Question

When monochromatic light of an unknown wavelength falls on a sample of silver, a minimum potential of 2.50 V is required to stop all of the ejected photoelectrons. Determine the (a) maximum kinetic energy and (b) maximum speed of the ejected photoelectrons. (c) Determine the wavelength in nm of the incident light. (The work function for silver is 4.73 eV.)

Answer 1

here ,

a) for the maximum kinetic energy

the maximum kinetic energy = e * stopping potential

maximum kinetic energy = 1.602 * 10^-19 * 2.50

maximum kinetic energy = 4 *10^-19 J

b) let the maximum speed is u

0.50 * 9.11 *10^-31 u^2 = 4 *10^-19

solving for u

u = 9.4 *10^5 m/s

the maximum speed is 9.4 *10^5 m/s

c) for the light

h * c/wavelength = 4.73 * 1.6 *10^-19 + 2.5 * 1.6 *10^-19

6.626 *10^-19 * 3 *10^8/wavelength = 4.73 * 1.6 *10^-19 + 2.5 * 1.6 *10^-19

solving for wavelength

wavelength = 1.72 *10^-7 m = 172 nm

the wavelength of the light needed is 172 nm