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Light of wavelength 5000 Å falls on a metal surface of work function 1.9 eV. Find (a) the energy of photos (b) kinetic energy of photoelectrons?

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Wavelength of light (λ)= 5000 Å=5000×10−10=5×10−7m
Work function (ℎv)=1.9eV =1.9×1.6×10−19J=3.04×10−19 J

Energy of photons 

E=ℎv=ℎc/λ=6.626×10−34×3×1085×10−7=3.97×10−19 J
Kinetic energy of photoelectrons =ℎv−ℎv=3.97×10−19−3.04×10−19 =9.3×10−20J

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