When the surface of a metal is illuminated with the light of wavelength 400 nm, the KE of the ejected photoelectrons is 1.68 eV. Then the work function of the metal is (hc = 1240 eV nm)

1.42 eV

The correct answer is Option (2) → 1.42 eV

The photoelectric effect can be described as -

$E_k=hv-\phi=\frac{hc}{λ}-\phi$

where,

$E_k$ = Kinetic energy of ejected photoelectron

$v$ = frequency of incident light

$λ$ = wavelength of incident light = 400 nm

$E_k=\frac{1240eVnm}{400nm}-1.68eV$

$=1.42e$