7-29. (a) Every increment of charge follows a circular path of radius R and encloses an
area
! R 2 , so the magnetic moment is the total current times this area. The entire
charge
Q rotates with frequency f = ! / 2" , so the current is
i = Qf = q! / 2"
µ = iA = (Q! / 2" )(" R 2 ) = Q! R 2 / 2
L = I! =
g=
1
MR 2!
2
2 M µ 2 MQ! R 2 / 2
=
=2
QL
QMR 2! / 2
(b) The entire charge is on the equatorial ring, which rotates with frequency
f = ! / 2" .
i = Qf = Q! / 2"
µ = iA = (Q! / 2" )(" R 2 ) = Q! R 2 / 2
g=
2 M µ 2 MQ! R 2 / 2
=
= 5 / 2 = 2.5
QL
QMR 2! / 5
7-33. (a) There should be four lines corresponding to the four mJ values −3/2, −1/2,
+1/2, +3/2.
(b) There should be three lines corresponding to the three ml values −1, 0, +1.
7-35. For
l = 2,
L = l (l + 1)h = 6 h = 2.45h,
j = l ± 1 / 2 = 3 / 2, 5 / 2 and J =
For j = 3 / 2,
J=
(3 / 2 )(3 / 2 + 1)h =
15 / 4 h = 1.94h
For j = 5 / 2,
J=
(5 / 2 )(5 / 2 + 1)h =
35 / 4 h = 2.96h
j ( j + 1)h
7-39. (a) L = L1 + L2
l = (l1 + l 2 ), (l1 + l 2 ! 1),..., l1 ! l 2 = (1 + 1), (1 + 1 ! 1), (1 ! 1) = 2, 1, 0
(b) S = S1 + S2
s = (s1 + s2 ), (s1 + s2 ! 1),..., s1 ! s2 = (1 / 2 + 1 / 2 ), (1 / 2 ! 1 / 2 ) = 1, 0
(c) J = L + S
j = (l + s ), (l + s ! 1),..., l ! s
For l = 2 and s = 1, j = 3, 2, 1
l = 2 and s = 0, j = 2
For l = 1 and s = 1, j = 2, 1, 0
l = 1 and s = 0, j = 1
For l = 0 and s = 1, j = 1
l = 0 and s = 0, j = 0
(d) J1 = L1 + S1
J 2 = L2 + S 2
(e) J = J1 + J 2
j1 = l1 ± 1 / 2 = 3 / 2, 1 / 2
j 2 = l 2 ± 1 / 2 = 3 / 2, 1 / 2
j = ( j1 + j2 ),
( j1 + j2 ! 1),...,
j1 ! j2
For j1 = 3 / 2 and j2 = 3 / 2, j = 3, 2, 1, 0
j1 = 3 / 2 and j2 = 1 / 2, j = 2, 1
For j1 = 1 / 2 and j2 = 3 / 2, j = 2, 1
j1 = 1 / 2 and j2 = 1 / 2, j = 1, 0
These are the same values as found in (c).
7-40. (a) E3 / 2 =
E3 / 2 =
hc
!
Using values from Figure 7-22,
1239.852eV nm
= 2.10505eV
588.99nm
E1 / 2 =
1239.852eV nm
= 2.10291eV
589.59nm
(b) "E = E3 / 2 ! E1 / 2 = 2.10505eV ! 2.10291eV = 2.14 # 10!3 eV
"E
2.14 # 10!3 eV
=
= 18.5T
(c) "E = 2 µ B B $ B =
2 µ B 2 5.79 # 10!4 eV / T
(
)
7-43. (a) For electrons: Including spin, two are in the n = 1 state, two are in the n = 2
state, and
one is in the n = 3 state. The total energy is then:
E = 2 E1 + 2 E2 + E3
where En =
n 2 h 2! 2
2mL2
(
) (
)
E = 2 E1 + 2 22 E1 + 32 E1 = 19 E1
where
2
E1
hc ) ! 2
(
=
2me c 2 L2
2
197.3) ! 2
(
=
2
2 (0.511 " 106 )(1.0 )
= 0.376eV
E = 19 E1 = 7.14eV
(b) Pions are bosons and all five can be in the n = 1 state, so the total energy is:
E = 5 E1 where E1 =
0.376eV
= 0.00142eV
264
E = 5 E1 = 0.00712eV
7-44. (a) Carbon: Z = 6; 1s 2 2 s 2 2 p 2
(b) Oxygen: Z = 8; 1s 2 2 s 2 2 p 4
(c) Argon: Z = 18; 1s 2 2 s 2 2 p 6 3s 2 3 p 6
7-45. Using Figure 7-34:
Sn (Z = 50)
1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 5s 2 5 p 2
Nd (Z = 60)
1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 5s 2 5 p 6 4 f 4 6 s 2
Yb (Z = 70)
1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 4 f 14 5s 2 5 p 6 6 s 2
Comparison with Appendix C.
