Identify the correct statements in the context of a photoelectric effect experiment with following parameters: work function of the metal = \( 1.8 \) eV and wavelength of incident radiation \( 400 \) nm.
A. Maximum kinetic energy of emitted electrons is \( \sim 1.3 \) eV
B. Stopping potential is \( \sim 1.3 \) V
C. The cut-off frequency is \( \sim 4.34\times10^{14} \) Hz
D. The cut-off frequency is \( \sim 8.6\times10^{15} \) Hz
Choose the correct answer from the following options.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, B and C only
Use Einstein’s photoelectric equation, \( K_{\max} = h\nu - \phi_0 \), with the photon energy found from \( E\,(\mathrm{eV}) = 1240/\lambda\,(\mathrm{nm}) \).
\( \displaystyle E = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{400\,\mathrm{nm}} = 3.1\,\mathrm{eV} \)
(A) Maximum kinetic energy of emitted electrons is \( \sim 1.3 \) eV — Correct.
\( K_{\max} = E - \phi_0 = 3.1 - 1.8 = 1.3\,\mathrm{eV} \)
(B) Stopping potential is \( \sim 1.3 \) V — Correct. The stopping potential satisfies \( eV_0 = K_{\max} \), so \( V_0 = 1.3 \) V.
(C) The cut-off frequency is \( \sim 4.34\times10^{14} \) Hz — Correct. At the threshold frequency the photon energy just equals the work function.
\( \displaystyle \nu_0 = \frac{\phi_0}{h} = \frac{1.8\times1.6\times10^{-19}}{6.63\times10^{-34}} \approx 4.34\times10^{14}\,\mathrm{Hz} \)
(D) The cut-off frequency is \( \sim 8.6\times10^{15} \) Hz — Incorrect. A photon of this frequency has \( h\nu \approx 35.6 \) eV, about twenty times the work function of 1.8 eV.