KEY TAKEAWAY: All matter exhibits wave-like properties, though these are only readily observable for very small objects like electrons.
The relationship is defined by the following equation:
$$\lambda = \frac{h}{p}$$
Where:
* $\lambda$ = de Broglie wavelength (m)
* $h$ = Planck’s constant (\$6.63 \times 10^{-34} \, \text{J s}$)
* $p$ = momentum of the particle (kg m/s)
Since momentum is the product of mass ($m$) and velocity ($v$), the equation can also be written as:
$$\lambda = \frac{h}{mv}$$
Where:
* $m$ = mass of the particle (kg)
* $v$ = velocity of the particle (m/s)
REMEMBER: The de Broglie equation links the wave nature ($\lambda$) to the particle nature ($p$ or $mv$) of matter.
Example:
Calculate the de Broglie wavelength of an electron moving at \$5.0 \times 10^6 \, \text{m/s}$. The mass of an electron is \$9.11 \times 10^{-31} \, \text{kg}$.
EXAM TIP: Always double-check your units before plugging values into the equation. Make sure mass is in kg, velocity is in m/s, and Planck’s constant is in J s.
Comparison of de Broglie Wavelengths:
| Object | Mass (kg) | Velocity (m/s) | de Broglie Wavelength (m) |
|---|---|---|---|
| Electron | \$9.11 \times 10^{-31}$ | \$1.0 \times 10^6$ | $\approx 7.3 \times 10^{-10}$ |
| Cricket Ball | 0.16 | 42 | $\approx 9.9 \times 10^{-35}$ |
| Running Person | 66 | 10 | $\approx 1.0 \times 10^{-36}$ |
APPLICATION: Electron microscopes utilize the wave nature of electrons (their de Broglie wavelength) to achieve higher resolution than optical microscopes, as electron wavelengths are much smaller than visible light wavelengths.
Electron Diffraction:
VCAA FOCUS: Be prepared to interpret electron diffraction patterns and relate them to the wave nature of matter. Understand how the de Broglie wavelength affects the extent of diffraction.
Comparison Table:
| Property | Photon | Matter (e.g., Electron) |
|---|---|---|
| Nature | Electromagnetic wave/particle | Matter wave/particle |
| Momentum | $p = \frac{h}{\lambda}$ | $p = mv = \frac{h}{\lambda}$ |
| Energy | $E = hf = pc$ | $E = \frac{1}{2}mv^2$ (Kinetic) |
| Mass | Zero (rest mass) | Non-zero |
| Wave-like Behavior | Diffraction, Interference | Diffraction, Interference |
COMMON MISTAKE: Confusing the energy equation for photons ($E=pc$) with the kinetic energy equation for matter ($E=\frac{1}{2}mv^2$). Remember to use the appropriate equation for each.
Free exam-style questions on de Broglie wavelength with instant AI feedback.
An electron is accelerated from rest through a potential difference of 500 V. Calculate the de Broglie wavelength of this electron. Account…
A newly discovered subatomic particle, the X particle, has a mass of $8.36 imes 10^{-28}$ kg. In an experiment, this X particle is accelerat…
State the de Broglie wavelength equation and define each of its terms.
A golf ball (mass 0.045 kg) is moving at a speed of 40 m/s. Which of the following best approximates its de Broglie wavelength?
Which of the following quantities is required to calculate the de Broglie wavelength of a particle?
A proton and an electron both have the same de Broglie wavelength. Which of the following statements correctly compares their momenta and ki…
A beam of electrons and a beam of protons are both accelerated through the same potential difference. Which of the following statements corr…