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:

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

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:

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

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

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:

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

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 1.6 × 10^-19

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

The Electronvolt (CIE A Level Physics)