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 A Level Physics)

Revision Note