And the lower energy level is since pointsinthe ˆdirection... 41.22: For the outer electrons, there are more inner electrons to screen the nucleus.. b This is the magnitude of the compon
Trang 12 34
34 2
s J 10 054 1
s J 10 4.716 )
1 ( ) 1 (
l l L
4 0
20 ) 1
41.2: a) 2, so 2
max
m b) l(l1) 62.45 c) The angle is arccos
, 6
arccos
L
L
and the angles are, for m l 2tom l 2,144.7,114.1,90.0,
3
35
,
9
65 The angle corresponding to m l will always be larger for larger l l
41.3: L l(l1) The maximum orbital quantum number l n1.Soif :
5 199 )
200 ( 199 199
200
49 19 ) 20 ( 19 19
20
41 1 2 )
1 ( 1
2
L l
n
L l
n
l l L l
n
The maximum angular momentum value gets closer to the Bohr model value the larger the value of n
41.4: The (l,m l) 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
25
eV 60 13
m 10 0 1
C) 10 60 1 ( 4
1
4
10
2 19 0
2 1 0
πε r
q q πε U
eV
4 14 eV J 10 60
1
J 10 3 2 19
18
41.6: a) As in Example 41.3, the probability is
0
2
0 2 3 2 2 3
2 2 1
4 2 2
4 4
|
|
a dr πr ψ
P
0803 0 2
5 1
1
b) The difference in the probabilities is
243 0 ) 2 )(
2 5 ( ) ) 2 5 ( 1 ( ) 5 1 ( e 2 e 1 e 1 e 2
Trang 241.7: a) |ψ|2ψ * ψ |R(r)|2|(θ)|2 (Aeim l)(Aeim l)
,
| ) (
|
| ) (
π A
πA d
A
0
2 2
0
2
2
1 1
2
| ) (
2 2
) 4 (
1
1 1
2
1 1 2 12 2 2
4 2 0
E E
E E E E n
e m πε
a) If m r m9.1110 31 kg
2 34
4 9 31
2 2 0
4
C Nm 10 988 8 s)
J 10 055 1 ( 2
C) 10 kg)(1.602 10
9.109 )
4
πε
e
m r
2.17710 18 J13.59eV
For 21 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
eV 59 13 2
eV) 59 13 ( )
4
0
4
m
m πε
e
m r
Similarly, the 21 transition, 5.095eV
2
eV 19
c) Ifm r 185 m.8 , using the result from part (a),
eV, 2525
185.8 eV) 59 13 ( )
4
0
4
m
m πε
e
m r
and the 21 transition gives (10.19 eV)(185.8)=1893 eV
2 34 0
2
2 0 1
C) 10 kg)(1.602 10
109 9 (
) s J 10 626 6 (
π
ε πme
h ε a m
m r
m 10 293
1
a
2
10 1
2
m r
185.8
1 m
8
1
m r
41.10: cos( ) sin( ),
l l
integer multiple of 2π so, m l must be an integer
Trang 341.11: a a r a
πa dV
ψ a
P
2 2
3
)
(
5 1 ) (
4 4
2 2 4
4 2 2
4 4
) (
2
0
3 2 3 3 3 3
0 2 2 2 2 3
2 2 3
e a
P
e
a e a a a a
e a r a ar a
dr e r a a P
a a r a
r a
o
41.12: a) (5.79 10 5 eV T)(0.400T) 2.32 10 5 eV, b) 2,
l
m B
μ
lowest possible value of m l
c)
41.13: a) g-statel 4#of statesis(2l1)9(m l 0,1, 2,3,4)
b) (5.79 10 5 eV T)(0.600T) 3.47 10 5 eV 5.56 10 24 J
B
U μ B
c) 8 8(5.79 10 5 eV T)(0.600T) 2.78 10 4 eV
B 4 4
4.4510 23 J
41.14: a) According to Fig 41.8 there are three different transitions that are consistent
with the selection rules The initial m l values are 0, 1; and the final m l value is 0 b) The transition from m l 0tom l 0 produces the same wavelength (122 nm) that was seen without the magnetic field
c) The larger wavelength (smaller energy) is produced from the m l 1 tom l 0 transition
d) The shorter wavelength (greater energy) is produced from the m l 1 tom l 0 transition
