1.

Computethe firstthreeenergylevelsof doublyionizedlithium. Whatis theionisationpotential

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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