Question

The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6ev the energies required inev to remove electron from the three lowest orbits of hydrogen atom are

Answer 1

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Answer 2

E = 13.6 z^2/n^2

E1 = 13.6 x 1/(1)^2 = 13.6  

E2 = 13.6 x (1)^2/(2)^2

13.6/4 = 3.4

E3 = 13.6 x (1)^2/(3)^2

= 13.6/9  

= 1.51

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