MASSACHUSETTS INSTITUTE OF TECHNOLOGY 5.76 Modern Topics in Physical Chemistry Spring, Problem Set #2 ANSWERS

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1 Reading Assignment: Bernath, Chapter 5 MASSACHUSETTS INSTITUTE OF TECHNOLOGY 5.76 Modern Topics in Physical Chemistry Spring, 994 Problem Set # ANSWERS The following handouts also contain useful information: C & S, page 7, radial expectation values of r k for -e atoms LS (j,j ) J Coupling Patterns Herzberg pp. 77-8, The Interval Rule: Analysis of Multiplets Problems -4 deal with material from my //94 lecture (Lecture 7). A lot of background material is provided. These problems illustrate non-text material dealing with secular equations, perturbation theory, transition probabilities, quantum mechanical interference effects, and atomic L-S-J vs. j j J limiting cases. C & S references are to Condon and Shortley The Theory of Atomic Spectra. Problems 5-9 are standard textbook problems, more basic, and much easier than -4 and. BACKGROUND MATERIAL FOR PROBLEMS -4 (i) Transition Amplitudes for np np n s Transitions in the L-S-J Limit µ e / R np r R n s dr C&S, p. 45. C&S, p. 47 gives all nonzero transition amplitudes: p S µ sp P () / µ p D µ sp P +µ p P µ sp P () / µ p P µ sp P () / µ p P µ sp P +(5) / µ p P µ sp P p P µ sp P 5µ 5µ p P µ sp P +(75) / µ. All other transition amplitudes are zero, most notably: p P µ sp P because there is no way to add one unit of photon angular momentum to an initial state with J to make a final state with J.

2 (ii) Energy levels for np and np n s in the L S J Basis Set In the L S J limit, for p (see C&S, pp. 98, 68): S F + F P F 5F H ee P F 5F P F 5F D F + F S / ζ P / ζ ζ H SO P ζ P ζ / ζ D / ζ So we have three effective Hamiltonians for (np) H() F + F +/ ζ +/ ζ F 5F ζ F + 5 F ζ + V V 5 +/ ζ F + ζ V ( ) F 5F ζ H F 5F + ζ / H() / ζ / ζ V F F + ζ + F + F 4 V + F 4 ζ V / ζ Similarly, for the sp configuration: P F G + ζ H P F G ζ / ζ P / ζ F + G P F G ζ and there are three effective Hamiltonians for (n s)(np) H() F G ζ Problem Set # ANSWERS Spring, 994 Page

3 H ( ) F 4 ζ + V V G + 4 ζ V / ζ H() F G + ζ (iii) Now we are ready to discuss the energy level diagram and relative intensities of all spectral lines for transitions between (np) (n s)(np) configurations. The relevant parameters are: F (np,np) F (n s,np) F ζ(np) F (np, np) (difference in repulsion energy for np by np vs. np by n s; F > if n n) (spin-orbit parameter for np; same for both configurations), ζ > by definition (quadrupolar repulsion between two np electrons) F >. G (n s, np) (exchange integral) G >. µ (np n s transition moment integral) All spectral line frequencies and intensities may be derived from these 5 fundamental electronic constants. Note that there are 5 L S J terms in np and 4 L S J terms in np n s, in principle giving rise to a transition array consisting of 5 4 transitions. The 5 parameters determine frequencies and intensities! We are not limited to the L S J or the j j J limit.. Construct level diagrams for the p and sp configurations at the L S J limit (ζ ), the j j limit (F for p, G for sp), and at several intermediate values of ζ/f or ζ/g. This sort of diagram is called a correlation diagram. For graphical purposes it is convenient to keep constant the quantity, which determines the splitting between highest and lowest levels of p, ( ), F 4 F ζ + 4 ζ E p + and a similar quantity for sp, G ζ + G ζ E(sp). Correlation diagrams for sp and p configurations can be found on pages 7 and 75 of Condon and Shortley. ζ For sp, χ 4G Problem Set # ANSWERS Spring, 994 Page

4 ε / G [ + χ ] / ε ζ 4 + χ ( ) / For p, χ 5F ζ ε / F [ + χ ] / ε ζ ( ) + χ /. Use the first order non-degenerate perturbation theory correction to the wavefunctions to compute the intensities for p sp transitions near the L S J limit (ζ F for p, ζ G for sp). For example, the nominal sp P level becomes H P P + / ζ sp P sp P sp P E E sp P + sp P. P P G + ζ The transition probability is the square of the transition amplitude, so the nominally forbidden transition p P sp P has a transition probability Problem Set # ANSWERS Spring, 994 Page 4

