Question

the maximum kinetic energy of photoelectrons ejected from a potassium surface by ultraviolet light of wavelength 200 NM is

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Answer 1

Answer:

Answer:

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

= hc / λ (h=planks constant,c=speed of light)

= 1242eV−nm / 188nm

=6.606eV.

Threshold λ=230nm Threshold energy=work function

(Φ)= 1242eV−nm / 230nm

     =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.