The Electronvolt

The electronvolt is a unit which is commonly used to express very small energies
This is because quantum energies tend to be much smaller than 1 Joule
The electronvolt is derived from the definition of potential difference:

$V = \frac{E}{Q}$

Where:
- V = potential difference (V)
- E = energy (J)
- Q = charge (C)

When an electron travels through a potential difference, energy is transferred between two points in a circuit or electric field
If an electron, with a charge of 1.6 × 10^-19 C, travels through a potential difference of 1 V, the energy transferred is equal to:

$E = Q V = (1.60 \times 10^{- 19}) C \times 1 V = 1.60 \times 10^{- 19} J$

Therefore, an electronvolt is defined as:

The energy gained by an electron travelling through a potential difference of one volt

$1 eV = 1.60 \times 10^{- 19} J$

To convert between eV and J:
- eV → J: multiply by 1.6 × 10^-19
- J → eV: divide by 1.6 × 10^-19

Relation to Kinetic Energy

When a charged particle is accelerated through a potential difference, it gains kinetic energy
If an electron accelerates from rest, an electronvolt is equal to the kinetic energy gained:

$e V = \frac{1}{2} m v^{2}$

Where:
- e = charge of an electron (1.60 × 10^–19 C)
- V = potential difference (V)
- m = mass of the particle (kg)
- v = velocity of the particle (m s^–1)
Rearranging the equation gives the speed of the electron:

$v = \sqrt{\frac{2 e V}{m}}$

Worked example

(a)

An electron has an energy of 2.4 eV.

Give the energy of the electron in joules.

(b)

A photon has an energy of 4.9 × 10⁻¹⁹ J.

Give the energy of the photon in electronvolts.

Answer:

Part (a)

$E = 2.4 eV = 2.4 \times (1.6 \times 10^{- 19})$

$E = 3.8 \times 10^{- 19} J$

Part (b)

$E = 4.9 \times 10^{- 19} J = \frac{4.9 \times 10^{- 19}}{1.6 \times 10^{- 19}}$

$E = 3.1 eV$

Worked example

Show that the photon energy of light with wavelength 700nm is about 1.8 eV.

Answer:

Step 1: Write the equation for photon energy

$E = \frac{h c}{λ}$

Step 2: Calculate the photon energy in Joules

$E = \frac{h c}{λ} = \frac{(6.63 \times 10^{- 34}) \times (3.0 \times 10^{8})}{700 \times 10^{- 9}} = 2.84 \times 10^{- 19} J$

Step 3: Convert the photon energy into electronvolts

$1 eV = 1.60 \times 10^{- 19} J$

J → eV: divide by 1.60 × 10^-19

$E = \frac{2.84 \times 10^{- 19}}{1.60 \times 10^{- 19}} = 1.78 eV$

Examiner Tip

Converting between electronvolts and joules is a skill that you will use a lot in exam questions on this topic. You do not need to remember the value of 1 eV as this will be given on your datasheet, but you do need to be confident in the conversion so that it doesn't slow you down.

The Electronvolt (AQA AS Physics)

Revision Note