The variation of the stopping potential ($V_0$) with the frequency of incident radiation ($\nu$) is plotted. If $\nu_0$ is the threshold frequency, $h$ is Planck's constant, and $e$ is the electronic charge, then the slope of the graph with the frequency axis is:
Show Hint
The slope of a stopping potential versus frequency graph is a universal constant ($\frac{h}{e}$). It remains exactly the same no matter what type of metal surface is chosen for the photoelectric experiment!
Understanding the Concept:
Einstein's Photoelectric Equation states that the maximum kinetic energy of an emitted photoelectron is equal to the total energy supplied by the incident photon minus the work function ($\phi_0$) of the metal surface:
\[
K_{\text{max}} = h\nu - \phi_0
\]
Since $K_{\text{max}} = eV_0$, where $V_0$ is the stopping potential, the equation can be written as a straight-line function.
Step 1: Rearrange the equation into the standard linear form ($y = mx + c$).
Dividing both sides of the relation by the electronic charge $e$:
\[
eV_0 = h\nu - h\nu_0 \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{h\nu_0}{e}
\]
Comparing this line function to $y = mx + c$, where stopping potential $V_0$ is on the vertical axis ($y$) and frequency $\nu$ is plotted on the horizontal axis ($x$):
Slope ($m$) = $\frac{h}{e}$
Intercept ($c$) = $-\frac{h\nu_0}{e}$
