# 2.3: Electron-volts

The *electron-volt* is a unit of *energy* or *work*. An electron-volt (eV) is the work required to move an electron through a potential difference of one volt. Alternatively, an electronvolt is equal to the kinetic energy acquired by an electron when it is accelerated through a potential difference of one volt. Since the magnitude of the charge of an electron is about \(1.602 × 10^{−19}\) C, it follows that an electron-volt is about \(1.602 × 10^{−19}\) J. Note also that, because the charge on an electron is negative, it requires work to move an electron from a point of high potential to a point of low potential.

*Exercise.* If an electron is accelerated through a potential difference of a million volts, its kinetic energy is, of course, 1 MeV. At what speed is it then moving?

First attempt.

\[\frac{1}{2}mv^2=eV\]

(Here \(eV\), written in italics, is not intended to mean the unit electron-volt, but *e* is the magnitude of the electron charge, and \(V\) is the potential difference (\(10^6\) volts) through which it is accelerated.) Thus \(v = \sqrt{2eV / m}\) . With \(m = 9.109 × 10^{−31}\) kg, this comes to \(v = 5.9 × 10^8 \text{m s}^{ −1}\) . Oops! That looks awfully fast! We’d better do it properly this time.

Second attempt.

\[(\gamma -1)mc^2=eV.\]

Some readers will know exactly what we are doing here, without explanation. Others may be completely mystified. For the latter, the difficulty is that the speed that we had calculated was even greater than the speed of light. To do this properly we have to use the formulas of special relativity. See, for example, Chapter 15 of the Classical Mechanics section of these notes.

At any rate, this results in \(\gamma = 2.958\), whence \(β = 0.9411 \text{ and }v = 2.82 × 10^8 \text{m s}^{ −1}\) .