Ultraviolet light of wavelength 280 nm is used in an experiment on photoelectric effect with lithium (work function = 2.5 eV) cathode. The maximum kinetic energy of the photoelectrons would be nearly (Planck's constant h = 6.63 × 10−34 Js)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.94 eV
By Einstein’s photoelectric equation, the maximum kinetic energy is the photon energy minus the work function.
\( \displaystyle K_{\max} = \frac{hc}{\lambda} - \phi_0 \)
\( \displaystyle \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\,\mathrm{J\,s} \times 3\times10^{8}\,\mathrm{m\,s^{-1}}}{280\times10^{-9}\,\mathrm{m}} = 7.10\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \frac{hc}{\lambda} = \frac{7.10\times10^{-19}\,\mathrm{J}}{1.6\times10^{-19}\,\mathrm{J\,eV^{-1}}} \approx 4.44\,\mathrm{eV} \)
\( \displaystyle K_{\max} = 4.44\,\mathrm{eV} - 2.5\,\mathrm{eV} = 1.94\,\mathrm{eV} \)
Shortcut: Photon energy in eV \( \approx 1240/\lambda \) (nm): \( 1240/280 \approx 4.43 \) eV, so \( K_{\max} \approx 1.93 \) eV, nearly 1.94 eV.