In an experiment on photoelectric effect, the stopping potential $V_{0}$ is plotted against the frequency $ν$ of the incident light. Considering $e$ as the electronic charge, the work function of the photoelectric surface is given by

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According to Einstein's photoelectric equation:

$h ν = Φ + e V_{0}$

Rearranging the equation for stopping potential $V_{0}$ :

$V_{0} = (\frac{h}{e}) ν - \frac{Φ}{e}$

This represents the equation of a straight line $y = m x + c$ , where the y-axis represents $V_{0}$ and the x-axis represents $ν$ .

Comparing the equations, the y-intercept is $c = - \frac{Φ}{e}$ .

From the given graph, the line intersects the negative y-axis at B, so the y-intercept is $- OB$ .

Equating the intercepts:

$- OB = - \frac{Φ}{e}$

$Φ = e \times (OB)$

Since $OB$ has the unit of potential (volts) and $e$ is the electronic charge (coulombs), their product gives the work function $Φ$ in joules.

Thus, the work function is given by the multiplication of $(OB)$ and $(e)$ in joules.

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