Ratio of Stopping Potential Versus Frequency Slopes for Silver and Sodium

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by: $K_{\text{max}} = h\nu - \phi$

where:

• $h$ is Planck's constant,
• $\nu$ is the frequency of the incident radiation,
• $\phi$ is the work function of the metal surface.

If $V_0$ is the stopping potential, then the maximum kinetic energy can also be expressed as: $K_{\text{max}} = e V_0$

where $e$ is the magnitude of the charge of an electron.

Equating the two expressions for $K_{\text{max}}$, we get: $e V_0 = h\nu - \phi$

Rearranging this equation to express the stopping potential $V_0$ as a function of frequency $\nu$: $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e}$

Comparing this with the equation of a straight line, $y = mx + c$, where $y = V_0$ and $x = \nu$:

• The slope of the plot of stopping potential ($V_0$) versus frequency ($\nu$) is: $m = \frac{h}{e}$
• The y-intercept is $c = -\frac{\phi}{e}$.

Since $h$ and $e$ are universal physical constants, the slope $m = \frac{h}{e}$ is independent of the work function $\phi$ of the metal and is identical for all photo-emissive materials.

Thus: $\text{Slope for Silver } (m_{\text{Ag}}) = \frac{h}{e}$ $\text{Slope for Sodium } (m_{\text{Na}}) = \frac{h}{e}$

The ratio of the slope for Silver to that for Sodium is: $\text{Ratio} = \frac{m_{\text{Ag}}}{m_{\text{Na}}} = \frac{h/e}{h/e} = 1$