By 100 nm photons kinetic energy 112x10 18 j 5 these

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by 100 nm photons Kinetic energy = 1.12x10 -18 J 5. These are one electron systems and ionization implies excitation of an electron to "n=infinity", that is, removal of the electron so that there is no attraction to the nucleus. The ionization energy is just the Rydberg energy. E = (+2.18x10 -18 J)Z 2 (must input energy to remove the electron) He + : E = (2.18x10 -18 J)(4) = 8.72x10 -18 J Li 2+ : E = (2.18x10 -18 J)(9) = 1.96x10 -17 J Be 3+ : E = (2.18x10 -18 J)(16) = 3.49x10 -17 J E = (87.2 x 10 -19 J) / (1.60 x 10 -19 J) = 54.5 eV For Li 2+ , Z = 3 and e = 0 – (–21.8 x 10 -19 (3 2 /1 2 ) J) = 196.2 x 10 -19 J in kJ mol -1 , E = (196.2 x 10 -19 J) (6.02 x 10 23 mol -1 ) = 11811 kJ/mol From the above equations we note that: ∆Ε Li / ∆Ε He = Ζ 2 Li / Ζ 2 Ηε = 3 2 /2 2 = 2.25 We see that the numerical values confirm this ratio. For Li 2+ , in eV, E = (2.25) (54.5)eV = 123 eV Using a similar approach for Be 3+ (Z = 4) we find: E = 348.8 x 10 -19 J = 21kJ/mol = 218 eV
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