Question

Draw a plot of frequency \( \nu \) of incident radiations as a function of stopping potential \( V_0 \) for a given photoemissive material. What information can be obtained from the value of the intercept on the stopping potential axis?

Hint:

The stopping potential intercept gives the work function of the material, and the slope provides the ratio \( \frac{e}{h} \).

Answer 1

The plot of frequency \( \nu \) vs stopping potential \( V_0 \) for a photoemissive material is a straight line. The equation governing the relationship is given by the photoelectric equation: \[ E_k = h\nu - \phi \] where \( E_k \) is the kinetic energy of the emitted photoelectron, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident radiation, and \( \phi \) is the work function of the material. The stopping potential \( V_0 \) is related to the kinetic energy of the emitted electron: \[ E_k = eV_0 \] where \( e \) is the charge of the electron. By equating the two expressions for \( E_k \), we get: \[ eV_0 = h\nu - \phi \] This equation represents a straight line of the form \( \nu = \frac{eV_0 + \phi}{h} \). Thus, the plot of \( \nu \) vs \( V_0 \) is a straight line with slope \( \frac{e}{h} \) and intercept \( \frac{\phi}{h} \), which represents the work function \( \phi \) of the material. The intercept on the \( V_0 \)-axis gives the value of the work function \( \phi \), and the slope is related to \( \frac{e}{h} \).