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In an experiment on the photoelectric effect, the calcium cathode of a photocell (work function: 3.20 eV) is hit by a narrow beam of light. The light hitting the calcium plate now has a wavelength of 355.1 nm. What is the momentum of a triggered photoelectron?

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Answer:

To find the momentum of a photoelectron ejected from a calcium cathode when hit by light of a given wavelength, we'll follow these steps:Determine the energy of the incident photons using the wavelength.Calculate the kinetic energy of the ejected photoelectrons using the photoelectric effect equation.Compute the momentum from the kinetic energy.Step 1: Calculate the Energy of the Incident PhotonsThe energy ( E ) of a photon can be found using the equation:[ E = \frac{hc}{\lambda} ]Where:( h ) is Planck's constant (( 6.626 \times 10^{-34} ) J·s)( c ) is the speed of light (( 3.00 \times 10^8 ) m/s)( \lambda ) is the wavelength of the light (355.1 nm or ( 355.1 \times 10^{-9} ) m)[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{355.1 \times 10^{-9}} ][ E \approx 5.60 \times 10^{-19} \text{ J} ]To convert this energy into electron volts (eV), knowing that ( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} ):[ E = \frac{5.60 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} ][ E \approx 3.50 \text{ eV} ]Step 2: Calculate the Kinetic Energy of the Ejected PhotoelectronsThe kinetic energy ( KE ) of the photoelectrons is given by the photoelectric effect equation:[ KE = E_{\text{photon}} - \phi ]Where:( E_{\text{photon}} ) is the energy of the incident photon (3.50 eV)( \phi ) is the work function of calcium (3.20 eV)[ KE = 3.50 \text{ eV} - 3.20 \text{ eV} ][ KE = 0.30 \text{ eV} ]Step 3: Compute the Momentum of the Ejected PhotoelectronsThe kinetic energy ( KE ) can also be expressed as:[ KE = \frac{p^2}{2m} ]Where:( p ) is the momentum of the photoelectron( m ) is the mass of the electron (( 9.11 \times 10^{-31} ) kg)First, convert ( KE ) into joules:[ KE = 0.30 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} ][ KE \approx 4.81 \times 10^{-20} \text{ J} ]Now solve for momentum ( p ):[ p = \sqrt{2mKE} ][ p = \sqrt{2 \times 9.11 \times 10^{-31} \text{ kg} \times 4.81 \times 10^{-20} \text{ J}} ][ p \approx \sqrt{8.75 \times 10^{-50} \text{ kg·J}} ][ p \approx 2.96 \times 10^{-25} \text{ kg·m/s} ]So, the momentum of a triggered photoelectron is approximately ( 2.96 \times 10^{-25} ) kg·m/s.

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