Photoelectric work function of a metal is 1 eV. Light of wavelength λ = 3000 Å falls on it. The photoelectrons come out with maximum kinetic energy of:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.13 eV
By Einstein’s photoelectric equation the maximum kinetic energy of a photoelectron is the photon energy minus the work function: \( K_{\max} = h\nu - \phi_0 = \frac{hc}{\lambda} - \phi_0 \).
With \( hc = 12400\,\mathrm{eV\,\mathring{A}} \):
\( \displaystyle E = \frac{hc}{\lambda} = \frac{12400\,\mathrm{eV\,\mathring{A}}}{3000\,\mathrm{\mathring{A}}} = 4.13\,\mathrm{eV} \)
\( K_{\max} = 4.13\,\mathrm{eV} - 1\,\mathrm{eV} = 3.13\,\mathrm{eV} \)