posted by Laila .
Photons of energy 12eV are incident on a metal. It is found that current flows from the metal until a stopping potential of 8.0V is applied.
If the wavelength of the incident photons is doubled, what is the maximum kinetic energy of the ejected electrons?
What is KEmax in eV.
Please someone help me on this...thank you.
With 12 eV photons, photoelectrons are ejected with energies of up to 8 eV. The metal work function must be 4 volts.
Doubling the wavelength of the incident light decreases the photon energy to 6 eV. The work function remains 4 V. The maximum photoelectron energy is then 2 eV.