41.15: a)
B B
, 1 , 3 3
μ
U B B μ U l
n
T) V e 10 (5.79
eV) 10 71 2 (
5
5
B
b) Three m l 0,1
Trang 441.16: a) B
m
e
2 2
) 00232 2
2
) 00232 2 (
eV 10 78 2
) T 480 0 )(
T V e 10 788 5 ( 2
) 00232 2 (
5
5
b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction But if n0
there could be since l < n allows for l 0
m
e B
2 ) 00232 2 (
2 where
) 00232 2
(
) ( 2 ) 00232 2
(
B B
m
e μ B
m μ U
B m m
e U
s
s
So the energy difference
2
1 2
1 ) 00232 2
U
U (2.00232)(5.78810 5eV T)(1.45T)
1.6810 4 eV And the lower energy level is (since pointsinthe ˆdirection)
2
1
z B
2
5 , 2 and 2
3 , 2 , 2
3 , 1 , 2
1 , 1 , 2
1 ,
41.19: j quantum numbers are either ,Soif 9 2and7 2,then 4
2
1 or 2
l
The letter used to describe l 4is “g”
λ , cm 21 eV)
10 (5.9
) s m 10 s)(300 eV
10 136 4 (
8
f E
hc
9
8
10 4 1 0.21m
) s m
10
00
3
(
Hz, a short radio wave
b) As in Example 41.6, the effective field is 2 5.1 10 2 T,for
than that found in the example
Trang 541.21: a) Classically 2
5
2 and
Iω
s
rad 10 2.5
4
3 m) 10 (1.0 kg) 10 2(9.11
s) J 10 5(1.054 4
3 2
5 4
3 5
2
30
2 17 31
34 2
2
ω mR ω
R mω L
b) vrω(1.010 17 m)(2.51030rad s)2.51013 m s 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: 2 (13.6eV)
2 eff
n
Z
electron sees Zeff 2.771andn5
eV
4.18 eV)
(13.6 5
(2.771) 2
2
E
41.24: However the number of electrons is obtained, the results must be consistent with
4 3
4
3
3s2 p6 s2 d10 p2
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
s l
s l
s l
m m
l n
m m
l n
m m
l n
41.26: For the s4 state, E 4.339eVand Zeff 4 (4.339) (13.6) 2.26
Similarly, Zeff 1.79 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
Trang 641.27: a) Nitrogen is the seventh element (Z = 7) N has two electrons removed, so 2 there are 5 remaining electrons electron configuration is 1s22s22p
2
4) (7 eV)
2
2
n
Z E
c) Phosphorous is the fifteenth element (Z = 15) P has 13 electrons, so the 2 electron configuration is 1s22s22p63s23p
d) The least tightly held electron: (13.6eV) 13.6eV
3
12) (15 2
2
E
4
e 6 13
eff
2 eff
becomes larger going down the columns in the periodic table
41.29: a) Again using 2 (13.6eV),
2 eff
n
Z
E n
the outermost electron of the BeL shell
(n = 2) sees the inner two electrons shield two protons so Zeff 2
eV
13.6 eV)
(13.6 2
2 2
2
E 2
b) For Ca, outer shell has n = 4, so (13.6eV) 3.4eV
4
2 2
2
E
41.30: E kx Z( 1)2(10.2eV)
, 0 28 eV
10.2
eV 10 7.46 1
3
Z
which corresponds to the element Nickel (Ni)
41.31: a) Z 20: f (2.481015 Hz)(201)2 8.951017 Hz
m
10 3.35 Hz
10 8.95
s m 10 3.00 λ
keV 3.71 Hz) 10 (8.95 s) eV 10 (4.14
10 17
8
17 15
f c
hf E
b) Z = 27:
m
10 1.79 λ
keV 6.96
Hz 10 1.68
10
18
E f
c) Z 48: f 5.481018Hz, E22.7keV,λ5.4710 11 m
Trang 741.32: See Example 41.3; , 2 (2 (2 2 )),
2 2 / 2 2 2
dr
ψ dr e
Cr ψ
maximum, r = a, the distance of the electron from the nucleus in the Bohr model.
