When light with a wavelength of 209 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.57 × 10^-19 j. determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

given,

light wavelength = 209 nm

maximum kinetic energy = 3.57 × 10⁻¹⁹ J

using photo electric equation

           =

         =5.94× 10⁻¹⁹ J

wavelength of light that should be used for double maximum K.E

              = 151.94 × 10⁻⁹ m = 151.94 m

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