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Computethe firstthreeenergylevelsof doublyionizedlithium. Whatis theionisationpotential |
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Answer» Solution :energy `E_n = (Z^2 mc^2 alpha ^2)/(2n^2 )` forhydrogen ,` z=1andE_n= (- 13.6eV )/( n^2)` forlithium, z= 3`thereforee_n= ( - 112.4 )/( n^2) eV ` ` thereforeE_1=- 122.4 ` eV Ionizationpotential`= 122.4 ` volts ` E_2=(-122.4 )/( 4 ) =- 30.8eV,E_3= (-122.4 )/(9) =- 13.6 eV , E_4= (-122.5 )/( 16)=- 7.65eV ` |
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