1、Atomic Physics (OUP 2005)C.J. Foot, Oxford, 9 February 2005Chapter 1(1.1) 0:18nm(1.2) Balmer series in hydrogen and transitions ton = 4 shell in He+. Energy / Z2=n2. Linesof similar wavelength show isotope shift: wave-length ratio H/He is 1:0004 (equal to ratio ofreduced masses given by eqn 1.13).(1
2、.3) From eqns 1.17 and 1.18:E = fi2n2E:The n = 2 shell has E = hcR1 = 3:4eV, henceE = 4:5105 eV which requires resolutiongreater than E=E = 75000 = 4=fi2.(1.5) K-absorption edges: Mn 6:54keV; Fe 7:11keV.Good contrast at 6:8keV. data from http:/www-structure.llnl.gov/xray/elements.html(1.6) www.physi
3、cs.ox.ac.uk/history.asp?page=Exhibit10(1.7) M-shellabsorptionisabout3:8keV(seeEx1.5)which implies M 32, but any reasonableguess is acceptable. Estimate relativistic ef-fects to be a few %, or higher.(1.9) 7:9104 K(1.10) BB = 14GHz for B = 1T. Light of wave-length = 600nm has f = 51014 Hz, hencef=f =
4、 3 105. Earths fleld is about5105 T (in the UK).Chapter 2(2.2) Apply ladder operator.(2.5) Isotope shift: 124GHz (see Ex 1.1); for flnestructure splitting and Lamb shift see Sec-tion 2.3.4. An etalon of length 1cm and flnesse100 has transmission peaks whose FWHM= 0:15GHz (assuming air between mirror
5、s).Thus it is easy to flnd an instrument withhigh enough resolution but in practice mea-surements are limited by Doppler broadening(eqn 6.38); (a) isotope shift is fully resolved,(b) flne structure is just resolved, and (c) theLamb shift cannot be resolved.(2.6) (a) vacuum ultraviolet, 0.45 ns, and
6、(b) nearinfra-red, 450 ns (using eqn 1.24).(2.10) (c) IradIang = 0:28a0 (to be checked).(d) Bulge in xy-plane rotating around the z-axis.(e) -transitions related to linear dipole oscil-lating along z-axis. -transitions related to cir-cular motion in xy-plane.(2.13) Excitation to n = 5;l = 4 conflgur
7、ation andsubsequent decay to n = 4;l = 3; n = 3;l = 2;n = 2;l = 1. (Lyman-fi not detected.)Chapter 3(3.1) (b) Binding energy of an electron, 4 13:6 =54:4eV. (c) For given separation the repulsiveenergy equals the binding energy so estimatedI.E. would be zero; this is not a small pertur-bation and th
8、e repulsive energy needs to becalculated more carefully as described in thetext. (Clearly the mean separation is greaterthan r.) (d) Ignoring repulsion, binding energyis 142 13:6 = 2667eV (which is 11% higherthan expt.) Including repulsion (proportionalto Z) gives 2286eV (which is a 5% lower thanexp
9、t.) The repulsion is less important for high-Z atoms (relative to the attraction to the nu-cleus).Chapter 4(4.1) 1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p67s(4.2) List in ascending order: 38541, 39299, 39795,40137, 40383, 40566, 40706, 40814. Plot graphof difierences in wavenumber between eac
10、h pairagainst wavenumber (highest value for eachpair); extrapolate to flnd where difierence goesto zero ( 41250cm1). I.E.(Na) given in Ta-ble 4.1, and quantum defect etc. in Table 4.2.1(4.3) See Table 4.2. The 3s conflguration has n =p13:606=5:14 = 1:63. Between 3s and 6s thequantum defect decreases
11、 by 1:5%. The 8s con-flguration has binding energy 0:31eV (assum-ing same quantum defect as for 6s), c.f. 0:21eVfor the n = 8 shell of hydrogen.(4.4) Quantum defect for 5s conflguration is 3:19(calculate, or look at Table 4.2); use this valueto estimate energy of 7s conflguration. Takingdifierence i
12、n energy between 5s and 7s, dividingby 2 gives an energy equivalent to a wavelengthof 767nm.(4.5) (a) 625nm from data given, c.f. 656nm forBalmer-fi. Outer electron in helium has sim-ilar energy to that in hydrogen. Actuallythe line in helium has a wavelength of 668nm; estimate not accurate because
13、of averag-ing singlet and triplets. (b) For the lowestconflguration n = pR1=35250 = 1:76 and = n n = 0:24. The given conflgurationshave = 0:24; 0:028; 0:23; 0:029; 0:003, i.e.s p d. (c) 6859cm1. (d) Binding en-ergy of 4f conflguration in Li+ is hcR1=4 =3:4eV. Answer given.(4.6) See previous Exercise
