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Homework#8

# Homework#8 - Chapter 7 The Hydrogen Atom 3 Assuming a trial...

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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 = AeZ-mlk 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 . _‘ 1932-1 = R32Y2—1 = W ((1-3) 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 10-5 eV/T) 26. From the text the magnitude of the spin magnetic moment is The the z-component 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 10-15 4 10-15 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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