Light corresponding to the transition `n = 4 to n = 2` in hydrogen atom falls on cesium metal (work function `= 1.9 eV)` Find the maximum kinetic energy of the photoelectrons emitted

Explanation

The energy of the photons emitted in transition `n = 4 to n = 2` is
`hv = 13.6eV[(1)/(2^(2)) - (1)/(4^(2))] = 2.55eV`
The maximum kinetic energy of the photonelectrons is
`= 2.55 eV - 1.9eV = 0.65 eV`