14、 and eqn 4.13.(4.7) Note error in Exercise: flne structure split-ting in neutral sodium (0.002eV) and hydro-gen (1:3105 eV). Fine-structure splitting inNa+10 is Z4 = 14641 times that of the sameconflguration in H, namely 0.2eV. Value forneutral atom is approximately the geometricmean of the other tw
15、o, i.e. flne structure ofneutral atoms scales as Z2.(4.8) (b) Ratio 1 : 20 : 14.(4.9) Note obvious error in Exercise: should be sumfrom ml = l to l.Chapter 5(5.3) Interval rule implies levels J = 0;1;2 belong-ing to a 3P term. Allowed transition to a 3Sterm that has no flne structure (its only level
16、has J = 1).First three wavenumbers listed obey an intervalrule that indicates levels J = 1;2;3 (3D term).Other levels must belong to 3P, 3D or 3F (fromselection rules). Sketching the energy levelsand allowed transitions shows that the same3P term as in the flrst part flts the data (withlevels J = 0;
17、1;2 and intervals between them of52 and 106cm1 respectively). The flne struc-tures of the terms in this example obey intervalrule to within a few %. Possible further ex-periments: observe the anomalous Zeeman ef-fect and count the number of components intowhich the line is split (c.f. Fig. 5.13) to
18、deduceJlower and Jupper; one could also measure gJ(see Exercise 5.8).(5.4) Lowest term 1S0. First three excited levels arean obvious triplet whose spacing obey the in-terval rule 3P with levels J = 0;1;2. Nextcome the terms 1P, 3S and 1S, none of whichhave flne structure (c.f. magnesium in Fig. 5.9)
19、.N.B. There is an numerical coincidence whichleads to the ratio of another pair of intervalsbeing almost exactly 2.(5.5) The LS-coupling scheme gives an accurate de-scription of the Mg atom, but is less good forthe Fe ion (ratio of intervals equals about 2.5,rather than 2 as expected from interval r
20、ule),hence S 6= 0 transitions observed in spectrumof the ion but such intercombination lines areNOT observable in Mg.(5.6) Hunds rules and magnetism are described inBlundell (2001): his Table 3.1 gives the mag-netic ground states for 3d ions.(5.7) The electrons in the low-lying conflgurationhave a r
21、esidual electrostatic energy muchgreater than the spin-orbit interaction (of the3p electron), hence the LS-coupling scheme isa good approximation. Electrons in the higherconflguration are further apart (smaller over-lap of their wavefunctions leading to a smallerexchange integral) and the residual e
22、lectrosta-tic interaction is smaller than the spin-orbitinteraction of the 3p-electron, thus the jj-coupling scheme is appropriate and the levelsare in two doublets. The J = 1 levels aremixed.(5.8) The change over from the LS- to jj-couplingscheme occurs because the spin-orbit interac-2tion increase
23、s relative to residual electrostaticinteraction (whereas in the previous Exercisethis arose because Er:e: decreased). For a J = 1level of a pure 1P term (i.e. a term for whichthe LS-coupling scheme is very well obeyed)has gJ = 1 (spin equals zero). This value isclose to the given g-factor hence assu
24、me thatit is this level and there is some mixing withthe wavefunction of 3P1.(5.9) (a) No (b) No (c) Yes (d) No (e) No.The 4d95s5p conflguration has a hole in the d-shell (that on its own would give a 2D5=2 level);coupling with the angular momentum of the5s and 5p electrons gives rise to many levels
25、including 2P3=2. (Other conflgurations mightbe involved but this is the one most likely tohave similar energy.)(5.10) N.B. No central component since MJ = 0 toMJ = 0 does not occur when J = 0. Sixcomponents whose relative separations (on adiagram similar to that in Exercise 5.12) wouldbe in the rati