5 P( P P ) sp P µ p P ζ µ (5). G + ζ Note that, for the transitions between either of the two sp J levels and either of the two p J or J levels, the transition probability includes two amplitudes which must be summed before squaring. This gives rise to quantum mechanical interference effects. In fact, it is because of these interference effects that the j j limit (/, /) (/, /) and (/, /) (/, /) transitions become rigorously forbidden. ψs corrected to st order p : D D + / ζ 6F ζ P P P / ζ 6F ζ D P P ζ P P + S 5F + ζ ζ P S S 5F + ζ sp: P P / ζ P P G + ζ / ζ P P + G + ζ P P P P Calculate intensities of selected forbidden transitions: Problem Set # ANSWERS Spring, 994 Page 5

6 D µ P / ζ D µ P G + ζ / ζ ζ D µ P P µ P P µ P + ( 4G +ζ)( ) ζ 6F ζ 6F µ 5µ / ζ G + ζ + 6F + ζ. Condon and Shortley (p. 94) give the transformations from the L S J to the j j J basis set. These transformed functions correspond to the functions that diagonalize H SO. p / / P + D / / P D P / / S P / / S + P sp P / / P + P / / P P P Construct the new p H(), H(), H() and sp H(), H(), H() matrices in the j j basis using the above transformations. The p and sp H(J) matrices are easily found in the j j J basis by determining the unitary transforms which transform the j j J basis functions into L S J basis functions and using the fact that: ~ H U HU Problem Set # ANSWERS Spring, 994 Page 6

7 p, J : Ux x a b α β S U U * P U ( ) T U U HU ( ) Tr H Tr H H H H H H + H, H H H + H, H H ( ) + H + H + H H H ( H H ) H H + H ( ) H + H checks p D µ sp P ζ 5 G + + ζ 6F ζ H H Problem Set # ANSWERS Spring, 994 Page 7 µ

8 G, F ζ in L S J limit p S µ sp P ζ S µ P P 5F ζ µ P / ζ G + ζ ζ 5F ζ µ ζ S µ P + F G P µ P ( ) p S µ sp P 4µ ζ 5F ζ + 4G + ζ L S J coupling F, G ζ ζ can be neglected in the denominator / ζ ( G + µ ) ζ ( ) H () H + H + H H + H H H + ( H H ) H + H H H F G ζ + F ( + G ) + ζ F + G + ζ H ζ + F + G F G ζ G H F G ζ + F + G F G ζ H F + G + ζ G () G F G ζ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ( ) ζ Problem Set # ANSWERS Spring, 994 Page 8

9 J, sp U H () F G + ζ Most of the work done on the last four pages can also be found in the notes for Lecture #7. H F + F H ζ H F 5F ζ H F + F F 5F ζ F ζ ( ) + ( ) + ( ) 5 F H ( 5F + ζ) + ζ H ( F + F ) + ( F 5F ζ ) F + 5F + ζ H () F ζ 5 F 5 F F + 5F + ζ ( ζ) ( ζ) p, J U H () F 5F ζ p, J U a b x α P β D x Problem Set # ANSWERS Spring, 994 Page 9

10 U H H H H H H H H + H H H H + H ( ) H H + H H H H ( H H ) H H + H + H H F 5F + ζ + F ζ ( + F ) F F ζ H 6F + ζ ζ F H F 5F + ζ + ( F + F ) + ζ F F + ζ ( ) F F H J sp configuration J, U ζ F F F F + ζ H ( ) F G ζ J, U P P Problem Set # ANSWERS Spring, 994 Page

11 H( ) H H H H 4. Use perturbation theory as in Problem to compute the transition intensities near the j j limit (F ζ or G ζ). You should discover that destructive interference starts to turn off the transitions that will become the forbidden P P and P P transitions in the L S J limit. Transition amplitudes in the j j basis [units of µ]. See also C + S, p. 65 sp \ p Perturbed j j basis functions (use result of Problem Set, #). p 5 F 5F ζ + 5 F 5F ζ + F 4F ζ F 4F ζ α Problem Set # ANSWERS Spring, 994 Page

