25.(a) Promoting one of the electrons (described in problem 23) to a not-fully occupied higher level, weFnd that the conFguration with the least total energy greater than that of the ground state has then= 1 and 2 levels still Flled, but now has only one electron in then= 3 level; the remaining twoelectrons are in then= 4 level. Thus,Efrst excited=2E1+2E2+E3E4µh28mL2¶(1)2µh28mL2¶(2)2+µh28mL2¶(3)2µh28mL2¶(4)2= (2+8+9+32)µh28mL2¶= 51µh28mL2¶.(b) Now, the conFguration which provides the next higher total energy, above that found in part (a),has the bottom three levels Flled (just as in the ground state conFguration) and has the seventhelectron occupying then= 5 level:Esecond excitedE1E2E3+E5µh28mL2¶(1)2µh28mL2¶(2)2µh28mL2¶(3)2
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