26、os 12 : 1 : 1 : 1 : 12 , where 1corresponds to the given frequency unit.(5.12) A 3P term has J = 0;1;2. Consideration of allpossibilities: J = 0 $ J0 = 1; J = 1 $ J0 = 1;J = 1 $ J0 = 2; J = 2 $ J0 = 2; J = 2 $J0 = 3 shows that only J = 1 $ J0 = 2 gives9 components. A diagram similar to Fig. 5.13,sho
27、ws that the Zeeman shifts of the compo-nents are (in units of BB=h): (gJ gJ0), gJ,gJ0 and 2gJ0 gJ but not necessarily in thatorder. (We assume that the pattern is sym-metric and that there is a central componentwith no shift.) These shifts must be in the ratio2 : 5 : 7 : 9. Now consider possible ter
28、ms:3P1 $ 3P2 or 3D23P2 $ 3S1; 3P1 or 3D1Note that 3P2 $3P1 appears twice so there areonly four possibilities. The g-factors are:term gJ3S1 23P1 3=23D1 1=23P2 3=23D2 7=6Guess that 3D2 with gJ = 7=6 is involvedsince a 7 appears in the ratios and indeed3P1 $3D2 flts data, otherwise check all pos-sibili
29、ties. Straightforward but rather long ifone does not stay on track.(5.13) (a) gJ = 2; 2=3; 4=3. (b) and (c) see books:Woodgate, Rae, Cohen-Tannoudji et al, etc.(d) Interval of 1700m1 = 510GHz henceB = 510=14 = 36T is the ux density of theorbital fleld. B 14GHzT1Chapter 6(6.1) 17T, 2T, 0.2T.(6.2) X =
30、 91:9MHz.(6.3) Hydrogen has a larger ground state h.f.s. thanlithium because of the relatively large magneticmoment of proton and the high strength of themagnetic fleld at the nucleus produced by a1s-electron (Exercise6.1).(6.4) Splitting proportional to AF (interval rule),where A / gI = I=I. The hy
31、perflne levelsin hydrogen have F = 1 and 2; in deuteriumF = 1=2 and 3=2. Thusf(H)f(D) =AHAD 3=2 =I=II0=I0 3=2 = 4:3The helium ion has the same nuclear spin ashydrogen and therefore the same angular mo-menta. The strength of the magnetic fleld atthe nucleus is proportional to the square ofthe electro
32、ns wavefunction at r = 0; sincej(0)j2 / Z3 and Z3 = 8 for helium, we flndf(H)f(He+) =AHAHe+ =I8I00 = 0:16(6.5) Same values of F as in Example 6.2, hence bothisotopes have same I. Since both isotopes havethe same hyperflne levels we can scale the value(as in Exercise 6.2) to flnd 9MHz.(6.6) I = 2.(6.
33、7) The two strong components arise from theabundant isotope; this is conflrmed by check-ing that the ratio of the total intensities: (70+342)=(5 + 3) = 14, is the same as the ratio ofabundances. The ground conflguration (4s)has a large hyperflne structure. The sum ofthe intensities to (or from) a gi
34、ven level is pro-portional to its statistical weight (2F + 1)|the same sum rule as in flne structure (Sec-tion 4.6.1). Both isotopes have intensity ra-tio of 5=3 and hence the levels are F = 1 andF = 2. Since J = 1=2 we deduce that I = 3=2for both isotopes. Ratio of their nuclear mag-netic moments (
35、39K to 41K): 1:6=0:9 = 1:8.(6.8) (e) 0:2T(6.9) Note correction.(6.10) About 1ppm (1 part per million).(6.13) Size of orbit inversely proportional to bothmass and nuclear charge Z (eqn 1.9):a0=(11 207) = 23fm (c.f. nuclear radius3.4fm). Energy proportional to Z2 and mass:hcR1 112 207 = 340keV. Volume
36、 shiftequals 4% of transition energy.Chapter 7(7.3) (a) See Ex.(13.5) with j1i!j0i and j2i!j1i.(c) cos( + ) = cos(). (f) When = 0the probabilities of being in j1i or j2i are un-changed by the pulse sequence. For a sys-tem that starts in j2i the flnal probabilitiesare the same as given in part (e); i
37、f initiallythe the state is j1i then these probabilities areswapped.(7.6) (b) and (c) Find the spontaneous decay rate forlevel i by summing Aij over all allowed transi-tions, e.g. for the 3p sum over 1s3p and 2s3p.PAij 1=PAijs1 ns3s 6:3106 1603p 1:9108 5.43d 6:5107 162p 6:3108 1.6For 2p, A21 equals
38、the reciprocal of its lifetime.Comparison of the 1s2p and 1s3p transitionsshows that the former has a higher Aij despitehaving a lower frequency; this arises because ofgreater overlap of the 2p and 1s wavefunctionsas compared to 3p and 1s. A simple sketchwould show this clearly.(d)Aij=106 g2=g1 zz !
