Light of wavelength 300 nm is used in an experiment on photoelectric effect with lithium having a work function of 2.7 eV. The maximum kinetic energy of ejected photo-electrons will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.43 eV
Einstein's photoelectric equation gives the maximum kinetic energy as the photon energy minus the work function.
\( \displaystyle K_{\max} = h\nu - \phi = \frac{hc}{\lambda} - \phi \)
Using \( hc = 1240\,\mathrm{eV\,nm} \), the photon energy is
\( \displaystyle E = \frac{1240\,\mathrm{eV\,nm}}{300\,\mathrm{nm}} = 4.13\,\mathrm{eV} \)
\( \displaystyle K_{\max} = 4.13\,\mathrm{eV} - 2.7\,\mathrm{eV} = 1.43\,\mathrm{eV} \)