12 sp + G G + ζ G G + ζ β p : P P sp: P ˆµ ˆµ +α ˆµ + β ˆµ + αβ ( ) ˆµ 5 5α ; α F 4F ζ > because ζ F near j-j limit Correlates to P P transition and approaches zero intensity in L S J limit. Problem Set # ANSWERS Spring, 994 Page

13 µˆ µˆ +α µˆ + β µˆ + (αβ) + α β (α β); β G G + ζ 5. Positronium is an atom-like system formed from an electron and a positron. Predict the energylevel pattern and the wavelengths of some of the electronic transitions of positronium. Predict the wavelengths of some electronic transitions in positronium: The energy level pattern will be hydrogenic. E ( µ,n) ; µ reduced mass. µn For µ m e, µ Mm e M + m e m e me For M m e, µ m m e e We can calculate the positronium energy levels by scaling the Rydberg Formula: ν R n m 5489 n m cm pos 5489 n m ; n principal quantum number of terminus state, m > n. λ Lyman series: λ 5489 m cm ; m λ 8.5 m Å m Balmer series; m cm ; λ 794. m λ 4 Å m Lyman Balmer 4.4Å near 5.4Å 9.Å 4 UV 945.Å 975.5Å Å Å Series Limit 8.5Å 794.Å near IR red Problem Set # ANSWERS Spring, 994 Page

14 6. Without using microstates, derive the ground-state terms and energy levels for the transition elements of the third row (Sc through Zn) of the periodic table. (Remember Cr and Cu are exceptions to the regular Aufbau filling of electrons into orbitals.) ) Hunds first rule: states with the highest spin multiplicity are lowest in energy. ) Hunds second rule: for states with identical S, highest L is lowest in energy. See Bernath, Chapter 5, pages and. Element Config. Ground state Sc s d D Ti s d F (L + ) V s d 4 F (L + + ) Cr sd 5 7 S / filled shell is particularly stable Mn s d 5 6 S (L + + ) Fe s d 6 5 D (L + + ) Co s d 7 4 F (like V) Ni s d 8 F (like Ti) Cu sd S full d shell is particularly stable 7. (a) What are r and /r for the s orbital of hydrogen? r and for Hydrogen. You can do the integrals yourself or look them up. See, for instance, n r n Condon and Shortley, page 7. a r n + r s a [ ( )] r a n r s a (b) What is the transition dipole moment in debye for the p z s transition of hydrogen? Calculate µ(p s) P er cos θ µ p P ( r,θ) s Using the equations and tables on page and of C & S ] / µ [ 8 7 Problem Set # ANSWERS Spring, 994 Page 4

15 8. In the atomic spectrum of neutral Ca there is a normal multiplet of six lines at, 4, 6, 6,, and 58 cm above the lowest frequency line of the multiplet. What are the quantum numbers of the states involved in the transition? Bernath, Chapter 5, #5 I first tried the problem very late at night and I found it completely intractable even though I knew it was easy. It seems much more straightforward at : AM. Check Lecture 8 notes, page for insight. A six line pattern is characteristic of a D P transition. This is plausible for Ca, because Ca has two e outside of a closed p shell, [Ar]. Step : Numerology One can make some educated guesses based on the separations between observed lines This could be the signature of J,, and of a D state This could be the signature of J,, of a P state. Assuming this is correct, are the ζs calculated plausible? For the D, ζ cm, ζ 4 cm ζ d 7. cm. For the P, ζ 6 cm, ζ 5 cm ζ p 5 cm. This is reasonable, as ζ decreases with increasing. Also, ζ decreases with increasing n, so if the D state is on top, then this is even better. Assume: J E/cm x + 6 D x + 4 x 58 P 5 Problem Set # ANSWERS Spring, 994 Page 5

16 Transition Line position/cm D P x D P x x 8 D P x x D P x 5 D P x x 44 D P x 58 listing transitions red blue: x 58 x 44 x x 5 x 8 x This assignment appears to be correct The following wavenumbers are listed in Moores tables for the n P S transitions of Na: (a) n J Wavenumber (cm ) 5.5 5, , , , , , , , , , and.5 4,7. Correct the line positions for the effect of spin-orbit coupling and determine ζ for each of the excited n P terms of Na. Bernath, Eq. (5.97) H SO ζ L S ζ [J(J + ) L(L + ) S(S + ) ] This is the np Rydberg series, so L, S Problem Set # ANSWERS Spring, 994 Page 6

