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 stopping potential \( V_0 \), also known as the cut-off voltage, is the voltage required to arrest the motion of the most energetic photoelectrons. This potential is directly linked to the maximum kinetic energy of these photoelectrons. Einstein's photoelectric equation states: \[h u = \phi + \frac{1}{2} m v^2\] where: - \( h \) is Planck's constant. - \( u \) is the frequency of the incident light. - \( \phi \) is the work function of the material. - \( \frac{1}{2} m v^2 \) represents the maximum kinetic energy of the emitted electron. The maximum kinetic energy of a photoelectron, \( K_{\text{max}} \), is also expressed in terms of the stopping potential \( V_0 \) as: \[K_{\text{max}} = e V_0\] where: - \( e \) is the elementary charge. - \( V_0 \) is the stopping potential. Equating the two expressions for kinetic energy yields: \[e V_0 = h u - \phi\] Rearranging this equation gives: \[V_0 = \frac{h}{e} u - \frac{\phi}{e}\] This equation follows the form of a straight line: \[V_0 = \left( \frac{h}{e} \right) u - \frac{\phi}{e}\] In this linear relationship, the slope is \( \frac{h}{e} \), and the \( V_0 \)-intercept is \( -\frac{\phi}{e} \). Conclusion: A plot of the cut-off voltage \( V_0 \) against the frequency \( u \) of the incident light results in a straight line. The slope of this line is \( \frac{h}{e} \), which confirms the photoelectric effect's consistency with a linear relationship between cut-off voltage and frequency, where the slope is defined by Planck's constant divided by the elementary charge.