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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
[...]... n 2 n 0 2 n 0 n 0 n 0 41. 53: The potential U ( x) ( N 1)( N ) N 2w w [ N 2 N N ] 2 2 k wN 2 N 2 m 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,... configuration would be 1s 4 2s 4 2p 3 41. 46: a) Z 2 (13.6 eV) (7) 2 (13.6 eV) 666 eV b) The negative of the result of part (a), 666 eV c) The radius of the ground state orbit is inversely proportional to the a nuclear charge, and (0.529 10 10 m) 7 7.56 10 12 m Z hc hc d) λ , where E 0 is the energy found in part (b), and λ 2.49 nm E E0 ( 112 212 ) 41. 47: a) The photon energy equals... 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, E hc λ1 λ 2 2πmc λ1 λ 2 B μ B λ1λ 2 μB e λ 1λ 2 λ λ Substitution of numerical values gives B 3.64 10 3 T, much smaller than that for sodium 41. 49: a) The minimum wavelength means the largest... 4.72 1011 m c (3.00 108 m s) The maximum wavelength is λ f (2.48 1015 Hz)(45 1) 2 λ 6.25 10 11 m e 2πmc hc B B 41. 50: a) E (2.00232) μB BS Z m λ λe 2π 9.11 10 31 kg (3.00 108 m s) b) B 0.307 T (0.0350 m)(1.60 10 19 C) 41. 51: a) To calculate the total number of states for the n th principle quantum number shell we must multiply all the possibilities The... (9.274 10 e ( 4.48210 3 c) B 5.00 T n1 2 n1 2 n1 2 n1 2 n1 2 n1 2 J T)B (1.3811023 J K) (300 K) T 1 ) B a) B 5.00 10 5 T b) B 0.500 T 24 0.9999998 0.9978 0.978 41. 44 Using Eq 41. 4 L mvr l (l 1), and the Bohr radius from Eq 38.15, we obtain the following value for v l (l 1) 2 (6.63 10 34 J s) v 7.74 10 5 m s 2π (9.11 10 31 kg) (4) (5.29 10.. .41. 42: a) The energy shift from zero field is U 0 ml B B For ml 2, U 0 (2) (5.79 10 5 e V T ) (1.40 T) 1.62 10 4 eV For ml 1, U 0 (1) (5.79 10 5 e V T) (1.40 T) 8.11 10... 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 E E1 ( Z 1) 2 (13.6 eV) For vanadium, Z = 23 E 6.58 103 eV 1.05 10 15 J hc (6.63 10 34 J s)(3.00... “moving” proton at the electrons position can be calculated from Eq 28.1 μ | q | v sin (1.60 10 19 C) (7.74 105 m s) sin(90) (10 7 A T m A) B 0 0.277 T 4π (4) 2 (5.29 10 11 m) 2 r2 41. 45: m s can take on 4 different values: ms 3 , 1 , 1 , 3 Each nlml state can 2 2 2 2 have 4 electrons, each with one of the four different m s values a) For a filled n 1 shell, the electron... 1)ml values So the total number of states is n 1 n 1 n 1 l 0 l 0 l 0 N 2 (2l 1) 2 1 4 l 4(n 1)(n) 2n 2n 2 n 2 2n 2 2 N 2n 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... 1.62 10 4 eV 8.11 10 5 eV 8.09 10 5 eV from part (a) Then, | λ | 2.81 10 11 m 0.0281 nm The wavelength corresponds to a larger energy change, and so the wavelength is smaller n 41. 43: From Section 38.6: 1 e ( E1 E0 ) kT We need to know the difference in energy n0 1 1 between the ms and ms states U μ z B 2.00232 μB ms B 2 2 So U 1 U 1 2.00232 μB B 2 . 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. 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