17 H SO ζ J(J + ) 4 E J+ E J ζ(j + ) ; ζ E (J +) J E is the degeneracy weighted average of J np and J : ( Ω J Ω 4, Ω are degeneracies ) E np 4E + E 6 E np cm ζ np cm n [Note: Ive defined E( S / ) ] (b) Devise an extrapolation procedure to determine the ionization potential and the quantum defect for this Rydberg series. Extrapolation of np Rydberg series to determine ionization potential. The energies of the np orbitals can be fit to the following formula: R E np IP ( n µ p ) IP Na Ionization Potential R Rydberg constant [974.7 cm for Na] µ p Quantum defect for the np series I tried several different fits using the Field groups non-linear least squares fitting program. I first tried a fit giving each E equal statistical weight. The resultant fit had systematic residuals. This comes as no surprise, as the Rydberg formula is more accurate at high n. A second fit was done with the 5p, 6p, and 7p levels de-weighted. The uncertainty in the fitted value of the IP decreased considerably. Fit # Fit # δ(e np ). cm δ(e np ) large for low n Problem Set # ANSWERS Spring, 994 Page 7

18 IP 445.7(9) cm IP 445.() cm µ p.86(6) µ p.858() A better fit could be obtained by systematically de-weighting the lower np levels and by observing and including higher np levels. The IP listed in Charlotte Moores tables is cm, over six standard deviations off from the value determined by fit #. Results of the two fits follow NUMBER OF DATA POINTS: 6 FIT # Na np quantum defect determination INITIAL PARAMETERS IP.4455D+5 cm Ryd.9747D+6 cm mu.867d+ OUTPUT FROM LAST LSQ PASS FINAL STANDARD NUMBER NAME VALUE DEVIATION IP D+5.9D+ cm Ryd.9747D+6 cm mu.8648d+.648d OBSERVED AND CALCULATED TRANSITION FREQUENCIES Line J RK BK J RK BK EXPT CALC EXPT-CALC SD units 5p cm 6p cm 7p cm 8p cm 9p cm p cm NUMBER OF DATA POINTS: 6 Na np quantum defect determination INITIAL PARAMETERS FIT # fit residual uncertainty in E IP Ryd mu.4455d d+6.867d+ cm cm OUTPUT FROM LAST LSQ PASS Problem Set # ANSWERS Spring, 994 Page 8

19 FINAL STANDARD NUMBER NAME VALUE DEVIATION IP.4456D+5.995D cm Ryd.9747D+6 cm mu.85887d+.8d OBSERVED AND CALCULATED TRANSITION FREQUENCIES Line J RK BK J RK BK EXPT CALC EXPT-CALC SD units 5p cm 6p cm 7p cm 8p cm 9p cm p cm Problem Set # ANSWERS Spring, 994 Page 9

20 cm Non-Linear Least Squares FIT Residuals: E calc IP Ry(n*) E obs E calc Fit # Fit # n* Effective Principal Quantum Number The spin-orbit operator for the hydrogen atom is 8 9 H SO ξ(r) ˆ ŝ ξ(r) V µ c r r Ze µ c 4πz r Bernath (5.46) For np orbitals of H, ζ np np ξ(r) np n Alkalis, like Na, are nearly one-electron atoms like H, so we might expect ζ np to scale similarly in Na; possibly as n. The n dependence of ζ np can be determined by doing a log-log plot of ζ np vs. n. Assume ζ np A p (n ) k lnζ np ln A k ln n p The p orbital cannot be included because the spin-orbit splitting is not resolved. I did one fit including the 9p and the other without it. n* Dependence of Spin-Orbit Constant Problem Set # ANSWERS Spring, 994 Page

