Question

Photons of wavelength \(350 \mathrm{nm}\) are incident on a metal plate in a photocell and electrons are ejected. A stopping potential of \(1.10 \mathrm{V}\) is able to just prevent any of the ejected electrons from reaching the opposite electrode. What is the maximum wavelength of photons that will eject electrons from this metal?

Short answer

Answer: The maximum wavelength of photons that will eject electrons from this metal is 506 nm.

Step-by-step solution

Calculate energy of incident photons

We need to determine the energy of the incident photons given their wavelength. We can calculate the energy using the Planck's formula: \(E = \dfrac{hc}{\lambda}\) Where E is the energy of the photons, h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), c is the speed of light (\(3 \times 10^8 \mathrm{m/s}\)), and λ is the wavelength. Here, \(\lambda = 350 \times 10^{-9} \mathrm{m}\). \(E = \dfrac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{350 \times 10^{-9} \mathrm{m}}\) \(E = 5.682 \times 10^{-19} \mathrm{J}\)

Calculate energy of ejected electrons

Next, we need to determine the energy of ejected electrons using the stopping potential V and electron charge e. \(E_{e} = eV\) Where e is the elementary charge (\(1.602 \times 10^{-19} \mathrm{C}\)), and V is the stopping potential (\(1.10 \mathrm{V}\)). \(E_{e} = 1.602 \times 10^{-19} \mathrm{C} \times 1.10 \mathrm{V}\) \(E_{e} = 1.762 \times 10^{-19} \mathrm{J}\)

Calculate work function

According to the photoelectric effect, the energy of incident photons is used to overcome the work function (Φ) of the metal and to provide the ejected electrons with kinetic energy. \(E = Φ + E_{e}\) Thus, we can calculate the work function: \(Φ = E - E_{e}\) \(Φ = 5.682 \times 10^{-19} \mathrm{J} - 1.762 \times 10^{-19} \mathrm{J}\) \(Φ = 3.92 \times 10^{-19} \mathrm{J}\)

Find maximum wavelength of photons

To find the maximum wavelength (λ_max) of photons that can eject electrons, we need to equate the energy of incident photons (calculate using Planck's formula) to the work function. \(\dfrac{hc}{\lambda_{max}} = Φ\) Rearrange the equation to solve for λ_max: \(\lambda_{max} = \dfrac{hc}{Φ}\) Substitute h, c, and Φ: \(\lambda_{max} = \dfrac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{3.92 \times 10^{-19} \mathrm{J}}\) \(\lambda_{max} = 5.06 \times 10^{-7} \mathrm{m}\) Thus, the maximum wavelength of photons that will eject electrons from this metal is \(506 \mathrm{nm}\).