A 600-nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of \(0.17 \mathrm{eV}\) are emitted. Determine (a) the work function and (b) the cutoff frequency of the surface. (c) What is the stopping potential when the surface is illuminated with light of wavelength \(400 \mathrm{nm} ?\)

Given information: - Wavelength of incident light, \( \lambda = 600 \ nm \). - Maximum kinetic energy of emitted electrons, \(K_{max} = 0.17 \ eV\). Relevant formulas for the photoelectric effect: - Energy of a photon, \(E = hf = \dfrac{hc}{\lambda}\), where \(h\) is Planck's constant, \(f\) is frequency, and \(c\) is the speed of light. - Photoelectric effect equation, \(K_{max} = hf - W\), where \(W\) is the work function.

For the calculations, it's better to use the wavelength in meters: \( \lambda = 600 \ nm = 600 \times 10^{-9} \ m\)

Using the formula for the energy of a photon, we can calculate the energy of the incident photon: \(E = \dfrac{hc}{\lambda} = \dfrac{(6.63 \times 10^{-34} \ Js)(3.0 \times 10^8 \ m/s)}{600 \times 10^{-9} \ m}\) \(E = 3.31 \times 10^{-19} \ J\) Now, convert this energy to electron volts (eV): \(E = \dfrac{3.31 \times 10^{-19} \ J}{1.6 \times 10^{-19} \ J/eV} = 2.07 \ eV\)

Using the photoelectric effect equation, we can calculate the work function of the surface: \(K_{max} = hf - W\) \(W = hf - K_{max} = E - K_{max} = 2.07 \ eV - 0.17 \ eV\) \(W = 1.9 \ eV\)

To find the cutoff frequency of the surface, use the equation relating the work function and the cutoff frequency: \(W = h f_c\) Solving for \(f_c\): \(f_c = \dfrac{W}{h}\) \(f_c = \dfrac{1.9 \ eV \times 1.6 \times 10^{-19} \ J/eV}{6.63 \times 10^{-34} \ Js}\) \(f_c = 4.57 \times 10^{14} \ Hz\)

Given the new wavelength: \( \lambda' = 400 \ nm = 400 \times 10^{-9} \ m\) First, calculate the energy of the photon corresponding to the new wavelength: \(E' = \dfrac{hc}{\lambda'} = \dfrac{(6.63 \times 10^{-34} \ Js)(3.0 \times 10^8 \ m/s)}{400 \times 10^{-9} \ m}\) \(E' = 4.97 \times 10^{-19} \ J\) Now, convert this energy to electron volts (eV): \(E' = \dfrac{4.97 \times 10^{-19} \ J}{1.6 \times 10^{-19} \ J/eV} = 3.11 \ eV\) We can calculate the new maximum kinetic energy: \(K'_{max} = E' - W = 3.11 \ eV - 1.9 \ eV\) \(K'_{max} = 1.21 \ eV\) At stopping potential, the kinetic energy of the emitted electrons is zero: \(eV_s = K'_{max}\) Solving for the stopping potential: \(V_s = \dfrac{K'_{max}}{e} = \dfrac{1.21 \ eV \times 1.6 \times 10^{-19} \ J/eV}{1.6 \times 10^{-19} \ C}\) \(V_s = 1.21 \ V\)

(a) The work function of the surface is \(W = 1.9 \ eV\). (b) The cutoff frequency of the surface is \(f_c = 4.57 \times 10^{14} \ Hz\). (c) The stopping potential when the surface is illuminated with light of wavelength 400 nm is \(V_s = 1.21 \ V\).