1. EEB, DII, S6, 2010 6) Atom spectroscopy: energy levels (see lab experiment no 3 & lecture 11.02.10) When analysing data it is often necessary to fit a functional expression to data points to derive some characteristic parameter prior to interpretations. This can be performed by using various computer programs. Instructions about how to do such analysis for specific examples by using IGOR are to be found on the internet (see HERE and HERE).lecture 11.02.10IGORHERE The problem below is related to your laborotory work and analysis of the H atom spectrum. By defining the ground state energy of atoms equal to zero, energy levels for the H atom (E H (n)) depend on the principal quantum number n asH atom spectrum E H (n) = IP H – R/(n) 2 (1) where IP H is the ionization potential for H and R is the Rydberg constant. For atoms (A) larger than the H atom, the energy levels (E A (n)) can be expressed asRydberg constant E A (n) = IP A – R /(n +  (l)) 2 (2) Where  (l) is an empirical parameter depending on the angular momentum quantum number (l) and IP A is the ionization potential for A and R is the Rydberg constant.Rydberg constant Below you will find energy levels for s and d orbitals of the sodium atom (Na) derived from spectra analysis of the Na atom. a-b) Fit the analytical expression (2) to the observed energy levels to determine  (l) for the s(l=0) and d(l=2) states. NB / hint: Alternatively you could rearrange expression (2) conveniantly and perfom a line fit. c) Explain the difference in the  (l) (l = 0,2) values (see lecture notes from 11.02.10).energy levels for s and d orbitals of the sodium atom (Na) derived from spectra analysislecture notes from 11.02.10

2. nE(cm -1 ) / s orbitalsE(cm -1 ) / d orbitals 329 172.889 425 739.99134 548.766 533 200.67537 036.774 636 372.62038 387.270 738 012.04439 200.93 838 968.5139 728.70 939 574.85 IP Na = 41449.44 cm -1. See fit procedure by IGOR at http://notendur.hi.is/agust/kennsla/ee10/eeb/Daemi/EEB-DIIs6-10.pxp

3. nE / s 425740 533200.7 636372.6 738012 838968.5 939574.8 Coefficient values ± one standard deviation K1 = -1.3562 ± 0.000842 =  (l=0) a)

4. nE / d 329172.9 434548.8 537036.8 638387.3 739200.9 839728.7 Coefficient values ± one standard deviation K1 = -0.010683 ± 0.000429  (l=2) b)

5.  (l=2) = -0.010683 <<  (l=0) = -1.3562: shielding effects: See next slides c)

6.    R 2 r Na: 1s 2 2s 2 2p 6 3s Na + : 1s 2 2s 2 2p 6 Skermun/ shielding: 3s Na*: 1s 2 2s 2 2p 6 3p 3p Na**: 1s 2 2s 2 2p 6 3d 3d

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