Question

In his experiment on photoelectric effect, Robert A. Millikan found the slope of the cut-off voltage versus frequency of incident light plot to be \( 4.12 \times 10^{-15} \, \text{Vs} \). Calculate the value of Planck’s constant from it.

Hint:

The slope of the cut-off voltage versus frequency plot provides a direct way to calculate Planck's constant using the equation \( eV_{\text{cut}} = h f - \phi \).

Answer 1

The relationship between the cut-off voltage \( V_{\text{cut}} \) and the frequency of light is given by: \[ eV_{\text{cut}} = h f - \phi \] where:
\( e \) is the charge of the electron,
\( V_{\text{cut}} \) is the cut-off voltage,
\( h \) is Planck's constant,
\( f \) is the frequency of the incident light, and
\( \phi \) is the work function of the metal.
The slope of the plot of \( V_{\text{cut}} \) versus frequency \( f \) is given by: \[ \text{Slope} = \frac{h}{e} \] Given the slope is \( 4.12 \times 10^{-15} \, \text{Vs} \), we can calculate Planck's constant \( h \) using: \[ h = \text{Slope} \times e \] Substituting the value of \( e = 1.6 \times 10^{-19} \, \text{C} \): \[ h = (4.12 \times 10^{-15} \, \text{Vs}) \times (1.6 \times 10^{-19} \, \text{C}) = 6.592 \times 10^{-34} \, \text{J·s} \]