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Transcript
2.3 Solutions 109 Iron: E K = 13.6(26 â 1)2 = 8,500 eV 1,241 = 0.146 nm = 1.46 AÌ Î»K = 8,500 2.25 The minimum wavelength of the photon will correspond to maximum frequency which will be determined by E = hvmax 2Ï c c hc hc = = = vmax hvmax E E 2Ï Ã 197.3 fm â MeV = 4.13 à 105 fm = 4.13 AÌ = 30 à 10â3 MeV λmin = hc eV 1.6 à 10â19 à 80 à 103 à 0.15 à 10â10 eV λC = h= c 3 à 108 â34 = 6.4 à 10 J â s hc 2.27 λc = eV 0.247 à 10â10 à 1.6 à 10â19 à 50,000 λc eV = h= c 3 à 108 = 6.59 à 10â34 J â s 2.26 λC = 2.28 According to Mosleyâs law 1 = A(Z â 1)2 λ 1 = A(26 â 1)2 λI 1 = A(29 â 1)2 λCu λCu 252 = 2 = 0.797 â λCu = 193 à 0.797 = 153.8 pm λI 28 2.29 λ K â λC = 84 pm = 0.84 AÌ 1,200 12.4 = 0.84 â 2 (28 â 1) V 12.4 hc = (V is in kV) where λC = eV V Solving for V in (2), V = 15.4 kV (1) (2) (3) 2.30 The L α line is produced due to transition n = 3 â n = 2. For the n = 2 shell the quantum numbers are l = 0 or l = 1 and j = l ± 12 , the energy states being 2 S1/2 , 2 P1/2 , 2 P3/2 . For n = 3 shell the energy states are 3 S1/2 , 3 P1/2 , 3 P3/2 , 3 d3/2 , 3 d5/2 The allowed transitions are 3 S1/2 â 2 P1/2 , 3 S1/2 â 2 P3/2 3 P1/2 â 2 S1/2 , 3 P3/2 â 2 S1/2
