Converting eV to V: Debunking the Myth and Simplifying the Process

[robwashere]

I've done some research on this, but some people say that you cannot convert these two things. They say that V is electric potential, and eV is energy.

What I was thinking was since eV = Q<sub>electron <sup>* V, can't you just divide eV by charge of electron to get V?

Sorry if this is a dumb question</sup></sub>

 

[Thundagere]

Umm...how to explain this.
OK, so put it this way. I can convert between meter and foot because they're both distance. Thus, the conversion values are equivalent.
But what you're saying is to convert between two things that have, effectively, little in common. in a conversion, the values must be equivalent. This isn't the case with what you have here.

 

[bbbeard]

I've done some research on this, but some people say that you cannot convert these two things. They say that V is electric potential, and eV is energy.

What I was thinking was since eV = Q_{electron ^{* V, can't you just divide eV by charge of electron to get V?}}

_{^{Sort of. If you have an energy expressed in electron volts, and you know the charge, you get a voltage (a potential difference) when you divide the energy by the charge. Charges are not always equal to the elementary electron charge. For example, an alpha particle is like a helium nucleus, so it has a charge of +2e. So to accelerate an alpha particle to an energy of 1 MeV, you have to have a potential of 500,000 volts.}}

 

[mathman]

Neutral particles such as neutrons or gamma rays are often described using electron-volts to define their energy - there is no charge.

 

[Pengwuino]

I've done some research on this, but some people say that you cannot convert these two things. They say that V is electric potential, and eV is energy.

What I was thinking was since eV = Q<sub>electron <sup>* V, can't you just divide eV by charge of electron to get V?

Sorry if this is a dumb question</sup></sub>

<sub><sup>You're trying to setup an algebraic equation with a unit on one side and a physical relationship on the other, which is not possible.

It would be like saying that $meter = vt + {{1}\over{2}}at^2$, which is clearly nonsensical.</sup></sub>