The work function for Aluminium surface is 4.2 eV. Find the threshold…

The correct answer is Option (1) → \( 2959\,\mathring{\mathrm{A}} \)

The threshold wavelength is the longest wavelength that can still eject photoelectrons. At this wavelength the photon energy just equals the work function \( \phi_0 \).

    \( \displaystyle \frac{hc}{\lambda_0} = \phi_0 \;\Rightarrow\; \lambda_0 = \frac{hc}{\phi_0} \)

    \( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
    \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
    \( \phi_0 = 4.2\,\mathrm{eV} = 4.2\times1.6\times10^{-19}\,\mathrm{J} = 6.72\times10^{-19}\,\mathrm{J} \)

    \( \displaystyle \lambda_0 = \frac{6.63\times10^{-34} \times 3\times10^{8}}{6.72\times10^{-19}} = \frac{1.989\times10^{-25}}{6.72\times10^{-19}} \approx 2.96\times10^{-7}\,\mathrm{m} \)

    \( \displaystyle \lambda_0 \approx 2.96\times10^{-7}\,\mathrm{m} \times \frac{10^{10}\,\mathring{\mathrm{A}}}{1\,\mathrm{m}} \approx 2959\,\mathring{\mathrm{A}} \)

Shortcut: With \( hc \approx 12400\,\mathrm{eV}\,\mathring{\mathrm{A}} \): \( \lambda_0 \approx 12400/4.2 \approx 2950\,\mathring{\mathrm{A}} \), which picks out Option (1) at once.