Question

the ionization energy of hydrogen atom is -13.6ev. calculate second ionization energy of helium?

please help me to solve this answer.

Answer 1

hell.. ooo!!

E = (-13.6xZ^2) / n^2 eV

where, n = number of orbit = 1 

Z = atomic number = 2 (He)

E = (-13.6x2^2) / 1^2

E = (-13.6x4) = -54.4 eV


hope this helps you....!!

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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