When a light of wavelength 150 nm is incident on silver surface a potential of 3.58 V is required to stop the photoelectrons from reaching the collector plate. Find the potential required to stop photoelectrons when a light of wavelength 200 nm is incident on it? (Work function of silver is 4.7 eV)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.51 V
By Einstein's photoelectric equation the photon energy goes partly into the work function and the rest becomes the maximum kinetic energy, which the stopping potential measures:
\( \displaystyle eV_0 = \frac{hc}{\lambda} - \phi_0 \)
For 150 nm the photon energy is therefore:
\( \displaystyle \frac{hc}{\lambda_1} = eV_1 + \phi_0 = 3.58\,\mathrm{eV} + 4.7\,\mathrm{eV} = 8.28\,\mathrm{eV} \)
The photon energy is inversely proportional to the wavelength, so for 200 nm:
\( \displaystyle \frac{hc}{\lambda_2} = 8.28\,\mathrm{eV}\times\frac{150}{200} = 6.21\,\mathrm{eV} \)
\( eV_2 = 6.21\,\mathrm{eV} - 4.7\,\mathrm{eV} = 1.51\,\mathrm{eV} \Rightarrow V_2 = 1.51\,\mathrm{V} \)
Shortcut: With \( hc \approx 1240\,\mathrm{eV\,nm} \), the photon energy at 200 nm is \( 1240/200 = 6.2\,\mathrm{eV} \), so \( V_2 \approx 6.2 - 4.7 = 1.5\,\mathrm{V} \), which points straight to Option (1).