Question

4. The wavelength of the photoelectric threshold for silver is 3,250 x10-10 m. Determine the velocity of electron ejected from a silver surface by ultraviolet light of wavelength 2,536 x1010 m 5. What is the maximum wavelength (in nm) of light required to produce any current via the photoelectric effect if the anode is made of copper, which has a work function of 4.7 eV?

Answer 1

If E is the energy of the incident photon of wavelength = lambda = 2536e - 10 m for the threshold wavelength of light = lambda^' implies electron kinetic energy = K = 0 and phi = hc/lambda^' and threshold wavelength = lambda^' = 3250.0e - 10 m and work function for silver = phi = hc/lambda^' = 3.8149 eV energy of incident ultraviolet photon = E = hc/lambda = 4.8890 eV E = hc/lambda = 4.8890 eV = K + phi phi = is the work function of silver: K = final kinetic energy of electron h = 6.626 times 10^-34 Js: c = 3 times 10^8 m/s K = mv^2/2 = hc/lambda - phi = 1.0741 eV electron mass = m = 9.1e - 31 kg v = squareroot 2K/m = 6.1467e + 05 m/s \r\n if E is the energy of the photon incident of wavelength lambda is E = hc/lambda = K