Photo-electrons are emitted from a metal surface when it is illuminated with UV light of wavelength 330 nm. The minimum amount of energy required to emit the electrons from the surface is 3.5 x 10-19 J. The kinetic energy of the photo electron is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.53 x 10-19 J
By Einstein’s photoelectric equation, the photon energy is used first to overcome the work function (the minimum energy needed to free an electron) and the rest appears as kinetic energy.
\( \displaystyle K_{\max} = \frac{hc}{\lambda} - \phi_0 \)
\( \displaystyle \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\,\mathrm{J\,s} \times 3\times10^{8}\,\mathrm{m\,s^{-1}}}{330\times10^{-9}\,\mathrm{m}} \)
\( \displaystyle \frac{hc}{\lambda} \approx 6.03\times10^{-19}\,\mathrm{J} \)
\( \displaystyle K_{\max} = 6.03\times10^{-19}\,\mathrm{J} - 3.5\times10^{-19}\,\mathrm{J} \)
\( \displaystyle K_{\max} \approx 2.53\times10^{-19}\,\mathrm{J} \)
Shortcut: With \( E(\mathrm{eV}) = 1240/\lambda(\mathrm{nm}) \): \( 1240/330 \approx 3.76 \) eV \( \approx 6.02\times10^{-19} \) J, and subtracting \( 3.5\times10^{-19} \) J leaves about \( 2.5\times10^{-19} \) J, which only Option (2) matches.