0 1
2 0 2
4 2 0 1
2 ) 4 (
1 then ), (
4
1 ) ( and 2 ) 4 (
1
me πε r
U E If r
e πε r
U
me πε
me
πε r r
e
2 4 )
4
(
1
2
2 0 2
0
3 2
2 2 1 2
2 1
a r a
s a
a π ψ
dr r ψ π dV ψ a
r
4 2
4 )
2
(
4 2 2
4 )
2
(
4 )
2
(
3 3 3 4 3
2
3 2 2 2
3
2
2 2 3
a a a e a a
r
P
a r a ar e
a a
r
P
dr e r a a
r
P
a
a r a
a r
13e 4 0.238
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 13.602eV
350
eV
0
2 350
4 2
2 3.22 10 eV, (650)
eV 13.60
r
m
10 2.24
m)
10 10 5
Trang 841.35: a) If normalized, then
4 4 8
1
2 32
4
1 4
4 3 2 3
2 2
2 0
2 2 0
2 2
dr e a
r a
r r a
dr e a
r r
πa
π I
dr r ψ π dV ψ
a r
a r
s s
0 n ax n n!1
dx e x
4
3 4(2) 4(6) 1 (24) 8
a a
a a
a a
I
8 24 24 1and isnormalized
8
1
2 3
3 3
a
b) We carry out the same calculation as part (a) except now the upper limit on the
integral is 4a, not infinity.
8
4 3 2
a
r a
r r a
Now the necessary integral formulas are:
) 24 24
12 4
(
) 6 6 3
(
) 2 2
(
5 4
3 2 2 3 4
4
3 3 2
2 3 3
3 2 2
2
a ra
a r a r a r e dr e r
a ra a
r a r e dr e r
a ra a r e dr e r
a r a
r
a r a
r
a r a
r
All the integrals are evaluated at the limits r 0and 4a After carefully plugging in the limits and collecting like terms we have:
)]
824 568 104 ( ) 24 24 8 [(
8
a I
(8 360 ) 0.176 Prob( 4 )
8
Trang 941.36: a) Since the given ψ(r)isreal,r2 |ψ|2r2ψ2 The probability density will be an extreme when
0 2
2 )
dr
dψ r ψ rψ dr
ψ ψ r rψ ψ
r dr d
This occurs at r 0,a minimum, and when ψ 0,also a minimum A maximum must correspond to ψ r dψ dr 0.Within a multiplicative constant,
, ) 2 2 (
1 ,
) 2 ( )
a dr
dψ e
a r r
and the condition for a maximum is
0 4 6 or ), 2 2 ( ) ( ) 2
( r a r a r a r2 ra a2
The solutions to the quadratic are r a(3 5) The ratio of the probability densities at these radii is 3.68, with the larger density at ra(3 5) b)ψ 0at r2a 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 r2a
41.37: a) arccos
L
L
L This is smaller for L z and as large as possible Thus L
1 and
) 1 (
1 arccos
1) ( 1) ( and
) 1 (
n n
n θ
n n l
l L n
m L
L
l
b) The largest angle implies l n1,m l l (n1)
(1 1 )
arccos
) 1 (
) 1 ( arccos
n
n n
n
θ L
41.38: a) L2x L2y L2 L2z l(l1)2 m l22 so L2x L2y l(l 1)m l2
b) This is the magnitude of the component of angular momentum perpendicular to the
z-axis c) The maximum value is l(l 1) L, when m l 0 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 m l .l
Trang 1041.39: r e r a
a r
5 24
1 )
a r
e a
r r a dr
3
24
, 4
; 0 4
when
a
r r dr
In the Bohr model,r n n2asor2 4a,which agrees
41.40: The time required to transit the horizontal 50 cm region is
ms
952 0 s m 525
m 500
x
v
x t
The force required to deflect each spin component by 0.50 mm is
N 10 98 1 ) s 10 952 0 (
) m 10 50 0 ( 2 mol atoms 10
022 6
mol kg 1079 0
2 3
3 23
2 z
z
t
z m
ma
F
According to Eq 41.22, the value of μ is z
m A 10 28 9
|
z
μ
Thus, the required magnetic-field gradient is
m
T 3 21 T J 10 28 9
N 10 98 1
24
22
z
z
z
μ
F
dz
dB
41.41: Decay from a d to 2 state in hydrogen means that p n3n2andm l
0 , 1 0
,
1
,
