A hydrogen atom in ground state obsebe a photon of ultraviolet radition of wavelength `50 nm` Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?

Answer

A, B, D

Explanation

Here `lambda = 50 nm`
Work function = Energy required to remove the electron from `n_1=1` to `n_2=oo`
`E=13.6(1/1-1/oo)=13.6`
`(hc)/lambda-13.6 = KE`
`1242/50-13.6 =KE`
implies `KE = 24.84 -13.6`
`=11.24 eV`