The work function for a metal surface is 8.28 eV. The threshold wavelength for this metal surface is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.5 \times 10^{-7}\,\mathrm{m} \)
The threshold wavelength is the longest wavelength that can eject electrons: the energy of one photon then just equals the work function.
\( \displaystyle \frac{hc}{\lambda_0} = \phi_0 \;\Rightarrow\; \lambda_0 = \frac{hc}{\phi_0} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
\( \phi_0 = 8.28\,\mathrm{eV} = 8.28 \times 1.6\times10^{-19}\,\mathrm{J} = 1.325\times10^{-18}\,\mathrm{J} \)
\( \displaystyle \lambda_0 = \frac{6.63\times10^{-34} \times 3\times10^{8}}{1.325\times10^{-18}} = 1.5\times10^{-7}\,\mathrm{m} \)
Shortcut: With \( hc = 1240\,\mathrm{eV\,nm} \): \( \lambda_0 = 1240/8.28 \approx 150\,\mathrm{nm} = 1.5\times10^{-7}\,\mathrm{m} \).