The de Broglie Wavelength

De Broglie proposed that electrons travel through space as a wave
- This would explain why they can exhibit behaviour such as diffraction
He therefore suggested that electrons must also hold wave properties, such as wavelength
- This became known as the de Broglie wavelength
However, he realised all particles can show wave-like properties, not just electrons
So, the de Broglie wavelength can be defined as:

The wavelength associated with a moving particle

The majority of the time, and for everyday objects travelling at normal speeds, the de Broglie wavelength is far too small for any quantum effects to be observed
A typical electron in a metal has a de Broglie wavelength of about 10 nm
Therefore, quantum mechanical effects will only be observable when the width of the sample is around that value

Calculating de Broglie Wavelength

Using ideas based upon the quantum theory and Einstein’s theory of relativity, de Broglie suggested that the momentum (p) of a particle and its associated wavelength (λ) are related by the equation:

$λ = \frac{h}{p}$

Since momentum p = mv, the de Broglie wavelength can be related to the speed of a moving particle (v) by the equation:

$λ = \frac{h}{m v}$

Using kinetic energy $E = \frac{1}{2} m v^{2}$ , momentum and kinetic energy can be related by:

Energy: $E = \frac{1}{2} m v^{2} = \frac{m v^{2}}{2} = \frac{p^{2}}{2 m}$

Momentum: $p = \sqrt{2 m E}$

Combining this with the de Broglie equation gives a form which relates the de Broglie wavelength of a particle to its kinetic energy:

$λ = \frac{h}{\sqrt{2 m E}}$

Where:
- λ = the de Broglie wavelength (m)
- h = Planck’s constant (J s)
- p = momentum of the particle (kg m s^-1)
- E = kinetic energy of the particle (J)
- m = mass of the particle (kg)
- v = speed of the particle (m s^-1)

Worked example

A proton and an electron are each accelerated from rest through the same potential difference.

Determine the ratio: $\frac{d e B r o g l i e w a v e l e n g t h o f t h e p r o t o n}{d e B r o g l i e w a v e l e n g t h o f t h e e l e c t r o n}$

Mass of a proton = 1.67 × 10^–27 kg
Mass of an electron = 9.11 × 10^–31 kg

Answer:

Step 1: Determine how the proton and electron can be related via their mass

The only information we are given is the mass of the proton and the electron
When the proton and electron are accelerated through a potential difference, their kinetic energy will increase
Therefore, we can use kinetic energy to relate them via their mass

Step 2: Write the equation relating the de Broglie wavelength of a particle to its kinetic energy

The de Broglie wavelength

$λ = \frac{h}{m v} = \frac{h}{p}$

Kinetic energy

$E = \frac{1}{2} m v^{2} = \frac{m v^{2}}{2} = \frac{p^{2}}{2 m}$

Kinetic energy in terms of momentum

$p = \sqrt{2 m E}$

Substitute the expression for momentum into the do Broglie wavelength equation

$λ = \frac{h}{\sqrt{2 m E}}$

Step 3: Find the proportional relationship between the de Broglie wavelength and the mass of the particle

$λ \propto \frac{1}{\sqrt{m}} \Rightarrow λ = k \frac{1}{\sqrt{m}}$

Step 4: Calculate the ratio

$\frac{d e B r o g l i e w a v e l e n g t h o f t h e p r o t o n}{d e B r o g l i e w a v e l e n g t h o f t h e e l e c t r o n} = \frac{1}{\sqrt{m_{p}}} \div \frac{1}{\sqrt{m_{e}}}$

$\sqrt{\frac{m_{e}}{m_{p}}} = \sqrt{\frac{9.11 \times 10^{- 31}}{1.67 \times 10^{- 27}}} = 2.3 \times 10^{- 2}$

This means the de Broglie wavelength of the proton is 0.023 times smaller than that of the electron OR the de Broglie wavelength of the electron is about 40 times larger than that of the proton

Examiner Tip

Particles with a greater mass, such as a proton, have a greater momentum. The greater the momentum, the smaller the de Broglie wavelength. Always perform a logic check on your answer to check that makes sense.

The de Broglie Wavelength (CIE A Level Physics)

Revision Note