E2-3. The de Broglie Wavelength and Wave-Particle DualityHL Only
1. Wave-Particle Duality
The principle of wave-particle duality dictates that light and matter can exhibit properties of both waves and particles.
- Light as a Particle: Light interacts with matter as discrete quantised packets of energy called photons. The photoelectric effect is the primary evidence for this behavior.
- Matter as a Wave: Matter propagates through space exhibiting wave-like behaviors. Interference and diffraction patterns (such as those seen in Young's Double Slit experiment performed with electrons) provide solid evidence for its wave nature.
2. The de Broglie Wavelength
Louis de Broglie hypothesized that if light waves can behave like particles, then matter particles (like electrons) can behave like waves. He proposed that all moving particles have a "matter wave" associated with them.
The de Broglie wavelength ($\lambda$) relates a particle's wavelength to its momentum ($p$):
Kinetic Energy & The de Broglie Wavelength:
Since the kinetic energy of a non-relativistic particle is $E_k = \frac{1}{2}mv^2$, we can express momentum in terms of kinetic energy: $$E_k = \frac{p^2}{2m} \implies p = \sqrt{2mE_k}$$ Substituting this into the de Broglie equation yields:
Evidence for the Wave Nature of Matter:
- For everyday objects, the de Broglie wavelength is far too small to observe any quantum effects.
- However, an electron accelerated through a potential difference has a wavelength of roughly $10 \text{ nm}$. When an electron beam is fired through a thin film of crystalline graphite, the atomic planes act as diffraction slits.
- The electrons spread out and create a circular diffraction pattern, proving that particles undergo interference and possess a wave nature.
3. Examples
Example 1: Electron Diffraction Dynamics
Problem: Electrons are accelerated from rest through a potential difference of $40 \text{ V}$ and directed onto a crystalline graphite film.
(a) Calculate the de Broglie wavelength of the accelerated electrons.
(b) If the accelerating voltage is increased to $100 \text{ V}$, explain qualitatively what happens to the diameter of the diffraction rings observed on a screen.
Solution:
Part (a): Determine kinetic energy and wavelength:
The kinetic energy gained by the electron is exactly equal to the electrical work done on it ($E_k = qV$):
$$E_k = (1.60 \times 10^{-19} \text{ C}) \times (40 \text{ V}) = 6.4 \times 10^{-18} \text{ J}$$
Now, apply the kinetic energy form of the de Broglie wavelength equation:
$$
\begin{aligned}
\lambda &= \frac{h}{\sqrt{2mE_k}} = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.11 \times 10^{-31}) \times (6.4 \times 10^{-18})}} \\
&= \frac{6.63 \times 10^{-34}}{\sqrt{1.166 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{3.41 \times 10^{-24}} \approx \mathbf{1.94 \times 10^{-10} \text{ m} \quad (0.194 \text{ nm})}
\end{aligned}
$$
Part (b): Evaluate the effect of higher voltage:
Increasing the potential difference to $100 \text{ V}$ increases the kinetic energy and momentum of the electrons. According to the de Broglie relation ($\lambda = h/p$), a larger momentum results in a smaller wavelength. Because diffraction angle is directly proportional to wavelength ($\sin \theta \propto \lambda$), a smaller wavelength causes less diffraction spreading. Therefore, the diameter of the diffraction rings will decrease (they will shrink closer together).
Example 2: The Macroscopic vs. Microscopic Divide
Problem: Wave-particle duality applies to all matter, yet we do not see macroscopic objects diffracting through doorways. To prove why:
(a) Calculate the de Broglie wavelength of a $0.020 \text{ kg}$ rifle bullet travelling at $500 \text{ m/s}$.
(b) Compare this to the wavelength of an electron travelling at $5.0 \times 10^6 \text{ m/s}$, and explain why only the electron exhibits observable wave behavior.
Solution:
Part (a): Bullet wavelength: $$\lambda_{\text{bullet}} = \frac{h}{mv} = \frac{6.63 \times 10^{-34} \text{ J s}}{(0.020 \text{ kg}) \times (500 \text{ m/s})} = \frac{6.63 \times 10^{-34}}{10} = \mathbf{6.63 \times 10^{-35} \text{ m}}$$
Part (b): Electron wavelength: $$ \begin{aligned} \lambda_{\text{electron}} &= \frac{h}{m_e v} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (5.0 \times 10^6)} = \frac{6.63 \times 10^{-34}}{4.555 \times 10^{-24}} \\ &= \mathbf{1.45 \times 10^{-10} \text{ m}} \ (\text{or } 0.145 \text{ nm}) \end{aligned} $$
Conclusion: Diffraction is only observable when the wave passes through a gap roughly equal to its wavelength. The electron's wavelength ($\sim 10^{-10}$ m) is the exact size of atomic spacing in crystals, allowing for clear diffraction patterns. The bullet's wavelength ($\sim 10^{-35}$ m) is billions of times smaller than the nucleus of an atom! There is no physical gap in the universe small enough to diffract a bullet. Hence, macroscopic objects behave purely as particles.
Example 3: De Broglie & Bohr Orbits
Problem: In the Bohr model of the hydrogen atom, an electron in the $n = 3$ excited state has an orbital speed of $v_3 = 7.29 \times 10^5 \text{ m/s}$ and an orbital radius of $r_3 = 4.76 \times 10^{-10} \text{ m}$. Prove quantitatively that the circumference of this specific orbit is exactly equal to three de Broglie wavelengths ($n=3$).
Solution:
Step 1: Calculate the circumference of the $n=3$ orbit: $$C = 2\pi r_3 = 2 \times \pi \times (4.76 \times 10^{-10} \text{ m}) = 2.99 \times 10^{-9} \text{ m}$$
Step 2: Calculate the de Broglie wavelength of the electron in this orbit: $$\lambda = \frac{h}{m_e v_3} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (7.29 \times 10^5)} = \frac{6.63 \times 10^{-34}}{6.64 \times 10^{-25}} = 9.98 \times 10^{-10} \text{ m}$$
Step 3: Compare the two values: Divide the circumference by the wavelength: $$\frac{C}{\lambda} = \frac{2.99 \times 10^{-9}}{9.98 \times 10^{-10}} \approx \mathbf{3.00}$$ This proves the condition $2\pi r = n\lambda$. The electron does not orbit like a planet; rather, it forms a continuous, stable standing wave mapping exactly $n$ wavelengths perfectly around the nucleus!