Question

(a) Define 'work function' of a metal. How can its value be determined from a graph between stopping potential and frequency of the incident radiation?
(b) The work function of a metal is 2.4 eV. A stopping potential of 0.6 V is required to reduce the photocurrent to zero, in a photoelectric experiment. Calculate the wavelength of light used.

Hint:

For photoelectric effect problems: - Use \( e V_s = h \nu - \phi \) to relate stopping potential and frequency. - For wavelength, use \( hc = 1240 \, \text{eV·nm} \) to convert energy to wavelength in nm.

Answer 1

(a): Define work function and its determination.
The work function (\( \phi \)) of a metal is the minimum energy required to remove an electron from its surface to infinity, typically measured in electron volts (eV).
To determine \( \phi \) from a graph of stopping potential (\( V_s \)) versus frequency (\( \nu \)), use the photoelectric equation: \[ e V_s = h \nu - \phi. \] Rearrange: \[ V_s = \frac{h}{e} \nu - \frac{\phi}{e}. \] This is a straight line with slope \( \frac{h}{e} \) and y-intercept \( -\frac{\phi}{e} \). The threshold frequency (\( \nu_0 \)) is where \( V_s = 0 \): \[ 0 = h \nu_0 - \phi \quad \Rightarrow \quad \phi = h \nu_0. \] Thus, \( \nu_0 \) is the x-intercept of the graph, and the work function is \( \phi = h \nu_0 \). (b): Calculate the wavelength.
Given: work function \( \phi = 2.4 \, \text{eV} \), stopping potential \( V_s = 0.6 \, \text{V} \). The photoelectric equation is: \[ e V_s = h \nu - \phi. \] In energy terms: \[ 0.6 = h \nu - 2.4 \quad \Rightarrow \quad h \nu = 3.0 \, \text{eV}. \] Use \( h \nu = \frac{hc}{\lambda} \), where \( hc = 1240 \, \text{eV·nm} \): \[ 3.0 = \frac{1240}{\lambda} \quad \Rightarrow \quad \lambda = \frac{1240}{3.0} = 413.33 \, \text{nm}. \] Thus, the wavelength of the light is approximately 413 nm.