# The de Broglie Equation ## Relating Wavelength to Momentum - Louis de Broglie proposed that all particles exhibit wave-like properties - He combined quantum theory with Einstein's relativity to connect a particle's momentum with its wavelength - The de Broglie equation establishes this relationship: $$\lambda = \frac{h}{p}$$

The de Broglie wavelength $\lambda$ of a particle is Planck's constant $h$ divided by its momentum $p$

**Where:** - $\lambda$ = de Broglie wavelength (m) - $h$ = Planck's constant (J s) - $p$ = momentum (kg m s$^{-1}$) ### Visualizing the Inverse Relationship The de Broglie equation shows that wavelength is inversely proportional to momentum. This means higher momentum particles have shorter wavelengths:

## Wave-Particle Duality Connection - The de Broglie equation links particle properties (momentum) with wave properties (wavelength) - This demonstrates wave-particle duality applies to all matter, not just electromagnetic radiation - Any moving object has an associated wavelength, though it's typically only observable for very small particles ## Alternative Forms of the Equation ### Using Velocity - Since momentum can be written as $p = mv$, the de Broglie equation becomes: $$\lambda = \frac{h}{mv}$$ **Where:** - $m$ = mass of the particle (kg) - $v$ = speed of the particle (m s$^{-1}$) ### Using Kinetic Energy - Kinetic energy relates to momentum through the equation: $$E_k = \frac{1}{2}mv^2$$ - Rearranging for momentum gives: $$E_k = \frac{p^2}{2m} \quad \text{or} \quad p = \sqrt{2mE_k}$$ - Substituting this into the de Broglie equation: $$\lambda = \frac{h}{\sqrt{2mE_k}}$$

For particles with the same kinetic energy, lighter particles have longer de Broglie wavelengths since $\lambda \propto \frac{1}{\sqrt{m}}$

**Where:** - $E_k$ = kinetic energy of the particle (J) - $m$ = mass of the particle (kg) ## Key Relationships | Quantity Increased | Effect on de Broglie Wavelength | |---|---| | Momentum | Decreases (inversely proportional) | | Speed | Decreases (inversely proportional) | | Mass | Decreases (inversely proportional) | | Kinetic Energy | Decreases (inversely proportional to square root) | ## Worked Example: Comparing Particle Wavelengths **Question:** An electron and a proton are both accelerated from rest through an identical potential difference. Calculate the ratio: $$\frac{\text{de Broglie wavelength of the electron}}{\text{de Broglie wavelength of the proton}}$$ Mass of electron = $9.11 \times 10^{-31}$ kg Mass of proton = $1.67 \times 10^{-27}$ kg **Solution:** **Step 1: Identify what's equal** - Both particles are accelerated through the same potential difference - Therefore, both gain identical kinetic energies **Step 2: Select the appropriate form** - Use the kinetic energy version of the de Broglie equation: $$\lambda = \frac{h}{\sqrt{2mE_k}}$$ - Since $E_k$ is constant for both particles: $$\lambda \propto \frac{1}{\sqrt{m}}$$ **Step 3: Form the ratio** $$\frac{\lambda_e}{\lambda_p} = \frac{\frac{1}{\sqrt{m_e}}}{\frac{1}{\sqrt{m_p}}} = \sqrt{\frac{m_p}{m_e}}$$ **Step 4: Calculate** $$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} = \sqrt{1833} = 42.8$$

The electron's de Broglie wavelength is approximately 43 times larger than the proton's wavelength when both have equal kinetic energy

### Visualizing Particle Wavelength Comparison Here's a comparison of typical de Broglie wavelengths for different particles accelerated through 100 V (gaining 100 eV of kinetic energy):

Notice how the lighter electron has a significantly longer wavelength compared to the heavier particles at the same kinetic energy. ## Practical Considerations - De Broglie wavelengths for macroscopic objects are incredibly small - Example: A 0.1 kg ball moving at 10 m s$^{-1}$ has $\lambda \approx 6.6 \times 10^{-34}$ m - This is far too small to produce observable diffraction effects - For subatomic particles, wavelengths are significant - Example: An electron accelerated through 100 V has $\lambda \approx 0.12$ nm - This is comparable to atomic spacing, making diffraction observable

Always check your data sheet for particle masses if not provided in the question

## Summary of Key Equations | Form | Equation | When to Use | |---|---|---| | Basic | $\lambda = \frac{h}{p}$ | When momentum is known | | With velocity | $\lambda = \frac{h}{mv}$ | When mass and speed are known | | With kinetic energy | $\lambda = \frac{h}{\sqrt{2mE_k}}$ | When kinetic energy is known |