Photons emitted when electrons in a H-atom make transition from a higher energy state to lower energy state, whose difference in angular momentum is  $\frac{\text{h}}{\text{π}},$ are made to incident on sodium metal (work function, W = 2.3 eV). The maximum possible kinetic energy of emitted photoelectrons is :

Difference in angular momentum  $=\frac{\text{h}}{\text{π}}$
$\therefore ({\text{n}}_{\text{2}}-{\text{n}}_{\text{1}})\frac{\text{h}}{\text{2π}}=\frac{\text{h}}{\text{π}}\therefore {\text{n}}_{\text{2}}-{\text{n}}_{\text{1}}=2$ (Difference in shell no.)
For photoelectric effect to be observed, Energy of photon > Work function (2.3 eV)
 Two photons are possible in H-atom where difference in shell number is 2 and energy > 2.3 eV
  $\therefore {\text{E}}_{\text{photon}}=12.09\text{\hspace{0.17em}eV\hspace{0.17em}}\left(\text{From\hspace{0.17em}}3\to \text{1\hspace{0.17em}transition}\right)$$\&2.55\text{\hspace{0.17em}eV}\left(\text{From\hspace{0.17em}}4\to \text{2\hspace{0.17em}transition}\right)$
Max KE of photoelectron will correspond to max energy of incident photon.
$\therefore {\left(\text{KE}\right)}_{\max }=12.09-2.3=9.79\text{\hspace{0.17em}\hspace{0.17em}eV}$