Photoelectric work function of a metal is 1 eV. Light of wavelength λ = 3000Å falls on it. The photoelectrons come out with maximum kinetic energy of:

3.13 eV

The correct answer is Option (3) → 3.13 eV

According to photoelectric equation,

$K.E.max=hv-\phi$

$=\frac{hc}{λ}-\phi$

$=\frac{(6.63×10^{-34})×(3×10^8)}{10^{18}×3000×10^{-10}×6.24}-1$

$=4.13-1$

$=3.13eV$