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: \[ K_{\text{max}} = hu - \phi \] where, \( K_{\text{max}} \) = maximum kinetic energy of emitted photoelectrons \( h \) = Planck’s constant \( u \) = frequency of incident light \( \phi \) = work function of the metal (a) If \( u<u_0 \), where \( u_0 = \frac{\phi}{h} \) is the threshold frequency, the photon energy \( hu \) is insufficient to overcome the work function. Consequently, no photoelectrons are emitted, irrespective of light intensity. (b) The kinetic energy, \( K_{\text{max}} = hu - \phi \), is dependent solely on frequency \( u \), not intensity. An increase in intensity raises the photon count and, thus, the photoelectron count, but not their kinetic energy. (c) The equation, expressed in terms of cut-off voltage, is: \[ e V_0 = hu - \phi \Rightarrow V_0 = \frac{h}{e} u - \frac{\phi}{e} \] This equation aligns with the form \( y = mx + c \), indicating a linear relationship between \( V_0 \) and \( u \), with a slope of \( \frac{h}{e} \).