Atomic Structure, Periodic Table, and Other Effects: Chapter 8 of Rex and T. Modern Physics

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1 Atomic Structure, Periodic Table, and Other Effects: Chapter 8 of Rex and T Modern Physics 11/16 and 11/19/2018 1

2 Introduction In Chapter 7, we studied the hydrogen atom. What about other elements, e.g., He, Li, Na, Cl, etc. How do we explain their behavior? We ll mainly cover Section 8.1 And a little bit of Section 8.2 Skipping 8.3, which you will cover in detail in PHY 440 It turns out that the exact solution exists only for the H-atom Even in He, with 2 e- s, we must account for electronelectron force Have to rely on numerical methods This chapter will be more conceptual than 7 2

3 Section 8.1 In Chapter 7 we saw that in the hydrogen atom, we specify the system with four quantum # s n, l, m l, m s To explain experimental data, Pauli proposed Exclusion Principle No two electrons in an atom may have the same set of quantum numbers n, l, m l, and m s So, in a He atom, which has 2 electrons, we can have the 2 electrons in the n = 1, l = 0 state BUT m s = +1/2, -1/2 In a Li atom, with 3 electrons, we cannot add another electron in the n = 1 state [it is filled or closed ] So, we have to look at the n = 2 state. Here l can be 0 to 1 According to rules, state with lower energy gets filled first 3

4 Pauli Exclusion Principle These two rules, Exclusion Principle, and lowest energy state is filled first, essentially explain the structure of the Periodic Table Hydrogen atom -> n=1, l=0, m l =0, m s =1/2 He atom -> n=1, l=0, m l =0, m s = ½ or m s =-1/2 Li atom -> As above + n = 2, l = 0, m l = 0, m s = +1/2 Be atom -> as above + m s = -1/2,... As we saw in Chap. 7, in a magnetic field V = -µ s B Implies that m s = +1/2 has lower energy We now get into using notation for short-hand n = 1, 2, 3, 4 Letter = K L M N think of it as a single electron system -> (screening) -> -> energy of system mainly depends on n 4

5 Periodic Table So, in ground state of H / He, electrons are in K shell If n 2, l can be 0, e.g., 1, 2, etc. These are known as sub-shells [denoted by nl] e.g., 1s, 2s, 2p H-atom -> electron is in 1s sub-shell of K shell He-atom -> 2 nd e- is also in 1s sub-shell of K shell Li -> 3 rd e- is in n = 2, l = 0, i.e., 2s sub-shell of L shell. Be -> 4 th electron also in 2s sub-shell of L shell B(oron) -> 5 th electron in 2p subshell of L shell ( n = 2, l = 1, m l = -1, m s = 1/2 ) C(arbon) -> 6 th electron has n=2, l=1, with different values of m l / m s ; and so on & so forth through table 5

6 Continued To avoid violating the Pauli exclusion principle, we have For each value of m l -> 2 values of m s For each value of l -> m l = -l to l in integer steps, i.e., (2l+1) values So, each sub-shell has 2 x (2l+1) electrons s sub-shell has 2 electrons ( l = 0 ) p sub-shell has 6 electrons (l = 1) And so on See Figure 8.1 and Table 8.1 (p. 271 of book) 6

7 Figures 8.2 and Onward As we fill these sub-shells and shells, i.e., n increases, two things will happen E n -> energy of state = -E 0 / n^2 becomes less negative <r> = ψ* n r ψ n (r^2 dr) will increase Recall ψ n (r) is the solution of the radial equation in Ch. 7 If E n becomes less negative => easier to remove the electron E = 0 => e - is free of electrostatic force of the nucleus Figure 8.3 -> Ionization E Figure 8.4 -> Atomic Radii See example 7.11 He and other inert gases have large negative E, because all shells are filled and two electrons form a pair Alkali metals, e.g., Li, Na, K have one electron in the outermost shell and easy to strip off -> very reactive Similarly Cl/Fl/Br have one missing e- in outermost shell 7

8 Total Angular Momentum This is a complex topic; I ll give a brief overview Essentially, total angular momentum, J, J = L + S, Is the sum of orbital & spin angular momentum L/L z & S/S z are quantized => J/J z are quantized J = h-bar [ j (j+1) ], J z = m j h-bar Where m j goes from j to j in integer steps and j = l ± s e.g., if l = 1 and s = ½, j can be ½ or 3/2 l = 0 then j = 1/2 one-electron case -> -> -> mj = -1/2, +1/2-3/2, -1/2, +½, +3/2 L & S are vectors, so you can do vector addition/ subtraction 8

9 Section 8.2 In this formalism, we use the notation nl j to show various states, Actually, this is modified a little for multi-e- states to n 2s+1 L j We use J (instead of L & S) because of a concept called spin-orbit coupling In the electron s frame of reference, it is the proton that is revolving around it The proton has charge and as it moves, you can think of it as a current loop, again => it generates a magnetic field. The e- spin couples (or, interacts ) to this internal -> -> magnetic field -> this B field is due to L of the e- -> -> -> -> Hence the interaction potential = -µ s B = S L [See Ch. 7] This effect is similar to when we have an external field 9

10 More Rules: Equation 8.8 Selection rules for transitions between various states are modified. Various levels split -> (Fine structure). These splittings are observable In sodium 3p -> 3s transition splits into two lines, and 3P 1/2 -> 3S 1/ nm 3P 3/2 -> 3S 1/ nm λ (Fig. 8.7) ΔE ~ 2 x 10^-3 ev [see Fig. on back cover of book] For completeness, there is something called -> -> HyperFine structure (nuclear spin couples to L, S of -> electrons, and also to external B fields). energy levels splits are less than in fine structure n I m going to do one multi-e- atom example 2s+1 L J We now use n, l, j, and mj to label states -> as the q # s 10

11 Problem #15 (Based) A H-atom is in the 4d state. Find all possible values for n, l, m l, s, m s, j, and m j, and label and count ALL of the states One group member coordinates all the others efforts One group member does n, another l, s, and j Group member who did n does m l and m s, other m j (note energy splitting between various sub-levels is small but visible) Extra question, if done (unrelated to above ques) Why no spin-orbit coupling in the ground state? 11

12 Another Example The simplest multi-electron state: Ex. 8.7 What are L, S, and J for the excited He atom? He has 2 electrons. In the ground state both are in the n = 1 shell -> this has l = 0 (i.e. L = 0) and The two electrons have opposite spins up and down We say that the total spin s is 0 ( = 2s + 1) L = 0, S = 0 => J = 0 => 1 1 S 0 (ground) n L=0 J For the first excited state, one of the two electrons goes to n = 2 state (l=0 or l=1). That is, the atom has 1s 1 2s 1 Or 1s 1 2p 1 # of electrons outside closed shells 1 electron in ground, 1 in n=2, l=0 OR: in n = 2, l = 1 In 1s 1 2s 1 -> L=0, but now S can be 0 or 1, i.e., J=0 or 1 1s 1 2p 1 -> L=1 & S can be 0 or 1 J = 1 or 0,1,2 (S = 0 or S=1). See Figure

13 Time Permitting What is the difference between L and l, m l and m s, and between J z and m j? (See Figure 7.3) Ques. 13 Describe in your own words 13

Many-Electron Atoms. Thornton and Rex, Ch. 8

Many-Electron Atoms. Thornton and Rex, Ch. 8 Many-Electron Atoms Thornton and Rex, Ch. 8 In principle, can now solve Sch. Eqn for any atom. In practice, -> Complicated! Goal-- To explain properties of elements from principles of quantum theory (without