Question

What is the maximum kinetic energy of the electrons ejected from a sodium surface by light of wavelength \(470 \mathrm{nm} ?\)

Short answer

The maximum kinetic energy of the ejected electrons is 0.36 eV.

Step-by-step solution

Write down the given information and constants

The given information and constants needed in this problem are: - Wavelength of light, \(\lambda = 470 \ \text{nm}\) - Work function of sodium, \(\phi = 2.28 \ \mathrm{eV}\) - Planck's constant, \(h = 6.626 \times 10^{-34} \ \mathrm{Js}\) - Speed of light, \(c = 3.00 \times 10^8 \ \mathrm{m/s}\) - Electronvolt, 1 eV \(= 1.602 \times 10^{-19} \ \mathrm{J}\)

Determine the energy of photons

First, convert the wavelength of the light to meters: \(\lambda = 470 \times 10^{-9} \ \mathrm{m}\). To find the energy of the photons with this wavelength, we use the Planck's equation: \[E = h \cdot f\] where \(E\) is the energy of the photons, \(h\) is Planck's constant, and \(f\) is the frequency of the light. To find the frequency, we use the relation between the speed of light \(c\), wavelength \(\lambda\), and frequency \(f\): \[c = \lambda \cdot f\] Rearrange the equation for \(f\): \[f = \frac{c}{\lambda}\] Now, plug in the values of \(c\) and \(\lambda\) to calculate the frequency \(f\): \[f = \frac{3.00 \times 10^8 \ \mathrm{m/s}}{470 \times 10^{-9} \ \mathrm{m}} = 6.383 \times 10^{14} \ \mathrm{Hz}\] Finally, we plug in the values of \(h\) and \(f\) to calculate the energy of the photons \(E\): \[E = (6.626 \times 10^{-34} \ \mathrm{Js}) \cdot (6.383 \times 10^{14} \ \mathrm{Hz}) = 4.231 \times 10^{-19} \ \mathrm{J}\]

Convert energy to electronvolts

To make the calculations easier, convert the photon energy in joules to electronvolts using the conversion factor: \[E = \frac{4.231 \times 10^{-19} \ \mathrm{J}}{1.602 \times 10^{-19} \ \mathrm{J/eV}} = 2.64 \ \mathrm{eV}\]

Calculate the maximum kinetic energy of the electrons

The maximum kinetic energy of the ejected electrons can be calculated using the photoelectric effect equation: \[K_{max} = E - \phi\] where \(K_{max}\) is the maximum kinetic energy of the electrons, \(E\) is the energy of the photons, and \(\phi\) is the work function of sodium. Plug in the values of \(E\) and \(\phi\) to calculate \(K_{max}\): \[K_{max} = 2.64 \ \mathrm{eV} - 2.28 \ \mathrm{eV} = 0.36 \ \mathrm{eV}\]

Present the final answer

The maximum kinetic energy of the electrons ejected from a sodium surface by light of wavelength 470 nm is \(0.36 \ \mathrm{eV}\).