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Unformatted text preview: 1 L=n 1 and k 2 L=n 2 . Hence k 1 =n 1 /L and k 2 =n 2 /L . c. E n1,n2 =(h 2 /8mL 2 )(n 1 2 + n 2 2 ) Lowest energy state has n 1 or 2 =1. Hence E 1,1 = (h 2 /4mL 2 ) d. 2 electrons can occupy the lowest energy state. The 3rd electron must go to the next highest level (2,1 or 1,2), giving 2x(h 2 /4mL 2 )+5(h 2 /8mL 2 ) = 9(h 2 /8mL 2 ). e. Bosons can all occupy th same energy state, so 3 E 1,1 = (3h 2 /4mL 2 ). IV. a. 1s 2 2s 2 2p 1 b. Z 2 E /n 2 = 25x13.6/4 eV = 85.0 eV c. Z screen 2 E /4=8.28, so Z screen =1.56 d. l=1, so =square root (2) B e. 4f > 3d > 2p so E(4f)E(3d)=Z screen 2 E /16 + Z screen 2 E /9 = 1.61 eV E(3d)E(2p)= Z screen 2 E /9 + Z screen 2 E /4 = 4.6 eV V. a. p E1 =Qr=1.60x1019 Coul x 2.36x1010 m=3.78x1029 Coul m b. measured dipole / theoretical dipole = 3.00/3.78 = 0.79 or 79%....
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This note was uploaded on 03/10/2008 for the course PHY 13 taught by Professor Garyr.goldstein during the Fall '04 term at Tufts.
 Fall '04
 GaryR.Goldstein
 Physics

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