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Unformatted text preview: Chapter 7 The Hydrogen Atom 3. Assuming a trial solution 9 = Aeikq’ (which is easily veriﬁed by direct substitution), and using
the boundary condition 9(0) = g(27r), we ﬁnd A80 = AeZmlk
which is only true if k is an integer.
6. As in the previous problem
R = Are—”2“o
11:3. 2A 1 L e—r/2ao
r 2&0
dB 7'3
2___ ,___ 2 _ _ —r/2ao
T clr (T 204)) e
d 2dR _A 2T_E+i e—r/2ao
dr r dr — a0 4a8 Substituting these into Equation (7.10) with Z = 1 and after substituting the Coulomb po
tential, we have I 1 2uE 1 2W2 1
— — —— 2 — 2 — = 0
(4(18 + 712 >T +< 2110 + 47reoh2 +( )1“ The 1/7' term vanishes, and the middle expression (without 7") reduces to 41reoh2
a0 = 2
ire
which is correct. From the 1' term we get
712 E
E = _ 2 = __0
Suao 4
which is consistent with the Bohr result.
* 17. 1 1
— — ._ — ﬂan ' 9 ‘4‘?
$214  RZIYI—l — ”W( )e ’ sm 6
Sﬁao ‘10
1 r _ 2
=R Y =———— —— e r/“°c050
$210 21 10 4x/ﬁa3/2 (an)
1 r2 _r . _‘
19321 = R32Y2—1 = W ((13) 5 ”“0 sm0cos 0e "’5
7ra0 ' h
* 21. Differentiating E = 7‘0 we ﬁnd h hc
dE = ngA or AE = X5 1AA! hc
In the Zeeman effect between adjacent mg states lAEl = HEB so MBB = (XE) IAAI or Ag/LBB AA: hc * 24. From Problem 21
‘ AELLBB AA: hc so the magnetic ﬁeld is _ hc AA __ (1240 eV  nm) (0.04 nm) : 1.99 T _ _ ________—._———————
B We (656.5 nm)2 (5.788 x 105 eV/T) 26. From the text the magnitude of the spin magnetic moment is The the zcomponent of the magnetic moment is (see Figure 7.9) The potential energy is V = —;I  E = —,usz and so the vertical component of force is
Fz = —dV/clz = [1,; (dB; / dz). From mechanics the acceleration is 29. For the 5f state 71  5 and Z = 3. The possible mg values are 0,:i:1,J:2, and 3:3 with
ms = :i:1/2 for each possible mg value. The degeneracy of the 5f state is then (with 2 spin
states per ml) equal to 2(7) = 14. munmuwmwwewam m... 34. a) AZ = 0 is forbidden b) allowed but with An = 0 there is no energy difference unless an external magnetic ﬁeld is
present c) AZ = —2 is forbidden
d) allowed with absorbed photon of energy 1 1
AE=E0 (52— 12):255 eV 38. The radial probability distribution for the ground state is Pm = r2 new = irwrao/
0 With 7‘ << a0 throughout this interval we can say e'QT/‘z0 z 1. Therefore the probability of
being inside a radius 10‘15 m is 1015 4 1015 47.3 10—15
/ P(r) dr m —g r2 dr = ———3 = 9.0 x 10‘15
0 “0 0 30'0 0
39.
P0") = 7'2 R(7')2 = —4;3r2e"2T/“°
“0 To ﬁnd the desired probability, integrate P(r) over the appropriate limits: 3
95% a0 0.95m, 1.05ao 4 1.05110
/ P(1‘) (17‘ = —— r2e_2r/“° dr
0 Letting m = r/ao 1.05% 1.05
/ P(7') dr = 4/ 33263—2": d9: = 0.056
0.95ao 0.95 where the deﬁnite integral was evaluated using Mathcad. ...
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 Spring '08
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