Initially, energy of electron =+3eV
finally, in 2nd excited state,
energy of electron =−32(13.6eV)
=−1.51eV
Loss in energy is emitted as photon,
So, photon energy λhc=4.51eV
Now, photoelectric effect equation
KEmax=λhc−ϕ=4.51−(λthhc)
=4.51eV−4000A12400eVA
=1.41eV