Find the binding anergy of an electron in the ground state of a hydrogen like atom in whose spectrum the thrid Balmer line is equal to 108.5 mm.

Options

A. `54.4 eV`
B. `13.6 eV`
C. `112.4 eV`
D. None of these

Answer

A

Explanation

`Delta E = (12375)/(1085) = 11.4 eV`
Third Balmer line is corresponding to the
transition, n =5 to n =2
`E_5 - E_2 =11.4`
`:. (E_1)/((5))^2 - (E_1)/((2))^2 11.4`
`E_1=-54.28 eV`
`:. |E_1| = 54.28 eV.`