21 ζ np A p (n*) k k fit.6 n* Dependence of Spin-Orbit Constant ζ np A p (n*) k k fit.. On the basis of first-order perturbation theory, the hyperfine structure of the ground electronic state of the H atom involves the interaction of the spins of the electron and proton with one another, and with any applied magnetic fields. It is possible to integrate out the spatial coordinates and to consider the system as two spins S I / governed by the spin Hamiltonian H spin b I S k + S S F z + k I Iz H hfs + H Zeeman, in which b F, k S, and k I are given by b F µ g e µ B g I µ I ψ s () k S g e µ B B k I g I µ N B and g e, g I, µ B, µ N are the g-factors and magnetons for the electron and the proton. The spin Hamiltonian can be split into two parts, b F I S / [referred to as the hyperfine structure (hfs) Hamiltonian], and Problem Set # ANSWERS Spring, 994 Page

22 (k S Sz + k I Iz ) (referred to as the Zeeman Hamiltonian). SI units are used and µ 4π 7 N A is the permeability of vacuum. (a) Calculate the values of b F, k S, and k I (the latter two as multiples of the field strength, B ) for the hydrogen s state. Calculate the values of b F, k s, and k I b () F µ g e µ B g I µ I ψ s µ 4π 7 NA µ B JT µ N JT g e. g I ψ s (r) π / r / a e a ; ψ s () πa Unit check: (NA )(Nm) (NA m ) m Nm J.6 9 J cm 9979 MHz/cm J MHz b F (.5977 ) b F 4.9 MHz MHz k I g I µ N B ( B / Tesla )MHz ( B / Tesla )MHz k S g e µ B B (B / Tesla )MHz B ( / Tesla )MHz Problem Set # ANSWERS Spring, 994 Page

23 (b) Now consider an isolated H atom (with no applied magnetic field). Show that the matrix of H hfs with respect to the m S m I basis is H hfs. 4 b F Find the energies and eigenstates in this basis and construct the matrix X that diagonalizes H hfs. What will be the eigenstates FM F of H hfs expressed in terms of the m S m I states? Give a discussion of this in terms of vector coupling. The eigenvectors of H hfs are listed on the next page. / / X / / The eigenvalues of H hfs are 4 bf and 4 bf. The triplet, F, lies at higher energy. The splitting between the F and F levels is b F 4.9 MHz. program matgen integer r double precision a(4,4), bf, ks, ki, h write(5,*) Magnetic field? [Units of Tesla] read(5,*) h bf 4.9 ki 4.6*h ks 86*h *** mi, ms > *** /, / >, /, / >, /, / >, /, /> *** *** a(,).5*bf +.5*ki +.5*ks a(,).d a(,).d a(,4).d a(,) a(,) a(,).5*bf +.5*ki.5*ks a(,).5*bf a(,4).d a(,) a(,) Problem Set # ANSWERS Spring, 994 Page

24 *** a(,) a(,) a(,).5*bf.5*ki +.5*ks a(,4).d a(4,) a(,4) a(4,) a(,4) a(4,) a(,4) a(4,4).5*bf.5*ki.5*ks write(5,) Magnetic Field / Tesla,h open(file matrix.dat, unit, status unknown) do r,4 write(,) a(r,), a(r,), a(r,), a(r,4) write(5,) a(r,), a(r,), a(r,), a(r,4) enddo close() format(4f.) format(a,f6.) end MATRIX BEFORE DIAGONALIZATION: Magnetic Field / Tesla EIGENVALUES & EIGENVECTORS: # # # # 4 Value: Vector: MATRIX BEFORE DIAGONALIZATION: Magnetic Field / Tesla EIGENVALUES & EIGENVECTORS: Problem Set # ANSWERS Spring, 994 Page 4

25 # # # # 4 Value: Vector: Problem Set # ANSWERS Spring, 994 Page 5

26 MATRIX BEFORE DIAGONALIZATION: Magnetic Field / Tesla EIGENVALUES & EIGENVECTORS: # # # # 4 Value: Vector: In the absence of an external field, the spin of the proton and the electron spin are strongly coupled when anti-parallel. (c) Determine (in terms of b F, k S, k I ) the matrices with elements m S m I H spin m Sm I and F M F H spin F M F in the general case when an applied field, B, is present. m I,m s 4 b F + k I + k s 4 b F + k I k s b F H b F 4 b F k I + k s 4 b F k I k s (d) From the results of part (c) show how the zero field FM F levels split in a weak magnetic field. In this case it is necessary to treat the magnetic field as a perturbation, namely H () b I S, H F () k S Sz + k I I. Give a plot of the splitting of these levels as calculated earlier for fields, B, from to. T (put your energy scale in MHz). See printouts of Fortran program on Athena and graph created, using program output, with Igor on a Mac. Problem Set # ANSWERS Spring, 994 Page 6

