recall and use λ = h / p
Wave‑Particle Duality 🌊✨
In physics, many objects behave like both waves and particles. Think of a water ripple (wave) and a ball (particle). This dual nature is key to understanding quantum mechanics.
Why It Matters
It explains why electrons can interfere like waves, yet also hit a detector as individual spots.
Key Formula
De Broglie wavelength: $$\lambda = \frac{h}{p}$$
Where h is Planck’s constant ($6.626\times10^{-34}\,\text{J·s}$) and p is momentum ($p = mv$).
Analogy: The “Bouncing Ball”
Imagine a ball thrown at a wall. If the ball is very light and fast, it behaves like a tiny wave that can bend around obstacles. If it’s heavy, it behaves like a classic particle that just bounces straight back.
Quick Example
Calculate the wavelength of an electron moving at $v = 1.0\times10^6\,\text{m/s}$.
- Find momentum: $p = mv$ ($m = 9.11\times10^{-31}\,\text{kg}$).
- Compute $p = 9.11\times10^{-31}\,\text{kg} \times 1.0\times10^6\,\text{m/s} = 9.11\times10^{-25}\,\text{kg·m/s}$.
- Use $\lambda = h/p$: $$\lambda = \frac{6.626\times10^{-34}}{9.11\times10^{-25}} \approx 7.3\times10^{-10}\,\text{m}.$$
That’s about 0.73 nanometres – smaller than a cell!
Exam Tips 📚
- Remember the formula: $\lambda = h/p$.
- Always check units – $h$ in J·s, $p$ in kg·m/s, giving $\lambda$ in metres.
- When given energy $E$, convert to momentum using $p = \sqrt{2mE}$ for non‑relativistic particles.
- Use the De Broglie wavelength to explain diffraction patterns in exams.
Quick Quiz
| Question | Answer |
|---|---|
| What is the De Broglie wavelength of a photon with energy $E = 2.0\,\text{eV}$? | Use $E = hu$ and $\lambda = c/u$. Result: $\lambda \approx 620\,\text{nm}$. |
| If an electron has $\lambda = 0.5\,\text{nm}$, what is its momentum? | $p = h/\lambda \approx 1.32\times10^{-24}\,\text{kg·m/s}$. |
Revision
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