Differing stopping voltage formula $V_0 = hf - \phi_{col}$
In a lecture, I was give the above formula for stopping voltage. It seemed a little simplified and weird to me as the LHS is in volts but the RHS is in electron volts. I did a bit of searching online, and figured that the proper formula for stopping voltage is: $eV_0 = h\nu - e\phi_{col}$. 
I think what my teacher did is to divide the RHS by $e$ to get the first formula. But in this case, it seems like $e = 1$ instead of the charge of an electron, $1.6*10^-19$. This is the bit I don't quite understand. I thought $U = qV$ and to get volts from electron-volts, I needed to divide by the electron charge. 
How is this so?
 A: If $V_0$ and $\phi_{col}$ are potential energies rather than electric potentials, then the first formula works. The symbol might be confusing, but the underlying meaning of the formula is the same - the potential energy of a freed electron is equal to the energy of the photon that freed it minus the "energy cost" of getting out of the metal (i.e. the work function). 
A: $\phi_{\rm col}$ is the work function energy and is the minimum amount of work which must be done to remove a free electron from a metal (surface).
It thus has the units of energy - joules but sometimes quoted in electron-volts or just volts with the energy then being $e\,\phi_{\rm col}$.  
So the right hand side of your equation gives the maximum energy of the emitted photoelectrons.  
The left hand side is the amount of electric potential energy those electrons could gain at the expense of their kinetic energy and so also has energy units.
The problem is that the same symbol $V_0$ is used in other text as the stopping potential and then the potential energy is $e\,V_0$ in such cases.  
Common variations of the equation are $eV_0 = hf - \phi_{col}$ where $V_0$ is in volts and $\phi_{\rm col}$ is in joules and $eV_0 = hf - e\phi_{col}$ where $V_0$ and $\phi_{\rm col}$ are both in volts.  
