8.69) and Table 8.1. 8.34 If you haven't already done so, do parts (a) and (b) of Problem 8.33, and then do part (c), but for the five spherical harmonics with/ 2. r momentum l expansion to . (This shows SECTIONS 8.7 and 8.8 (The Energy Levels of the Hydrogen Atom and Hydrogenic Wave Functions) m mechanics el.) r the special 8.35 Prove that the degeneracy of the nth level in the hy- drogen atom is n'; that is, verify the result (8.77). (But be aware that this number gets doubled because of the electron's spin, as we describe in Chapter 9.) constant d solution is ow that this acceptable). r differential mbination of n, and prove 8.36 .It is known that a certain hydrogen atom has a defi- nite value of l. (a) What does this statement tell you about the angular momentum? (b) What are the allowed energies consistent with this information? nstant. 8.37 The mean value (or expectation value) of 1/r for any state is (1/r) the 1s state of hydrogen. Comment. [Hint: See the integrals in Appendix B.] = J (1/r) P (r) dr. Find (1/r) for r the case ution. (Any sin 0 is the e complete ependence m = -1. (a) It is known that a certain hydrogen atom has 8.38 2. How many different states are n = 5 and m consistent with this information? (b) Answer the same question (in terms of n and m) for arbitrary values ofn and m leave it as 12 a maxi- Even when n and / are specified, there are still (21 +1) distinct corresponding to the (2/ + 1) orientations m = 1, l - 1, - . . , -l. For s s (=0), there is just one orientation; for p states ( = 1) there (2 X 1)+1 = 3; for d states (I = 2) there are (2 x 2) + 1 = 5, and so These numbers are shown in paren theses on the right of each horizontal be Fig. 8.16. The total degeneracy of any level can be found by adding al numbers for the level in question. For example, the n = 1 level is nondegee ate; the n 2 level has degeneracy 4; the n = 1 = 0,1,., (n -1), and hence has degeneracy* (Problem 8.35) single momentum of a multielectron atom. 9 For the case n (Problem 8.39) 3 level 9. The nth level This wave functio for the electron, maximum at the zero angular mon state with / # 0. classically: A cla momentum is zer I>0 has import Chapter 10. It als dent on the spat 1+3+5+ .. +(2n - 1) = n (8.77 *Actually, the total degeneracy is twice this answer. This is because the electron has other degree of freedom, called spin, which can be thought of as the angular m tum due to its spinning on its own axis (much as the earth spins on its north-sou This spin can have two possible orientations, and for each of the states describeu there are really two states, one for each orientation of the spin. This will be discuss Chapter 9. energies of atom radii.

I. PROBLEM 1. CHARGE AND CURRENT DISTRIBUTIONS. Problem 10.3 (a) Find the fields, and the charge and current distributions, corresponding to V (r, t) = 0, A(r, t) = 4T €, r2 Don't forget that the potentials are not necessarily given in Loren gauge and the equations you should use are 10.4 and 10.5. The potential representation (Eqs. 10.2 and 10.3) automatically fulfills the two homogeneous Maxwell equations, (ii) and (iii). How about Gauss's law (i) and the Ampère/Maxwell law (iv)? Putting Eq. 10.3 into (i), we find that a 1 v?v +(v · A) = --p; (10.4) at €0 this replaces Poisson's equation (to which it reduces in the static case). Putting Eqs. 10.2 and 10.3 into (iv) yields a?A V x (V x A) = HoJ – Hoco V at or, using the vector identity V x (V x A) = V(V · A) – V²A, and rearranging the terms a bit: a²A av V - A + Ho€0- at v²A :-HoJ. (10.5) Equations 10.4 and 10.5 contain all the information in Maxwell's equations.