Find the maximum velocity of photoelectrons ejected by an 80-nm radiation, if the work function of photoelectrode is \(4.73 \mathrm{eV}\).

We can find the energy of the incident photons using the formula: \[ E = h \cdot f \] Where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.63 \times 10^{-34} \mathrm{Js}\)), and \(f\) is the frequency of the radiation. We can find the frequency using the wavelength (\(\lambda\)) and the speed of light (\(c\)): \[ f = \frac{c}{\lambda} \] Now we can plug in the given wavelength \(\lambda = 80 \times 10^{-9} \mathrm{m}\) and the speed of light \(c = 3 \times 10^{8} \mathrm{m/s}\): \[ f = \frac{3 \times 10^{8}\ \mathrm{m/s}}{80 \times 10^{-9}\ \mathrm{m}} = 3.75 \times 10^{15}\ \mathrm{Hz} \] Now plug the frequency into the energy equation using Planck's constant: \[ E = 6.63 \times 10^{-34} \mathrm{Js}(3.75 \times 10^{15}\ \mathrm{Hz}) = 2.49 \times 10^{-18} \mathrm{J} \]

To compare the energy of the incident photons with the work function of the photoelectrode, we need to convert the energy to electron volts (eV). The relation between Joules and electron volts is: \[ 1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J} \] So, convert the photon energy to electron volts: \[ E = \frac{2.49 \times 10^{-18}\ \mathrm{J}}{1.6 \times 10^{-19}\ \mathrm{J/eV}} = 15.56\ \mathrm{eV} \]

Using the work function (\(\phi = 4.73\ \mathrm{eV}\)), we can calculate the kinetic energy \(KE\) of the ejected photoelectrons: \[ KE = E - \phi \] \[ KE = 15.56\ \mathrm{eV} - 4.73\ \mathrm{eV} = 10.83\ \mathrm{eV} \]

Now, we can use the kinetic energy to find the maximum velocity of the photoelectrons. We first convert the kinetic energy back to Joules: \[ KE = 10.83\ \mathrm{eV} \cdot 1.6 \times 10^{-19}\ \mathrm{J/eV} = 1.73 \times 10^{-18}\ \mathrm{J} \] Use the kinetic energy equation to find the velocity: \[ KE = \frac{1}{2}mv^2 \] Where \(m\) is the mass of an electron (\(9.11 \times 10^{-31}\ \mathrm{kg}\)) and \(v\) is the velocity of the ejected photoelectron. Rearrange the equation and solve for \(v\): \[ v = \sqrt{\frac{2 \times KE}{m}} \] \[ v = \sqrt{\frac{2 \times 1.73 \times 10^{-18}\ \mathrm{J}}{9.11 \times 10^{-31}\ \mathrm{kg}}} = 7.96 \times 10^{5}\ \mathrm{m/s} \] So, the maximum velocity of the photoelectrons ejected by the 80-nm radiation is about \(7.96 \times 10^{5}\ \mathrm{m/s}\).