The threshold frequency of a photoelectric metal is ${\mathrm{\nu }}_0$. If light of frequency $4{\mathrm{\nu }}_0$ is incident on this metal, then the maximum kinetic energy of emitted electrons will be

According to Einstiens photoelectric, equation $hv=h{v}_0+(KE{)}_{max}$
where hv is the energy of the incident photon v is the frequency of photon, v₀ is the threshold frequency and hv₀ is the work function of the material.
Now $(KE{)}_{max}=hv-h{v}_0$
Here,$v=4v_{0}so,(KE{)}_{max}=3h{v}_0$
So maximum kinetic energy of the photoelectron is 3hv₀.