The surface a metal is illuminated with the light of 400 nm. The kientic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is ($hc = 1240\, eV-nm$)

1.42 eV

$K_{max} = \frac{hc}{\lambda} - \phi$

$\Rightarrow \phi = \frac{hc}{\lambda} - K_{max} = \frac{1240}{400} - 1.68 eV = 3.1 eV - 1.68 eV = 1.42 eV$