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Compute the first three energy levels of doubly ionized lithium. What is the ionisation potential ? |
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Answer» Solution :ENERGY `E_n = -(Z^2 mc^2 alpha^2)/(2n^2)` For hydrogen `Z = 1 and E_n = (-13.6 eV)/(n^2)` For lithium , `Z = 3 :. E_n = (-122.4)/(n^2) eV` `:. E_1 = -122.4 eV` Ionizaiton POTENTIAL = 122.4 VOLTS `E_2 = (-122.4)/4 = -30.8 eV, E_3 = (-122.4)/9 = -13.6 eV, E_4 = (-122.4)/(16) = -7.65 eV`. |
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