and
as large as possible. Thus
1and1  nlmnl
l
.
)1(
1
arccos
1)(1)(and)1(













nn
n
θ
nnllLnmL
L
lz

b) The largest angle implies
)1(,1  nlmnl
l
 
.)11(arccos
)1(
)1(
arccos
n
nn
n
θ
L













41.38: a)
.)1(so)1(
2222222222

lyxlzyx
mllLLmllLLLL 
b) This is the magnitude of the component of angular momentum perpendicular to the
z
-axis. c) The maximum value is
,)( L1ll  
when
.0
l
m
That is, if the electron is
known to have no
z
-component of angular momentum, the angular momentum must be
perpendicular to the
z
-axis. The minimum is
l

when
.lm
l

41.39:
ar
er
a
rP
24
5
24
1
)(








ar
e
a
r
r
a
dr
dP

2
4
3
5
)4(
24
1









,4;04when0
4
3
ar
a
r
r
dr
dP

In the Bohr model,
,4so
2
2

aranr
n

which agrees.
41.40:
The time required to transit the horizontal 50 cm region is
ms.952.0
sm525
m500.0



x
v
x
t
The force required to deflect each spin component by 0.50 mm is
N.1098.1
)s10952.0(
)m1050.0(2
molatoms10022.6
molkg1079.02
22
23
3
232
zz



















t
z
mmaF
According to Eq. 41.22, the value of
z
μ
is
.mA1028.9||
224


z
μ
Thus, the required magnetic-field gradient is
m.T3.21

TJ1028.9
N1098.1
24
22






z
zz
μ
F
dz
dB
41.41: Decay from a
d3
to
p2
state in hydrogen means that




l
mnn and23
.0,10,1,2 
l
m

However selection rules limit the possibilities for decay. The
emitted photon carries off one unit of angular momentum so
l
must change by 1 and
hence
l
m
must change by 0 or
.1

The shift in the transition energy from the zero field
value is just
)(
2
)(
2323
B llll
mm
m
Be
B
μmmU 

where
3
l
m
is the
l
md3

value and
2
l
m
is the
l
mp2
value. Thus there are only three
different energy shifts. They and the transitions that have them, labeled by
,
m
are:
12,01,10:
2
11,00,11:0
10,01,12:
2



m
Be
m
Be


41.42: a) The energy shift from zero field is






0B0
,2For. UmBmU
ll

eV.108.11T)(1.40T)Ve1079.5(
)1(,1For.eV101.62T)40.1()TVe1079.5()2(
55
0
45




Um
l
b)
,λ|λ|
0
||
0
E
E


where




R
E
1
5
36
00
λ)),91()41((eV6.13(
(a).partfromeV1009.8eV1011.8eV1062.1andm10563.6
5547 
 E
Then,
nm0281.0m1081.2|λ|
11


. The wavelength corresponds to a larger
energy change, and so the wavelength is smaller.
41.43: From Section 38.6:
.
)(
0
1
01
kTEE
e
n
n


We need to know the difference in energy

between the
2
1

s
m
and
.00232.2states.
2
1
B
BmμBμUm
szs

So
BμUU
B
00232.2
2
1
2
1


kTBμ
B
e
n
n
)00232.2(

21
21




B
e
e
)T10482.4(
K)(300K)J10(1.381T)BJ10(9.274(2.00232)
13
2324






a)
.9999998.0T1000.5
21
21
5



n
n
B

b)
9978.0T500.0
21
21


n
n
B
c)
978.0T00.5
21
21


n
n
B
41.44
Using Eq. 41.4
,)1(  llmvrL
and the Bohr radius from Eq. 38.15, we obtain the following value for
v
s.m1074.7
)m1029.5()4(kg)10(9.112
s)J1063.6(2
)(
)1(
5
1131

34
0
2








πanm
ll
v

The magnetic field generated by the “moving” proton at the electrons position can be
calculated from Eq. 28.1
T.277.0
m)1029.5()4(
)sin(90s)m10(7.74C)1060.1(
A)mT10(
sin||
4
2112
519
7
2
0








A
r
vq
π
μ
B

41.45:
s
m
can take on 4 different values:
.,,,
2
3
2
1
2
1
2
3






s
m
Each
l
nlm
state can
have 4 electrons, each with one of the four different
s
m
values.
a) For a filled
1

n
shell, the electron configuration would be
;s1
4
four electrons
and
.4

Z
For a filled
2

n
shell, the electron configuration would be
;2p2ss1
1244
twenty electrons and

.20

Z
b) Sodium has
;11

Z
11 electrons. The ground-state electron configuration would
be
.2p2s1s
344
41.46: a)
eV.666eV)6.13((7)eV)6.13(
22
Z
b)
The negative of the result of
part (a), 666 eV. c) The radius of the ground state orbit is inversely proportional to the
nuclear charge, and
m.107.567m)10529.0(
1210 

Z
a
d)
,
)(
λ
22
2

1
1
1
0




E
hc
E
hc
where
0
E
is the energy found in part (b), and
nm.49.2λ

41.47:
a) The photon energy equals the atom’s transition energy. The hydrogen atom
decays from
1,to2


nn
so:
m.101.22
J101.63
s)m10(3.00s)J10(6.33
λ

J1063.1
eV)J10(1.60eV)2.10(
)1(
1
(2)
1
eV60.13
7
18
834-
18
19
22






















E
hc
E
b) The change in an energy level due to an external magnetic field is just
.
B
BμmU
l

