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)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.42 eV
By Einstein's photoelectric equation, the photon energy is used partly to free the electron (the work function \( \phi_0 \)) and the rest appears as the electron's kinetic energy.
\( h\nu = \phi_0 + K_{\max} \)
\( \displaystyle h\nu = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{400\,\mathrm{nm}} = 3.1\,\mathrm{eV} \)
\( \phi_0 = h\nu - K_{\max} = 3.1 - 1.68 = 1.42\,\mathrm{eV} \)