Imagine that we have two parallel plates at the two ends of a vacuum tube. We can shine a light with sufficient energy ($hf\geq\phi$) and cause photoelectrons to travel from one plate onto the next.
This will result in a photoelectric current. We can then apply a potential difference across the two plates such that we oppose the photoelectric current produced.
Image Credits: Cyberphysics
Increasing the voltage to stop the current
As we increase this potential difference until the photoelectric current eventually stops. The stopping potential, $V_s$ is the voltage that we must apply in order to just stop the photoelectric current from flowing. This voltage when multiplied by the charge on an electron will give use the work done in bringing the photoelectrons to a stop: $$ \begin{equation}\begin{aligned} W&=QV\\ &=eV_s\\ \end{aligned}\end{equation} $$ where $e$ is the charge of an electron, $1.6\times 10^{-19}C$.
Because the electrons are still being ejected but fail to reach the other plate, we can say the work function is being exceeded by the kinetic energy is being reduced to zero by the work done by the stopping potential: $$ \begin{equation}\begin{aligned} E_k&=W\\ \frac12mv^2&=eV_s\\ \end{aligned}\end{equation} $$
Electron velocity from stopping potential
We can thus find the velocity of the electrons as they cross the vacuum if we know the mass of an electron and the stopping potential: $$ \begin{equation}\begin{aligned} \frac12mv^2&=eV_s\\ v^2&=\frac{2eV_s}{m}\\ v&=\sqrt{\frac{2eV_s}{m}}\\ \end{aligned}\end{equation} $$
The electron-volt as a unit of energy
We have already seen the relation between the work done (measured in Joules) and the stopping potential: $$ \begin{equation}\begin{aligned} W&=QV\\ &=eV_s\\ \end{aligned}\end{equation} $$ We can deduce that another unit for energy other than the joule is the electron-volt: $$ \begin{equation}\begin{aligned} W&=QV\\ &\rightarrow eV\\ \end{aligned}\end{equation} $$
The electron-volt is thus the work done on an electron to move it through a potential difference of 1 volt: $$ \begin{equation}\begin{aligned} eV&\rightarrow 1.6\times 10^{-19}C \times 1V\\ &=1.6\times 10^{-19}J\\ \end{aligned}\end{equation} $$
$eV$ to $J$ calculator
Conversion factor: 1.6e-19
We will use this as a unit to represent very small energy values such as binding energy.