The ground state has
,0

l
m
and it is not shifted. The n = 2 state has
,1


l
m
so it is
shifted by
J1004.2
)T(2.20)TJ10274.9()1(
23
24





U
and since
m.101.53
J101.63
J102.04
m)1022.1(
λλ
λ
λ
12
18
23
7






















E
E
E
E
Since the n = 2 level is lowered in energy (brought closer to the n =
1 level) the change in
energy is less, and the photon wavelength increases due to the magnetic field.
41.48:
The effective field is that which gives rise to the observed difference in the energy
level transition,
.
λλ
λλ
2
λλ
λλ
21
21
21
21
BB























e
πmc
μ
hc
μ
E
B
Substitution of numerical values gives
,T1064.3

3
B
much smaller than that for
sodium.
41.49:
a) The minimum wavelength means the largest transition energy. If we assume
that the electron makes a transition from a high shell, then using the screening
approximation outlined in Section 41.5, the transition energy is approximately the
ionization energy of hydrogen. Then
eV).6.13()1(
2
1
 ZEE
For vanadium, Z = 23.
m.1089.1
J1005.1
)sm1000.3(s)J1063.6(
λ
J101.05eV1058.6
10
15
834
153












E
hc
E
For the longest wavelength, we need the smallest transition energy, so this is the
12



nn
transition
).(

K
So we use Moseley’s Law:
m.1050.2λ
Hz101.201)(23Hz)1048.2(
10
18215



f
c
f
b) The rhenium, Z = 45, the minimum wavelength is
m.1072.4λ

)eVJ10(1.60eV)6.13()44(
)sm10(3.00s)J1063.6(
)eV6.13()1(
λ
11
192
834
2








Z
hc
The maximum wavelength is
215
8
1)(45Hz)1048.2(
)sm1000.3(
λ



f
c
m.1025.6λ

11

41.50: a)
.
λ
2
λ
)00232.2(
B
e
πmc
B
hc
B
m
e
SB
μE
Z


b)


T.307.0
C)10(1.60m)(0.0350
)sm10(3.00kg109.112
19
831







π
B
41.51: a) To calculate the total number of states for the
th
n
principle quantum number
shell we must multiply all the possibilities. The spin states multiply everything by 2. The
maximum l value is (n –1), and each l value has
l
ml )12( 
values.
So the total number of states is
.2
222
2
)()1(4
2
412)12(2
2
2
1
0
1
0
1

0
nN
nnn
nn
n
llN
n
l
n
l
n
l





  






b) The n = 5 shell (O-shell) has 50 states.
41.52:
a)
Apply Coulomb’s law to the orbiting electron and set it equal to the centripetal
force. There is an attractive force with charge +2e a distance r
away and a repulsive force

a distance 2r away. So,
.
)2(4
)()(
4
)()2(
2
2
0
2
0
r
mv
r
ee
r
ee






But, from the quantization of angular momentum in the first Bohr orbit,
vmvrL 
So


3
2

2
2
2
0
2
2
0
2
)4(44
2
mrr
m
r
mv
r
πε
e
r
πε
e
mr




















2
2
0
3
2
0
2
2
4
7
4
4
4
7
me
πε
r
mr
πε
r

e

m.103.02m)10529.0(
7
4
7
4
1110
0

 a
And
.sm1083.3
m)10kg)(0.52910(9.11
s)J10054.1(
4
7
4
7
6
1031
34
0









mamr
v

b)
eV.5.83J1034.1)sm10(3.83kg1011.9
2
1
2
1726312









mvKE
c)
)2(44
4
)2(44
2
2
0
2
0
2

0
2
0
2
rπE
e
r
πε
e
r
πε
e
r
πε
e
PE
















eV.166.9J1067.2
42
7
17
0
2












rπε
e
d)
eV,83.4eV]83.5eV9.166[






E

which is only off by about 5% from the
real value of 79.0 eV.
41.53: The potential
2
2
1
)(
xkxU


is that of a simple harmonic oscillator. Treated
quantum mechanically (see Section 40.4) each energy state has energy
).(
2
1
 nωE
n

Since electrons obey the exclusion principle, this allows us to put two electrons (one for
each
)
2
1

s
m
for every value of neach quantum state is then defined by the ordered
pair of quantum numbers
).,(
s

mn
By placing two electrons in each energy level the lowest energy is then
.
][
22
)()1(
2
2
1
2
2
1
22
22
2
1
0
1
0
1
0
1
0
m
k
NwN
NNNw
NNN
w
nwnwE

N
n
N
n
N
n
N
n
n






















































Here we used the hint from Problem 41.51 to do the first sum, realizing that the first
value of n is zero and the last value of n is N – 1, giving us a total of N energy levels
filled.
41.54:
a) The radius is inversely proportional to Z, so the classical turning radius is
.2 Za
b) The normalized wave function is
arZ
s
e
Z
πa
rψ


33
1
1
)(
and the probability of the electron being found outside the classical turning point is
.
4
4
22
22
33
2
2
1

drre
Za
dr
πrψP
aZr
ZaZa
s




Making the change of variable
duZadraZru )(, 
changes the integral to
,4
22
2
duueP
u



which is independent of Z. The probability is that found in Problem 41.33, 0.238,
independent of
Z.