A light of wavelength 500 nm falls on a metal surface, whose work function is 2.14 eV. The kinetic energy of the emitted electrons is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.34 eV
By Einstein's photoelectric equation, the photon energy goes partly into freeing the electron (the work function \( \phi_0 \)) and the rest appears as the maximum kinetic energy of the emitted electron. The kinetic energy asked for is this maximum value.
\( \displaystyle K_{\max} = \frac{hc}{\lambda} - \phi_0 \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
\( \lambda = 500\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{m} \)
\( \displaystyle \frac{hc}{\lambda} = \frac{6.63\times10^{-34} \times 3\times10^{8}}{5\times10^{-7}} = 3.98\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \frac{3.98\times10^{-19}\,\mathrm{J}}{1.6\times10^{-19}\,\mathrm{J/eV}} \approx 2.48\,\mathrm{eV} \)
\( \displaystyle K_{\max} = 2.48 - 2.14 = 0.34\,\mathrm{eV} \)
Shortcut: With \( hc \approx 1240\,\mathrm{eV\,nm} \), the photon energy is \( 1240/500 = 2.48\,\mathrm{eV} \) at once, and \( K_{\max} = 2.48 - 2.14 = 0.34\,\mathrm{eV} \).