Unformatted text preview: The last electron will thus have energy E=(N/2) 2 (h 2 /8mL 2 )=(h 2 /32m)(N/L) 2 . f. Bosons can all occupy the same energy state, so n=1 and E= (h 2 /8mL 2 ). IV. a. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Note highest is in 4s not 3d. b. Z 2 E /n 2 = 19 2 x13.6/4 2 eV = 307 eV c. Z screen 2 E /16=4.34 eV, so Z screen =2.26 d. Must add 4.34 eV to ionize K. 3.6 eV is released when Cl absorbs electron. So net of 0.74 eV must be added to pair of atoms. V. a. 1s state is allowed (with opposite m s to other electron). E 1s = Z screen 2 E /1 = 24.8 eV. b. 3d > 2p > 1s, each step must have  ∆ l=1. c. E(3d)E(2p)=Z screen 2 E /9 + Z screen 2 E /4 = 3.4 eV E(2p)E(1s)= Z screen 2 E /4 + Z screen 2 E /1 = 18.6 eV Then λ 3,2 =hc/(3.4 eV) = 3.6x107 m and λ 2,1 =hc/(18.6 eV) = 6.7x108 m....
View
Full Document
 Fall '04
 GaryR.Goldstein
 Physics, Atom, Photon, Atomic orbital, Fundamental physics concepts, ev

Click to edit the document details