A light of wavelength λ = 400 nm illuminates a photosensitive surface of work function 1.4 eV. What is the maximum kinetic energy of the emitted photoelectrons?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.7 eV
By Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons is the photon energy minus the work function. The photon energy in eV is quickly found from \( hc = 1240 \) eV nm.
\( \displaystyle E = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{400\,\mathrm{nm}} = 3.1\,\mathrm{eV} \)
\( K_{\mathrm{max}} = E - \phi_0 = 3.1\,\mathrm{eV} - 1.4\,\mathrm{eV} = 1.7\,\mathrm{eV} \)