Question

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

Answer 1

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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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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