An electron and proton are separated by a large distance. The...

Initially, energy of electron $= + 3eV$

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

energy of electron $= - \dfrac{( 13 . 6eV )}{{3}^{2}}$

$= - 1 . 51eV$

Loss in energy is emitted as photon,

So, photon energy $\dfrac{hc}{\lambda} = 4 . 51eV$

Now, photoelectric effect equation

{KE}_\max = \dfrac{hc}{\lambda} - ϕ = 4 . 51 - \left(\dfrac{hc}{{\lambda}_{th}}\right)

$= 4 . 51eV - \dfrac{12400eVA}{4000A}$

$= 1 . 41eV$