The work function of a certain metal is $3.31× 10^{-19} J$. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 Å is

0.42 eV

Here, work function, $\phi_0=3.31× 10^{-19} J$

Wavelength, $λ=5000 Å=500010^{-10}m$

Energy of the incident photon, $E=\frac{hc}{λ}=\frac{6.63× 10^{-34}×3×10^8}{5000×10^{-10}}=3.98×10^{-19}J$

According to Einstein’s photoelectric equation

$K_{max}=hv-\phi_0=3.98×10^{-19}J-3.31×10^{-19}J=0.67×10^{-19}J$

$=\frac{0.67×10^{-19}}{1.6×10^{-19}}eV=0.42eV$