Question

A doubly ionized lithium atom is hydrogen like with atomic number `3`. Find the electron in `(Li^(+ +))` from the first to thired Bohr orbit . The ionization energy of the hydrogen atom is `13.6 eV`.

Answer 1

`E_(n) = - (13.6 Z^(2))/(n^(2)) e V. E_(1) = - (13.6 xx 3^(2))/(1^(2)) = - 122.4 e V`.
` E_(3) = - (13.6 xx 3^(2))/(3^(2)) = - 13.6 e V`.
The corresponding wavelength is `lambda = (12375)/(E_(3) - E_(1)) = 113.74 Å`