Ask Question
18 February, 20:25

When 7.9*1014 Hz light shines on a plate of unknown material, it is determined that the stopping potential is 1.1 V. Determine the work function (in eV) & the cutoff frequency. Wo = eV fo =

+4
Answers (1)
  1. 18 February, 20:50
    0
    W = 2.158 eV

    fo = 5.23 x 10^14 Hz

    Explanation:

    f = 7.9 x 10^14 Hz, Vo = 1.1 V

    Let W be the work function.

    Use the Einstein equation

    Energy = W + eVo

    hf = W + eVo

    where, h is the Plank's constant and e be the electronic charge.

    W = hf - eVo

    W = (6.6 x 10^-34 x 7.9 x 10^14) - (1.6 x 10^-19 x 1.1)

    W = 5.214 x 10^-19 - 1.76 x 10^-19 = 3.454 x 10^-19 J

    W = (3.454 x 10^-19) / (1.6 x 10^-19) = 2.158 eV

    Let fo be the cut off frequency

    W = h fo

    fo = W / h = (3.454 x 10^-19) / (6.6 x 0^-34) = 5.23 x 10^14 Hz
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “When 7.9*1014 Hz light shines on a plate of unknown material, it is determined that the stopping potential is 1.1 V. Determine the work ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers