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InterviewSolution
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For a certain hypothetical one electron atom, the wavelength `(in Å)` for the spectral lines for transitions originating at n=p and terminating at n=1 are given by `lambda = (1500 p^2)/(p^2 - 1), where p = 2,3,4` (a)Find the wavelength of the least energetic and the most energetic photons in this series.(b) Construct an energy level diagram for this element showing the energies of the lowest three levels. (c ) What is the ionization potential fo this elelment? |
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Answer» Correct Answer - A::B::C::D (a) `lambda =1500 ((1)/(1 -1/p^2))` `lambda_(max)` corresponds to least energetic photon with p = 2. `:. lambda_(max) = 1500((1)/(1-1//4)) = 2000 Å` `lambda_(min)` corresponds ot most energetic photon `with p = 00` `lambda_(min) = 1500 Å` (b) `lambda_(oo -1) =1500 Å` `...........E_3 = - 0.95 eV` `........... E_2 = -2.05 eV` `........... E_1 = -8.25 eV` `:. E_(oo) - E_1 = (12375)/(1500) eV` =8.25 eV `:. E_1 = - 8.25 eV (as E_(oo) = 0)` `lambda_(2 -1) = 2000 Å` `:. E_2 -E_1 = (12375)/(2000) eV`h =6.2 eV `:. E_2 = -2.05 eV` Similarly, `lambda_(31) = 1500((1)/(1-1//9))` =1687.5 Å `:. E_3 - E_1 = (12375)/(1687.5) eV = 7.3 eV` `:. E_3 = - 0.95 eV` (c ) lonization potential = 8.25 V. |
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