Question

In a photoelectric experiment the incident photons have frequency \( \frac{3}{2} \nu \), where \( \nu \) is the threshold frequency of the material. What is the kinetic energy of the emitted electrons?

• A. \( \frac{h \nu}{2} \)
• B. \( h \nu \)
• C. \( \frac{3h \nu}{2} \)
• D. \( 2 h \nu \)

Hint:

In photoelectric experiments, the kinetic energy of the emitted electrons is the difference between the energy of the incident photons and the work function of the material. Be sure to use the equation \( E_{\text{kinetic}} = h f - h \nu \).

Answer 1

The Correct Option is C

In a photoelectric effect experiment, the energy of the incident photons is given by \( E_{\text{photon}} = h f \), where: - \( h \) is Planck's constant, - \( f \) is the frequency of the incident photons. For this question, we are told that the frequency of the incident photons is \( \frac{3}{2} \nu \), where \( \nu \) is the threshold frequency of the material. Therefore, the energy of the incident photons is: \[ E_{\text{incident}} = h \times \frac{3}{2} \nu = \frac{3 h \nu}{2} \] According to the photoelectric equation: \[ E_{\text{kinetic}} = E_{\text{incident}} - E_{\text{work}} \] where: - \( E_{\text{kinetic}} \) is the kinetic energy of the emitted electrons, - \( E_{\text{work}} \) is the work function (the minimum energy required to eject an electron) and is given by \( E_{\text{work}} = h \nu \). Substituting into the equation: \[ E_{\text{kinetic}} = \frac{3 h \nu}{2} - h \nu = \frac{3 h \nu}{2} - \frac{2 h \nu}{2} = \frac{h \nu}{2} \] Thus, the kinetic energy of the emitted electrons is \( \frac{h \nu}{2} \). So the correct answer is (C).