The light emitted in the transition `n = 3 to n= 2` in hydrogen is called `H_(alpha)` light .Find the maximum work fonction a metel one have so that `H_(alpha)` light can emit photoelectrons from it

Answer

A

Explanation

Here `n_1=2`,`n_2=3`
Energy possessed by `H_alpha` light
`=13.6((1)/(n_1^2)-(1)/(n_2^2))`
`=13.6 xx (1/4-1/9)`
`=(13.6 xx 5)/(36) = 1.89 eV`.
For `H_alpha` light to be able to emit photoelectrons from a metal the work function must be greater than or equal to `1.89 eV`.