Sn: agrees
Nd: 5 p 6 and 4 f 4 are in reverse order
Yb: agrees
2
7-46. Both Ga and In have electron configurations (ns ) (np ) outside of closed shells
2
6
10
(n ! 1, s ) (n ! 1, p ) (n ! 1, d )
. The last p electron is loosely bound and is more
easily removed than one of the s electrons of the immediately preceding elements
Zn and Cd.
7-48.
En = !
Z eff2 E1
Z eff = n
(Equation 7-25)
n2
! En
5.14eV
=3
= 1.84
E1
13.6eV
7-49. (a) Fourteen electrons, so Z = 14. Element is silicon.
(b) Twenty electrons, so Z = 20. Element is calcium.
7-50. (a) For a d electron, l = 2, so Lz = !2h, ! 1h, 0, 1h, 2h
(b) For an f electron, l = 3, so Lz = !3h, ! 2h, ! 1h, 0, 1h, 2h, 3h
2
2
7-56. (a) E1 = !13.6eV (Z ! 1) = !13.6eV (74 ! 1) = !7.25 " 104 eV = !72.5keV
2
2
(b) E1 (exp ) = "69.5keV = "13.6eV (Z " ! ) = "13.6eV (74 " 1)
(
74 " ! = 69.5 # 103 eV / 13.6eV
1/ 2
)
= 71.49
! = 74 " 71.49 = 2.51
7-58. (a) "E = hc / !
E (3P1 / 2 ) ! E (3S1 / 2 ) =
1240eV nm
= 2.10eV
589.59nm
E (3P1 / 2 ) = E (3S1 / 2 ) + 2.10eV = !5.14eV + 2.10eV = !3.04eV
E (3D ) ! E (3P1 / 2 ) =
1240eV nm
= 1.52eV
818.33nm
E (3D ) = E (3P1 / 2 ) + 1.52eV = !3.04eV + 1.52eV = !1.52eV
(b) For 3P :
Z eff = 3
3.04eV
= 1.42
13.6eV
For 3D :
Z eff = 3
1.52eV
= 1.003
13.6eV
(c) The Bohr formula gives the energy of the 3D level quite well, but not the 3P
level.
7-59. (a) !E = gm j µ B B
(Equation 7-72) where s = 1 / 2, l = 0 gives j = 1 / 2 and
(from Equation 7-73) g = 2.
m j = ±1 / 2.
(
)
"E = (2 )(±1 / 2 ) 5.79 # 10!5 eV / T (0.55T ) = ±3.18 # 10!5 eV
The total splitting between the m j = ±1 / 2 states is 6.37 " 10!5 eV .
(b) The m j = 1 / 2 (spin up) state has the higher energy.
(c) "E = hf # f = "E / h = 6.37 $ 10!5 eV / 4.14 $ 10!15 eV s = 1.54 $ 1010 Hz
This is in the microwave region of the spectrum.
7-61. (a) "E =
eh
B = 5.79 # 10!5 eV / T
2m
(
)(0.05T ) = 2.90 # 10
!6
eV
2
(579.07nm ) (2.90 # 10"6 eV )
!2
(b) $! =
$E =
= 7.83 # 10"4 nm
hc
1240eV nm
(c) The smallest measurable wavelength change is larger than this by the ratio
0.01nm 7.83 " 10!4 nm = 12.8. The magnetic field would need to be
increased by this
same factor because B " #E " #!. The necessary field would be 0.638T.
7-66. !min = cos"1 # ml h
%
l (l + 1) h $ with ml = l.
&
cos !min = l l (l + 1) . Thus, cos2 !min = l 2 "%l (l + 1)#& = 1 $ sin2 !min
or, sin2 !min = 1 "
l (l + 1) " l 2 l 2 + l " l 2
l2
=
=
l (l + 1)
l (l + 1)
l (l + 1)
1/ 2
And, sin!min
" 1 #
=$
%
& l +1'
For large l, !min is small.
1/ 2
Then sin!min $ !min
" 1 #
=%
&
' l +1(
$
1
1/ 2
(l )