The work function for a metal is 4.14 eV. The threshold wavelength for this metal surface is: \( (h = 6.62\times10^{-34}\,\mathrm{J\,s},\ c = 3\times10^{8}\,\mathrm{m\,s^{-1}}) \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3000 Å
At the threshold wavelength the photon energy just equals the work function, \( \phi_0 = hc/\lambda_0 \), so \( \lambda_0 = hc/\phi_0 \).
\( \phi_0 = 4.14\,\mathrm{eV} = 4.14\times1.6\times10^{-19}\,\mathrm{J} = 6.62\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \lambda_0 = \frac{hc}{\phi_0} = \frac{6.62\times10^{-34}\times3\times10^{8}}{6.62\times10^{-19}} = 3\times10^{-7}\,\mathrm{m} = 3000\,\mathrm{\mathring{A}} \)
Shortcut: For a photon, \( \lambda\,(\mathrm{\mathring{A}}) \approx 12400/E\,(\mathrm{eV}) \). So \( \lambda_0 \approx 12400/4.14 \approx 3000\,\mathrm{\mathring{A}} \) without any powers of ten.