1.

the maximum kinetic energy of photoelectron eject from a potassium surface by ultraviolet light of wavelength 200nm is photoelectric thereshold wavelength for potassium is 440nm​

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.



Discussion

No Comment Found

Related InterviewSolutions