9. Let the quantum numbers of the pair in question be nand n+ 1, respectively. We note that EEnhmLmLmLnn+−=++122222222188218bgbgTherefore, En+1– En= (2n+ 1)E1. Now EEEnE+−====+152111521bg, which leads to 2n+ 1 = 25, or n= 12. Thus, (a) The higher quantum number is n+1 = 12+1 = 13. (b) The lower quantum number is
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.