If the work function of the target material is 1.25 eV and the wavelength of the incident radiation is 2460 Å, the stopping potential required to stop the emitted electrons is:
(Given: h = \( 6.6 \times 10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.75 V
By Einstein's photoelectric equation the maximum kinetic energy of the photoelectrons is the photon energy minus the work function, and the stopping potential satisfies \( eV_0 = K_{\max} \). The speed of light \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \) and \( e = 1.6\times10^{-19}\,\mathrm{C} \) are taken as the standard values.
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34} \times 3\times10^{8}}{2460\times10^{-10}} = 8.05\times10^{-19}\,\mathrm{J} = 5.03\,\mathrm{eV} \)
\( \displaystyle eV_0 = E - \phi_0 = 5.03 - 1.25 = 3.78\,\mathrm{eV} \;\Rightarrow\; V_0 \approx 3.78\,\mathrm{V} \)
The nearest option is 3.75 V; the options differ by 1 V, so the small difference comes only from rounding the constants.