3.2.2.4
The de Broglie wavelength: $\lambda = h/mv$
Energy Levels & Wave-Particle Duality — AQA A-Level Physics
$$\begin{aligned}
\lambda &= \frac{h}{mv} \\
&= \frac{h}{p}
\end{aligned}$$
- $λ$: de Broglie wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).The wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it... (m)
- $h$: Planck's constant (6.63 × 10⁻³⁴ J s)
- $m$: mass of the particle (kg)
- $v$: speed of the particle (m s⁻¹)
- $p$: momentum of the particle (kg m s⁻¹)
- Every moving particle has a de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelength; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it.... Larger mass or faster speed gives a shorter wavelength.
- For electrons accelerated through a p.d. $V$: $eV = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2eV}{m}}$.
- Substituting into the de Broglie equation: $\lambda = \frac{h}{\sqrt{2meV}}$.
- For everyday objects, the wavelength is negligibly small — wave behaviour is undetectable.
- For electrons and neutrons, the wavelength is comparable to atomic spacings, so diffraction is observable.
Worked Example
An electron is accelerated from rest through a potential differenceThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). of 150 V. Calculate its de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelength; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it....
Show Solution
1
Find the electron's speed
Using $eV = \frac{1}{2}mv^2$:
$$v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 150}{9.11 \times 10^{-31}}}$$ $$= \sqrt{5.27 \times 10^{13}} = 7.26 \times 10^{6} \text{ m s}^{-1}$$2
Calculate de Broglie wavelength
$$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (7.26 \times 10^{6})}$$
$$= \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.0 \times 10^{-10} \text{ m}$$3
Compare to atomic spacing
$\lambda = 0.10$ nm, which is similar to the spacing between atoms in a crystal ($\sim 0.1$ nm). This is why electron diffraction works.
Answer
$\lambda = 1.0 \times 10^{-10}$ m (0.10 nm)
Examiner Tips and Tricks
- To find the de Broglie wavelength of an accelerated electron: first find speed from $eV = \frac{1}{2}mv^2$, then substitute into $\lambda = h/mv$.
- Or use the shortcut $\lambda = h/\sqrt{2meV}$ directly.
Common Mistake
MEDIUM
Students often: Using the wrong mass in the de Broglie equation (e.g. using proton mass for an electron).
Instead: Check which particle is being discussed. The electron mass is $9.11 \times 10^{-31}$ kg. The proton mass is $1.67 \times 10^{-27}$ kg. A proton has a much shorter de Broglie wavelength than an electron at the same speed.
Instead: Check which particle is being discussed. The electron mass is $9.11 \times 10^{-31}$ kg. The proton mass is $1.67 \times 10^{-27}$ kg. A proton has a much shorter de Broglie wavelength than an electron at the same speed.