Question

In a photoelectric experiment using sodium, when incident light of wavelength \(570 \mathrm{nm}\) and intensity \(1.0 \mathrm{W} / \mathrm{m}^{2}\) is used, the measured stopping potential is \(0.28 \mathrm{V} .\) (a) What would the stopping potential be for incident light of wavelength \(400.0 \mathrm{nm}\) and intensity \(1.0 \mathrm{W} / \mathrm{m}^{2} ?\) (b) What would the stopping potential be for incident light of wavelength 570 nm and intensity $2.0 \mathrm{W} / \mathrm{m}^{2} ?(\mathrm{c})$ What is the work function of sodium?

Short answer

Answer: The stopping potential for incident light of wavelength 400 nm and intensity 1.0 W/m^2 is 0.925 V, and the work function of sodium is 3.46 x 10^-19 J.

Step-by-step solution

a) Stopping potential for incident light of wavelength 400 nm and intensity 1.0 W/m^2

First, we will need to calculate the energy of the incident photons using Planck's equation: \(E = hf = \dfrac{hc}{\lambda}\) where \(E\) is the energy of the photon, \(h\) is Planck's constant \((6.63 \times 10^{-34} Js)\), \(f\) is the frequency of the light, \(c\) is the speed of light \((3 \times 10^8 m/s)\), and \(\lambda\) is the wavelength. For the given \(\lambda=400 \mathrm{nm}\), we can find the energy of the photon: \(E = \dfrac{6.63 \times 10^{-34} Js \times 3 \times 10^8 m/s}{400 \times 10^{-9}m} = 4.972 \times 10^{-19} J\) Next, we will use Einstein's photoelectric equation to find the stopping potential \(V\): \(E = eV + \phi\) where \(e\) is the elementary charge \((1.6 \times 10^{-19} C)\), \(\phi\) is the work function, and \(V\) is the stopping potential. We know the stopping potential for the case with \(\lambda=570 \mathrm{nm}\) is \(V_{570} = 0.28 V\), so we can rewrite the previous equation as: \(eV_{570} + \phi = E_{570}\) We can now substitute the values of \(E_{570}\) and \(V_{570}\) into this equation: \(1.6 \times 10^{-19} C \times 0.28 V = 4.972 \times 10^{-19} J - eV\) Solving for \(V\): \(V = \dfrac{4.972 \times 10^{-19} J - 0.28V \times 1.6 \times 10^{-19} C}{1.6 \times 10^{-19} C} = 0.925 V\) Therefore, the stopping potential for incident light of wavelength \(400.0\mathrm{nm}\) and intensity \(1.0 \mathrm{W} / \mathrm{m}^{2}\) is \(0.925 V\).

b) Stopping potential for incident light of wavelength 570 nm and intensity 2.0 W/m^2

The stopping potential depends only on the wavelength of incident light and not the intensity. The intensity of light affects the number of emitted electrons, but it does not affect their kinetic energy. Therefore, the stopping potential will be the same as the one given in the problem which is 0.28 V.

c) Work function of sodium

We already have a relation that determines the work function, which is: \(eV_{570} + \phi = E_{570}\) We can substitute the given values and solve for \(\phi\): \(\phi = E_{570} - eV_{570} = \dfrac{6.63 \times 10^{-34} Js \times 3 \times 10^8 m/s}{570 \times 10^{-9}m} - 1.6 \times 10^{-19} C \times 0.28 V = 3.46 \times 10^{-19} J\) Therefore, the work function of sodium is \(3.46 \times 10^{-19} J\).