According to Bohr's theory,the electronic energy of a atom in the oth orbit is given by `E_(n) = (-2.17 xx 10^(-18))/(n^(2)) J`
Calculate the longest wavelength of light that will be needed in remove an electron the third Bohr orbit of `He^(o+)

Explanation

`(1)/(lambda) = 1.09 xx 10^(5)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)`
For longest wave length transition would accur from `n = oo`
or `= 1.09 xx 10^(5)((1)/(3^(2)) - (1)/(oo^(2)))Z^(2)`
`= 1.09 xx 10^(5) xx (4)/(9) = (4.36)/(9) xx 10^(5)`
`= 0.484 xx 10^(5)cm^(-1)`
`or lambda =- (9)/(4.36) xx 10^(-5)`
`= 2.06 xx 10^(-5)cm`
`= 2.06 xx 10^(-5) xx 10^(8) Å = 2060 Å`