How does wavelength affect the photoelectric effect?

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A08
Mar 12, 2016

Explanation:

Borrowing some material for this answer:

A single photon, whose wavelength is lower than the threshold wavelength for a specific metal, has the required energy to eject one electron thus creating the observed photoelectric effect.

Einstein used Planck's famous expression #E = hnu#, to explain other experimental results of photoelectric effect.

#E# being energy, #nu#, the frequency of radiated energy and #h# is Planck's Constant. We know that speed of light in vacuum #c# is related to its frequency #nu# and wavelength #lambda#

#c=nulambda#

Einstein put forward the following equation:

#(hc)/lambda_text{ incident}=phi+KE_max#

where #h# is Planck's constant, #lambda_text{ incident}# wavelength of incident light, #phi# work function of the metal and #KE_max# the maximum kinetic energy of the ejected electron(s).

chemistry-batz.wikispaces.com
For photoelectric effect to occur, the energy of the photon must be greater than the work function.
#phi=(hc)/lambda_text{ cut-off}#

As the wavelength of the incident light decreases but is lower than the cut-off wavelength, the maximum kinetic energy of the photo electrons increases.

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