Question

The threshold wavelength of a photosensitive metal is 662.5nm if this metal is irrediated with a radiation of wavelength 331.3nm find the maximum kinetic energy of the photoelectrons if the wave length of the radiation is increased to 496.5nm calculate the change in maximum kinetic energy of the photoelectrons

Answer 1

Explanation:

wavelength(λ) of UV light =188nm Energy of incident rays

=λhc (h=planks constant,c=speed of light)

=188nm1242eV−nm

=6.606eV.

Threshold λ=230nmThreshold energy=work function

(Φ)=230nm1242eV−nm=5.4eV. 

By Einstein photo-electric effect, Energy Incident =Φ+MaximumK.E. Thus the maximum kinetic energy of emitted electrons would be

=6.606−5.4=1.206eV.