91Ó°ÊÓ

The photoelectric effect equation is given by: $$K_{max} = hf - \Phi$$ Where \(K_{max}\) is the maximum kinetic energy of ejected electrons, \(h\) is the Planck's constant, \(f\) is the frequency of the incident radiation, and \(\Phi\) is the work function of the material. The energy (\(E\)) of a photon is related to its wavelength (\(\lambda\)) and frequency (\(f\)) by the following equations: $$E = hf$$ $$c = \lambda f$$ Where \(c\) is the speed of light. Now we'll solve each part using these equations and the given values.

First, we need to calculate the energy of the incident radiation using the given wavelength (\(\lambda = 413 \mathrm{nm}\)). We can find the frequency (\(f\)) using the second equation: $$f = \frac{c}{\lambda}$$ Substitute given values (converting wavelength to meters): $$f = \frac{3.0 \times 10^8 \, m/s}{(413 \times 10^{-9}) \, m}$$ Then calculate the frequency \(f\): $$f \approx 7.266 \times 10^{14} \, Hz$$ Now, calculate the energy (\(E\)) of the incident radiation using the first equation: $$E = h \times f$$ Substitute given values (Pay attention to the units): $$E = (6.626 \times 10^{-34} \, Js) \times (7.266 \times 10^{14} \, Hz)$$ Then calculate the energy \(E\): $$E \approx 4.815 \times 10^{-19} \, J$$

Now, using the photoelectric effect equation, we can calculate the maximum kinetic energy of ejected electrons: $$K_{max} = E - \Phi$$ Convert the work function (\(\Phi\)) from electron volts to joules: $$\Phi = 2.16 \, eV \times \frac{1.6 \times 10^{-19} \, J}{1 \, eV}$$ $$\Phi \approx 3.456 \times 10^{-19} \, J$$ Substitute values for \(E\) and \(\Phi\) into the equation: $$K_{max} = (4.815 \times 10^{-19} \, J) - (3.456 \times 10^{-19} \, J)$$ Calculate \(K_{max}\): $$K_{max} \approx 1.359 \times 10^{-19} \, J$$ Convert the maximum kinetic energy back to electron volts: $$K_{max} \approx \frac{1.359 \times 10^{-19} \, J}{1.6 \times 10^{-19} \, J/eV}$$ $$K_{max} \approx 0.849 \, eV$$ The maximum kinetic energy of ejected electrons is about \(0.849 \, eV\).

To find the threshold wavelength, we can use the energy equation and the photoelectric effect equation. Set the maximum kinetic energy (\(K_{max}\)) to zero: $$0 = hf - \Phi$$ Solve for the frequency (\(f\)): $$f = \frac{\Phi}{h}$$ Substitute values for work function (\(\Phi\)) and the Planck's constant (\(h\)): $$f = \frac{3.456 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, Js}$$ Calculate \(f\): $$f \approx 5.217 \times 10^{14} \, Hz$$ Now, use the wavelength-frequency relation to find the threshold wavelength (\(\lambda\)): $$\lambda = \frac{c}{f}$$ Substitute values for the speed of light (\(c\)) and frequency (\(f\)): $$\lambda = \frac{3.0 \times 10^8 \, m/s}{5.217 \times 10^{14} \, Hz}$$ Then calculate \(\lambda\): $$\lambda \approx 5.754 \times 10^{-7} \, m$$ Convert the threshold wavelength to nanometers: $$\lambda \approx 575.4 \, nm$$ The threshold wavelength for this rubidium surface is approximately \(575.4 \, nm\).