Question

The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope of \( \frac{h}{e} \). Explain this observation.

Hint:

The cut-off voltage \( V_0 \) is linearly related to the frequency \( \nu \) of the incident light. The slope of the line is \( \frac{h}{e} \), where \( h \) is Planck's constant and \( e \) is the charge of the electron.

Answer 1

The photoelectric effect can be analyzed by considering the stopping potential (cut-off voltage) \( V_0 \) required to stop the most energetic photoelectrons. The stopping potential is related to the maximum kinetic energy of the photoelectrons. From Einstein's photoelectric equation: \[ h \nu = \phi + \frac{1}{2} m v^2 \] where: - \( h \) is Planck's constant, - \( \nu \) is the frequency of the incident light, - \( \phi \) is the work function of the material, - \( \frac{1}{2} m v^2 \) is the maximum kinetic energy of the emitted electron. The kinetic energy of the photoelectron is also related to the stopping potential \( V_0 \) by the equation: \[ K_{\text{max}} = e V_0 \] where: - \( e \) is the charge of the electron, - \( V_0 \) is the stopping potential. Now, equating the two expressions for kinetic energy, we get: \[ e V_0 = h \nu - \phi \] which can be rewritten as: \[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \] This equation is of the form of a straight line: \[ V_0 = \left( \frac{h}{e} \right) \nu - \frac{\phi}{e} \] where the slope of the line is \( \frac{h}{e} \), and the intercept on the \( V_0 \)-axis is \( -\frac{\phi}{e} \). Conclusion: The plot of the cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light is a straight line with a slope of \( \frac{h}{e} \). This confirms that the photoelectric effect is consistent with the linear relationship between the cut-off voltage and frequency, with the slope directly related to Planck's constant divided by the charge of the electron.