In the photoelectric effect, it is found that incident photons with 4.70 eV of energy will produce electrons with a maximum kinetic energy of 3.90 eV. What is the threshold frequency of this material? Planck’s constant is 6.62607 × 10−34 J · s. Answer in units of Hz.
Thank you for the opportunity to help you with your question!
Let the frequency is n then we know that photon energy is sum of work and Kinetic energy of ejected electron. therefore we can write
4.7 = (h * n) + 3.9
to be consistent in units
h = 6.62607 x 10^(-34) J.s = 4.135667 x 10^(-15) eV-s
substituting values we get, 4.7 =3.9+ 4.135667 x 10^(-15) n
hence n = 0.8/ [4.135667 x 10^(-15)] = 1.9343 x 10^(14) Hz
Please find the solution enclosed here with. In case of any doubt please feel free to ask … If you need help in any assignment of math/ science … any online exam / discussion, Please contact for quick & quality services.