The threshold wavelength of silver is \( 3250\times10^{-10}\,\mathrm{m} \). The velocity of the electron ejected from the silver surface by ultra violet light of wavelength \( 2536\times10^{-10}\,\mathrm{m} \) is approximately:
(Given \( h = 4.14\times10^{-15}\,\mathrm{eV\,s} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 6\times10^{5}\,\mathrm{m\,s^{-1}} \)
By Einstein’s photoelectric equation the maximum kinetic energy of the ejected electron is the photon energy minus the work function, \( K = h\nu - \phi_0 \), where \( \phi_0 = hc/\lambda_0 \) is fixed by the threshold wavelength.
With \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \):
\( hc = 4.14\times10^{-15}\,\mathrm{eV\,s}\times3\times10^{8}\,\mathrm{m\,s^{-1}} = 1.242\times10^{-6}\,\mathrm{eV\,m} \)
\( \displaystyle E_{\mathrm{photon}} = \frac{hc}{\lambda} = \frac{1.242\times10^{-6}}{2.536\times10^{-7}}\,\mathrm{eV} = 4.90\,\mathrm{eV} \)
\( \displaystyle \phi_0 = \frac{hc}{\lambda_0} = \frac{1.242\times10^{-6}}{3.25\times10^{-7}}\,\mathrm{eV} = 3.82\,\mathrm{eV} \)
\( K = 4.90 - 3.82 = 1.08\,\mathrm{eV} = 1.08\times1.6\times10^{-19}\,\mathrm{J} = 1.7\times10^{-19}\,\mathrm{J} \)
Putting \( K = \frac{1}{2}mv^{2} \) with \( m = 9.1\times10^{-31}\,\mathrm{kg} \):
\( \displaystyle v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2\times1.7\times10^{-19}}{9.1\times10^{-31}}}\,\mathrm{m\,s^{-1}} = \sqrt{3.8\times10^{11}}\,\mathrm{m\,s^{-1}} \approx 6\times10^{5}\,\mathrm{m\,s^{-1}} \)
Shortcut: An electron of kinetic energy 1 eV moves at about \( 5.9\times10^{5}\,\mathrm{m\,s^{-1}} \), and speed grows as \( \sqrt{K} \). With \( K \approx 1.08\,\mathrm{eV} \), \( v \approx 5.9\times10^{5}\times\sqrt{1.08} \approx 6\times10^{5}\,\mathrm{m\,s^{-1}} \).