Question

Explain the following observations using Einstein’s photoelectric equation:
(a) Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value.
(b) It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons.
(c) The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope \( \frac{h}{e} \).

Hint:

Einstein’s equation clearly shows the quantum nature of light: energy comes in discrete packets (photons), and only photons with energy higher than the work function can emit electrons.

Answer 1

Einstein's Photoelectric Equation: \[ K_{\text{max}} = h\nu - \phi \] where,
\( K_{\text{max}} \) = maximum kinetic energy of emitted photoelectrons
\( h \) = Planck’s constant
\( \nu \) = frequency of incident light
\( \phi \) = work function of the metal
(a) When \( \nu < \nu_0 \), where \( \nu_0 = \frac{\phi}{h} \) is the threshold frequency, the photon energy \( h\nu \) is insufficient to overcome the work function. Therefore, no photoelectrons are emitted regardless of light intensity.
(b) The kinetic energy \( K_{\text{max}} = h\nu - \phi \) depends only on frequency \( \nu \), not intensity. Increasing intensity increases the number of photons and thus number of photoelectrons, but not their kinetic energy.
(c) Rearranging the equation in terms of cut-off voltage: \[ eV_0 = h\nu - \phi \Rightarrow V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \] This is of the form \( y = mx + c \), showing a straight-line graph between \( V_0 \) and \( \nu \), with slope \( \frac{h}{e} \).