Question Detail
Question
The stopping potential in a photoelectric chamber is 3 V. The work function for the metal is 2.2 eV. Calculate the incident photon frequency. (Take Planck's constant h = 6.626 x 10-34 Js; electronic charge e = 1.602 x 10-19 C: 1 eV = 1.602 x 10-19 J).
Correct Answer
Option A is the correct answer.
Detailed Explanation
According to the photoelectric effect equation, the energy of the incident photon (Ephoton) is equal to the work function (φ) plus the maximum kinetic energy of the emitted electrons (KEmax): Ephoton = φ + KEmax.First, convert the work function and kinetic energy to Joules:Work function φ = 2.2 eV = 2.2 * 1.602 x 10-19 J = 3.5244 x 10-19 J.Maximum kinetic energy KEmax = eVs = 1.602 x 10-19 J/V * 3 V = 4.806 x 10-19 J.Now, calculate the photon energy:Ephoton = 3.5244 x 10-19 J + 4.806 x 10-19 J = 8.3304 x 10-19 J.Finally, the frequency (f) can be found using Ephoton = hf, where h is Planck's constant:f = Ephoton / h = 8.3304 x 10-19 J / 6.626 x 10-34 Js = 1.257 x 1015 Hz.Rounding to three significant figures, the incident photon frequency is approximately 1.26 x 1015 Hz.
Hint
Remember the photoelectric effect equation: Ephoton = φ + KEmax, and KEmax = eVs.