Sodium has a work function of 2 eV. Calculate the maximum energy and speed of the emitted electrons, when sodium is illuminated by radiation of wavelength 150 nm:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10.1\times10^{-19}\,\mathrm{J} \) and \( 1.5\times10^{6}\,\mathrm{m/s} \)
Einstein’s photoelectric equation gives the maximum kinetic energy of the emitted electrons as \( K_{\max} = h\nu - \phi = \frac{hc}{\lambda} - \phi \), and the speed follows from \( K_{\max} = \frac{1}{2}mv_{\max}^{2} \).
Energy of the photon
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^{8}}{150\times10^{-9}}\,\mathrm{J} = 1.326\times10^{-18}\,\mathrm{J} = 13.3\times10^{-19}\,\mathrm{J} \)
Work function
\( \phi = 2\,\mathrm{eV} = 2\times1.6\times10^{-19}\,\mathrm{J} = 3.2\times10^{-19}\,\mathrm{J} \)
Maximum kinetic energy
\( K_{\max} = (13.3 - 3.2)\times10^{-19}\,\mathrm{J} = 10.1\times10^{-19}\,\mathrm{J} \)
Maximum speed (electron mass \( m = 9.11\times10^{-31}\,\mathrm{kg} \))
\( \displaystyle v_{\max} = \sqrt{\frac{2K_{\max}}{m}} = \sqrt{\frac{2\times10.1\times10^{-19}}{9.11\times10^{-31}}} = \sqrt{2.2\times10^{12}}\,\mathrm{m/s} \)
\( v_{\max} \approx 1.5\times10^{6}\,\mathrm{m/s} \)
Shortcut: Use \( E\,(\mathrm{eV}) = 1240/\lambda\,(\mathrm{nm}) \). Here \( E = 1240/150 = 8.27\,\mathrm{eV} \), so \( K_{\max} = 8.27 - 2 = 6.27\,\mathrm{eV} = 6.27\times1.6\times10^{-19}\,\mathrm{J} \approx 10\times10^{-19}\,\mathrm{J} \). Only Option (2) has an energy of this size, so the speed need not even be calculated.