The threshold frequency of the metal of the cathode in a photoelectric cell is \( 1\times10^{15}\,\mathrm{Hz} \). When a certain beam of light is incident on the cathode, a stopping potential of 4.144 V is required to reduce the current to zero. The frequency of the incident radiation is (take \( h=6.63\times10^{-34}\,\mathrm{J\,s} \) and \( e=1.6\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\times10^{15}\,\mathrm{Hz} \)
The stopping potential measures the maximum kinetic energy of the photoelectrons, \( K_{\mathrm{max}}=eV_0 \). With the work function written as \( \phi_0=h\nu_0 \), Einstein’s photoelectric equation becomes \( eV_0=h\nu-h\nu_0 \). So the incident frequency is \( \nu=\nu_0+eV_0/h \).
The frequency added to the threshold value is:
\( \begin{aligned} \frac{eV_0}{h} &= \frac{1.6\times10^{-19}\,\mathrm{C}\times4.144\,\mathrm{V}}{6.63\times10^{-34}\,\mathrm{J\,s}} \\ &= \frac{6.63\times10^{-19}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{J\,s}} \\ &= 1\times10^{15}\,\mathrm{Hz} \end{aligned} \)
The frequency of the incident radiation is then:
\( \begin{aligned} \nu &= \nu_0+\frac{eV_0}{h} \\ &= 1\times10^{15}\,\mathrm{Hz}+1\times10^{15}\,\mathrm{Hz} \\ &= 2\times10^{15}\,\mathrm{Hz} \end{aligned} \)