Unformatted text preview: 2. H: 1s1, He: 13?, Li: 1s22s1, Be: 132252, B: 1322322121, C: 1322522192, N: 1322322133, 0: 1822822p4, F: 1322322p5, Ne: 1522522196
* 7. From Figure 8.4 we see that the radius of Na is about 0.19 nm. We know that for single 1 t toms
e ec ron a E .— Ze2
SWEQ'I‘
Th f
ere ore 871507‘E 2 (0.19 nm) (—5.14 eV) _ 1 36
25 _____ __ :: "_——_.—————_—— — I e
e 1.44 V  nm 9. a) 5 electrons, B
b) 11 electrons, Na
0) ﬁlled 3p: Ar 12. J ranges from IL — S] to IL + S] or 2,3,4. Then in spectroscopic notation 25+1LJ, we have
three possibilities: 3F2, 3F3, or 3F4. The ground state has the lowest J value, or 3F2. With n = 4 the full notation is 43172. * 15. In the 3d state E = 2 and s =1/2,soj= 5/2 or 3/2. As usual mg = 0,:L1,:l:2. The value
of 7m ranges from —j to j, so its possible values are i1/2, i3/2, and :l:5/2. As always m5 = i1 / 2. The two possible term notations are 3D5/2 and 3D3/2. * 21. '
ha ha 1 1
AE=———=  ———————_—_—_— =7.33 1—3V
A1 A2 (1240 W m“) (766.41 nm 769.90 mm) 4 X 0 e As in Example 8.8 the internal magnetic ﬁeld is B mAE _ (9.109 x 1031 kg) (7,334 x 103 eV) _ 63 4 T
" efi ’ (1.602 x 1019 c) (6.582 x 10—16 eV . s) —  23. The 25 to Is transition is forbidden by the AL = :i:1 selection rule. The two lines result from
the transitions from the two 2p levels to the ls level. * 24. As in Example 8.8 * 26. The minimum angle corresponds to the maximum value of J2 and hence the maximum value
of m,, which is mj = j. Then
{i _ j a: ___T'_J'h___=_____
“5 J i/j(j+1)h \/j0'+1) Solving for j we ﬁnd 60859—1
28. Using the fact that g = 2 we have
~ _ £435?
Ila92 Tm
l ~ Z2
. B: e s L ...
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