27 Notice that at. Tesla ( Gauss) I and s are almost entirely decoupled. MATRIX BEFORE DIAGONALIZATION: Magnetic Field / Tesla EIGENVALUES & EIGENVECTORS: # # # # 4 Value: Vector: MATRIX BEFORE DIAGONALIZATION: Magnetic Field / Tesla EIGENVALUES & EIGENVECTORS: # # # # 4 Value: Vector: Problem Set # ANSWERS Spring, 994 Page 7

28 MATRIX BEFORE DIAGONALIZATION: Magnetic Field / Tesla EIGENVALUES & EIGENVECTORS: # # # # 4 Value: Vector: Hydrogen s Levels in a Magnetic Field F,m F m s,m 4 /, /,,, /, /, /, / 4 /, / Magnetic Field / Tesla Problem Set # ANSWERS Spring, 994 Page 8

29 (e) Determine the energy levels in a strong magnetic field of T, regarding the hyperfine interaction as a small perturbation, that is, H () k S S z + k I Iz, H () b I S. F In this case show explicitly that the first-order perturbation spin functions are () () () () ψ φ, ψ 4 φ 4, b F (), () () ψ φ + g e µ B B + g I µ N B ) φ ( () () b ψ F () φ g e ) φ, ( µ B B + g I µ N B while the second-order energies corresponding to these four functions are E ge µ B B gi µ N B + 4 bf, E ge µ B B + gi µ N B bf + 4 4( ge µ B B b+ g I µ N B ), E ge µ B B gi µ N B b F 4 4 g e µ B B + g I µ N B ), E 4 ge µ B B + gi µ N B + bf. 4 ( F b F m s, m I The electron spin resonance (ESR) spectrum for the hydrogen atoms has only two equally intense lines, because the magnetic moment of the proton is too small to contribute to the intensity, and because the mixing of the m S m I states in the strong field is small. Show explicitly with numerical results that this is indeed the case for the problem that you are considering. Calculate the splitting of the two ESR lines in MHz, and compare your result with the experimentally observed value of 4.4 MHz. What is the corresponding wavelength? How could you use this calculation to substantiate the existence of interstellar clouds of atomic hydrogen? E (k + k I ) B (g µ B g I µ s e N ) Problem Set # ANSWERS Spring, 994 Page 9

30 E (k k I ) B (g µ B + g I µ s e N ) E (k k I ) B (g µ B + g I µ s e N ) 4 (k s + k I ) B (g e µ B g I µ N ) E 4 E E B (g µ B + g I µ e N ) () H b F () all other Hij () H ψ i ψ i + E o j i i bf + B (g e µ + g I µ N ) bf B (g µ + g I µ N ) 4 4 e B B H ij E o j ψ j () () E E + H ; H () ij are the elements of H hfs given in part b) B (g e µ B g I µ N ) + b 4 F () E + H () E [ () H + ] E o E o b F e B (g µ B + g I µ N ) 4 b F + 4B (g µ B + g I µ N ) e Problem Set # ANSWERS Spring, 994 Page

31 () ] o o [ H E () + H () + E E E b F e B (g µ B + g I µ N ) 4 b F 4B (g µ B + g I µ N ) e () () E 4 E 4 + H 44 B b (g e µ B g I µ N ) + 4 F The calculated position of the F F transition is b F 4.9 MHz. The transitions are from and 4. ν E E B (g e µ B g I µ N ) + B (g e µ B + g I µ N ) + bf + 4B (g µ ν E E 4 B (g e µ B + g I µ N ) 4 b + b F 4B (g µ + b F B (g e µ B g I µ N ) 4 F B + g I µ N ) Splitting ν (E E ) (E E 4 ) bf bf b F 4.9 MHz. e e b F B + g I µ N ) ν 4.9 MHz.49 GHz.49 9 s λ [.49 9 s.9979 cm s ].7 cm. At zero field, one observes a single line, F F. In the presence of a modest B field, the line will split and the doublet will shift to the blue (doublet will be seen in a range of a few to a few tens of GHz). The observed splitting of the doublet will change depending upon whether the interstellar cloud is moving towards or away from us, and this is due to the Doppler shift. Problem Set # ANSWERS Spring, 994 Page