m l However selection rules limit the possibilities for decay The emitted photon carries off one unit of angular momentum so lmust change by 1 and hence m l must change by 0 or 1 The shift in the transition energy from the zero field value is just
) (
2 )
(
2 3 2
m
B e B μ m m
where
3
l
m is the d m l value and
2
l
m is the 2p m l value Thus there are only three different energy shifts They and the transitions that have them, labeled by m are:,
1 2 , 0 1 ,
1
0
:
2
1 1 , 0 0 ,
1
1
:
0
1 0 , 0 1 ,
1
2
:
2
m
B
e
m
B
e
Trang 1141.42: a) The energy shift from zero field is U0 m lBB.Form l 2,U0
eV
10 8.11 T) (1.40 T) V e
10
79
5
(
) 1 ( ,
1 For eV 10 1.62 T) 40 1 ( ) T V e 10
79
5
(
)
2
(
5 5
0 4
5
b) | λ| λ ,
0
|
|
0 EE
5
36 0
0 (13.6eV((14) (19)),λ
(a) part from eV 10 09 8 eV 10 11 8 eV 10 62 1 and
m
10
563
Then, |λ|2.8110 11 m0.0281nm The wavelength corresponds to a larger energy change, and so the wavelength is smaller
41.43: From Section 38.6: ( )
0
1 e E1 E0 kT
n
n We need to know the difference in energy
between the
2
1
s
2
1
Bm B μ B
μ U
So U U 2.00232μBB
2 2
kT B
μ B
e n
n (2.00232)
2
B
e
e
) T 10 482 4 (
K) (300 K) J 10 (1.381 T)B J 10 (9.274 (2.00232)
1 3
23 24
a) 5.00 10 T 0.9999998
2
2
n
n B
b) 0.500T 0.9978
2
2
n
n B
c) 5.00T 0.978
2
2
n
n B
41.44 Using Eq 41.4
, ) 1
L
and the Bohr radius from Eq 38.15, we obtain the following value for v
s
m 10 74 7 ) m 10 29 5 ( ) 4 ( kg) 10 (9.11 2
s) J 10 63 6 ( 2 )
(
)
1
11 31
34 0
π a
n
m
l
l
The magnetic field generated by the “moving” proton at the electrons position can be calculated from Eq 28.1
T 277 0 m)
10 29 5 ( ) 4 (
) sin(90 s) m 10 (7.74 C) 10 60 1 ( A) m T 10 ( sin
|
|
5 19
7 2
r
v
q
π
μ
Trang 1241.45: m can take on 4 different values: s , , , 23.
2 1 2 1 2
s
have 4 electrons, each with one of the four different m values s
a) For a filled n1 shell, the electron configuration would be 1s4; four electrons and Z 4 For a filled n2 shell, the electron configuration would be 1s42s42p12; twenty electrons and Z 20
b) Sodium has Z 11; 11 electrons The ground-state electron configuration would
be 1s42s42p3
41.46: a) Z2 (13.6eV)(7)2 (13.6eV)666eV 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 (0.52910 10 m) 77.5610 12 m
Z a
) (
λ
2
2 2
1 1
1
E
hc E
hc
where E is the energy found in part (b), and 0 λ2.49nm
41.47: a) The photon energy equals the atom’s transition energy The hydrogen atom
decays from n2ton1, so:
m
10 1.22 J
10 1.63
s) m 10 (3.00 s) J 10 (6.33 λ
J 10 63 1
eV) J 10 (1.60 eV) 2 10 ( ) 1 (
1 (2)
1 eV 60 13
7 18
8 34
-18
19 2
2
E hc E
b) The change in an energy level due to an external magnetic field is just U m l μBB The ground state has m l 0,and it is not shifted The n = 2 state has m l 1, so it is shifted by
J 10 04 2
) T (2.20 ) T J 10 274 9 ( 1 (
23
24
U
and since
m
10 1.53 J
10 1.63
J 10 2.04 m) 10 22 1 ( λ
λ λ
λ
12 18
23
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,
λ λ
λ λ 2
λ λ
λ λ
2 1
2 1 2
1
2 1 B
e
πmc μ
hc μ
E B
Substitution of numerical values gives B3.6410 3 T, much smaller than that for sodium