91Ó°ÊÓ

The photoelectric effect equation is: \(K_{max} = E - W\), where: - \(K_{max}\) is the maximum kinetic energy of the ejected electrons (\(eV\) or \(J\)) - \(E\) is the energy of the incident light (\(eV\) or \(J\)) - \(W\) is the work function of the material (\(eV\) or \(J\)) The energy of the incident light can be found using the Planck's equation: \(E = hf\), where: - \(h\) is the Planck's constant (\(6.626 \times 10^{-34}\,\mathrm{Js}\)) - \(f\) is the frequency of the light (\(Hz\)) Since we're given the wavelength \(\lambda\), we can find the frequency using the speed of light equation: \(f = \frac{c}{\lambda}\), where: - \(c\) is the speed of light (\(3.00 \times 10^8\,\mathrm{m/s}\)) - \(\lambda\) is the wavelength of the light (\(470\,\mathrm{nm}\))

First, convert the wavelength from nanometers to meters: \(\lambda = 470\,\mathrm{nm} = 470 \times 10^{-9}\,\mathrm{m}\) Now, calculate the frequency of the light: \(f = \frac{c}{\lambda} = \frac{3.00 \times 10^8\,\mathrm{m/s}}{470 \times 10^{-9}\,\mathrm{m}} = 6.38 \times 10^{14}\,\mathrm{Hz}\) Finally, calculate the energy of the incident light using Planck's constant: \(E = hf = (6.626 \times 10^{-34}\,\mathrm{Js})(6.38 \times 10^{14}\,\mathrm{Hz}) = 4.23 \times 10^{-19}\,\mathrm{J}\)

The work function of sodium is given as \(W = 2.28a\) eV, where \(a = 1\) eV. To calculate the work function of sodium in joules, we'll use the conversion factor \(1\,\mathrm{eV} = 1.602 \times 10^{-19}\,\mathrm{J}\). So, we have: \(W = (2.28\,\mathrm{eV})(1.602 \times 10^{-19}\,\mathrm{J/eV}) = 3.65 \times 10^{-19}\,\mathrm{J}\)

Now, we can use the photoelectric effect equation to find the maximum kinetic energy of the ejected electrons: \(K_{max} = E - W = (4.23 \times 10^{-19}\,\mathrm{J}) - (3.65 \times 10^{-19}\,\mathrm{J}) = 5.8 \times 10^{-20}\,\mathrm{J}\) Therefore, the maximum kinetic energy of the ejected electrons from the sodium surface is \(5.8 \times 10^{-20}\,\mathrm{J}\).