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): Work Function Definition and Determination.
The work function (\( \phi \)) of a metal is the minimum energy, typically measured in electron volts (eV), required to liberate an electron from its surface.
To determine \( \phi \) graphically from stopping potential (\( V_s \)) versus frequency (\( u \)) data, utilize the photoelectric equation:\[e V_s = h u - \phi.\]Rearranging yields:\[V_s = \frac{h}{e} u - \frac{\phi}{e}.\]This linear relationship has a slope of \( \frac{h}{e} \) and a y-intercept of \( -\frac{\phi}{e} \). The threshold frequency (\( u_0 \)) is the frequency at which \( V_s = 0 \):\[0 = h u_0 - \phi \quad \Rightarrow \quad \phi = h u_0.\]Therefore, \( u_0 \) represents the x-intercept of the graph, and the work function is calculated as \( \phi = h u_0 \).(b): Wavelength Calculation.
Given: work function \( \phi = 2.4 \, \text{eV} \), stopping potential \( V_s = 0.6 \, \text{V} \). The photoelectric equation applies:\[e V_s = h u - \phi.\]In terms of energy:\[0.6 = h u - 2.4 \quad \Rightarrow \quad h u = 3.0 \, \text{eV}.\]Using \( h u = \frac{hc}{\lambda} \) with \( hc = 1240 \, \text{eV·nm} \):\[3.0 = \frac{1240}{\lambda} \quad \Rightarrow \quad \lambda = \frac{1240}{3.0} = 413.33 \, \text{nm}.\]Consequently, the wavelength of the incident light is approximately 413 nm.