How does wavelength affect the photoelectric effect?

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).

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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.