37.6.2 The Electron Volt

The joule is a unit of appropriate size in mechanics and thermodynamics, where we dealt with macroscopic objects, but it is poorly matched to the needs of atomic physics. It will be very useful to have an energy unit appropriate to atomic and nuclear events.

FIGURE 37.19 shows an electron accelerating (in a vacuum) from rest across a parallel-plate capacitor with a 1.0 V potential difference. What is the electron’s kinetic energy when it reaches the positive plate? We know from energy conservation that $K_{f} + q V_{f} = K_{i} + q V_{i}$ , where $U = q V$ is the electric potential energy. $K_{i} = 0$ because the electron starts from rest, and the electron’s charge is $q = - e$ . Thus

\begin{matrix} K_{f} & = & - q (V_{f} - V_{i}) = - q Δ V = e Δ V = (1.60 \times 10^{- 19} C) (1.0 V) \\ = & 1.60 \times 10^{- 19} J \end{matrix}

The figure illustrates an electron accelerating across a 1 volt potential difference gains 1 electron volt of kinetic energy.

FIGURE 37.19 Full Alternative Text

Let us define a new unit of energy, called the electron volt, as

1 electron volt = 1 eV \equiv 1.60 \times 10^{- 19} J

With this definition, the kinetic energy gained by the electron in our example is

K_{f} = 1 eV

In other words, 1 electron volt is the kinetic energy gained by an electron (or proton) if it accelerates through a potential difference of 1 volt.

NOTE The abbreviation eV uses a lowercase e but an uppercase V. Units of keV $(10^{3} eV)$ , MeV $(10^{6} eV)$ , and GeV $(10^{9} eV)$ are common.

The electron volt can be a troublesome unit. One difficulty is its unusual name, which looks less like a unit than, say, “meter” or “second.” A more significant difficulty is that the name suggests a relationship to volts. But volts are units of electric potential, whereas this new unit is a unit of energy! It is crucial to distinguish between the potential V, measured in volts, and an energy that can be measured either in joules or in electron volts.

NOTE To reiterate, the electron volt is a unit of energy, convertible to joules, and not a unit of potential. Potential is always measured in volts. However, the joule remains the SI unit of energy. It will be useful to express energies in eV, but you must convert this energy to joules before doing most calculations.

EXAMPLE 37.5 The speed of an alpha particle

Alpha particles are usually characterized by their kinetic energy in MeV. What is the speed of an 8.3 MeV alpha particle?

SOLVE Alpha particles are helium nuclei, having $m = 4 u = 6.64 \times 10^{- 27} kg$ . The kinetic energy of this alpha particle is $8.3 \times 10^{6} eV$ . First, we convert the energy to joules:

K = 8.3 \times 10^{6} eV \times \frac{1.60 \times 10^{- 19} J}{1.00 eV} = 1.33 \times 10^{- 12} J

Now we can find the speed:

\begin{array}{l} K = \frac{1}{2} m v^{2} = 1.33 \times 10^{- 12} J \\ v = \sqrt{\frac{2 K}{m}} = 2.0 \times 10^{7} m/s \end{array}

This was the speed of the alpha particle in Example 37.4.

EXAMPLE 37.6 Energy of an electron

In a simple model of the hydrogen atom, the electron orbits the proton at $2.19 \times 10^{6} m / s$ in a circle with radius $5.29 \times 10^{- 11} m$ . What is the atom’s energy in eV?

MODEL The electron has a kinetic energy of motion, and the $electron + proton$ system has an electric potential energy.

SOLVE The potential energy is that of two point charges, with $q_{proton} = + e$ and $q_{elec} = - e$ . Thus

E = K + U = \frac{1}{2} m_{elec} v^{2} + \frac{1}{4 π ϵ_{0}} \frac{(e) (- e)}{r} = - 2.17 \times 10^{- 18} J

Conversion to eV gives

E = - 2.17 \times 10^{- 18} J \times \frac{1 eV}{1.60 \times 10^{- 19} J} = - 13.6 eV

REVIEW The negative energy reflects the fact that the electron is bound to the proton. You would need to add energy to remove the electron.