39、=1015 Ds1 rad s1 a02p3s 6:3 1/3 5/36 2.9 0:541s3p 170 3 8/9 18.4 0:522s3p 22 3 5/36 2.9 3:02p3d 65 5/3 5/36 2.9 3:91s2p 630 3 3/4 15.5 1:3Column zz contains 1n2j 1n2i.! = 2cR1( 1n2j 1n2i).(e)Isat=Wm22p3s 4:61s3p 370002s3p 1402p3d 491s2p 72000Chapter 8(8.1) 2:4GHz, 0:4GHz.(8.2) The Doppler width is 2
40、:3GHz and the twolines havea frequency difierence of 10GHz (flnestructure). To be resolved the Zeeman split-ting must be greater than the linewidth, butthis would make it comparable with the flnestructure and therefore observation of a trueweak-fleld efiect is not really possible.(8.3) Linewidth 28M
41、Hz.Collimation angle 0:014rad.(8.4) Hyperflne levels are F = 3;4 and F0 = 2;3;4;5.The selection rule F = 0;1 leads to sixallowed electric dipole transitions. Analy-sis of the frequency difierences shows thatB;D;E;b;d and e are cross-over resonances(and also ff fc = fC fA = 201:5MHz).(a) Interval rul
42、e:E5;4=5 = 251:4=5 = 50:3E4;3=4 = 201:5=4 = 50:4E3;2=3 = 151:5=3 = 50:5hence A6P3=2 = 50:4MHz.4The interval between the hyperflne levels of theground conflguration is4A6S1=2 = fc fA or ff fC= (fc fa)+(fF fA)+(fa fF)= 151:5+452:9+fa fFwhere fa fF = 8588:2MHz is given on thespectrum, so that A6S1=2 =
43、2298:2MHz.(b) Estimate the temperature from theDoppler width of the absorption, i.e. ignorethe Doppler-free peaks and take the FWHMof the dip to be fF fA 450MHz.fD = 2pln2u = 2pln2r2kBRTM) T = MkB(fD)28ln2 400KAnything from 300-500K is an acceptable an-swer. The underlying Doppler absorption pro-fll
44、e is broadened because of hyperflne splittingof the lines and in fact these data were takenat room temperature.(8.5) Splitting proportional to 1=n3 (eqn 6.10).(8.6) (a) 0:5MHz. (b) Dicult to guess cross-sectionfor collisions between an atom in the groundstate and an atom in the excited state thatcau
45、ses de-excitation of the excited atom (in-elastic collision)|assuming pressure broaden-ing of 30GHz/bar, as in Example 8.3, wouldimply a contribution of 9MHz to the linewidthat the transition frequency (and half this valueat the frequency of the radiation), but see be-low. (c) 2MHz. (d) Zero contrib
46、ution toflrst order. Second order Doppler broadening 1MHz (c.f. eqn 8.23).Measured width 17MHz. Pressure shift is thedominant contribution and so it must be about15MHz (three times larger than in hydrogenwhich is not unreasonable).(8.7) 0:7mbar. = 1 1018 m2. p = = 6 1010 m (c.f. the radius of the Bo
47、hr orbit forn = 2 is 21010 m); metastable hydrogen isdelicate because the 2s 2S1=2 and 2p 2P1=2 lev-els lie close together in energy so that it takesonly a weak perturbation to mix them.Chapter 9(9.1) Solar irradiance 1:4kWm2 (some of which isabsorbed, or re ected, before reaching groundlevel but ig
48、nore this). Frad = 3:3107 N.(9.2) qphoton =!=c =k.(9.3) Consider conservation of energy for the emis-sion of a photon of frequency ! in rest frameof the atom:Mc2 +!0 =! + Mc2Here = 1v2=c21=2 1 + 12(v2rec=c2),where vrec is the recoil velocity (related to therecoil energy by Erec = 12Mv2rec). Hence(!
49、!0) = (1 )Mc2! !0 Erec=For an atom moving with velocity v0 before theemission, the frequency in the laboratory frameis related to ! (the frequency in the rest frameof the atom) by a Lorentz transformation; weuse the transformation from the lab. frame tothe rest frame so that the angle is betweenv0 and the wavevector kem (which gives thedirection of photon in the laboratory):! = 0!0(1 v0c cos )Sinc
