An electron and proton are separated by a large distance. The electron starts approaching the proton with…

Initially, energy of electron $=+3\mathrm{eV}$

finally, in ${\mathrm{2}}^{\mathrm{nd} }$ excited state,

energy of electron $=-\frac{(13.6\mathrm{eV})}{3^2}$

$=-1.51\mathrm{eV}$

Loss in energy is emitted as photon,

So, photon energy $\frac{\mathrm{hc}}{\lambda }=4.51\mathrm{eV}$

Now, photoelectric effect equation

${\mathrm{KE}}_{\max }=\frac{\mathrm{hc}}{\lambda }-\varphi =4.51-\left(\frac{\mathrm{hc}}{{\lambda }_{\mathrm{th}}}\right)$

$=4.51\mathrm{eV}-\frac{12400\mathrm{eVA}}{4000\mathrm{A}}$

$=1.41\mathrm{eV}$