Electrons ejected from the surface of a metal, when light of a certain frequency is incident on it, are stopped fully by a retarding potential of 3 V. The photoelectric effect in this metal begins at a frequency of \( 6\times10^{14}\,\mathrm{s^{-1}} \). The frequency of the incident light (in \( \mathrm{s^{-1}} \)) is (Planck’s constant \( h=6.4\times10^{-34}\,\mathrm{J\,s} \), charge on the electron \( e=1.6\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 13.5\times10^{14} \)
By Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons is \( K_{\mathrm{max}}=h\nu-\phi_0 \). It equals \( eV_0 \), where \( V_0 \) is the stopping potential. The work function is \( \phi_0=h\nu_0 \), where \( \nu_0 \) is the threshold frequency at which emission begins.
\( eV_0=h(\nu-\nu_0) \)
\( \begin{aligned} \nu-\nu_0 &= \frac{eV_0}{h} \\ &= \frac{1.6\times10^{-19}\,\mathrm{C}\times3\,\mathrm{V}}{6.4\times10^{-34}\,\mathrm{J\,s}} \\ &= 7.5\times10^{14}\,\mathrm{s^{-1}} \end{aligned} \)
\( \begin{aligned} \nu &= \nu_0+7.5\times10^{14}\,\mathrm{s^{-1}} \\ &= 6\times10^{14}\,\mathrm{s^{-1}}+7.5\times10^{14}\,\mathrm{s^{-1}} \\ &= 13.5\times10^{14}\,\mathrm{s^{-1}} \end{aligned} \)