E.2 · IB Physics
Electron diffraction and de Broglie wavelength
Accelerate electrons through a potential difference, fire them at a crystal and watch how their momentum and de Broglie wavelength set the size of the diffraction rings on the screen.
- Kinetic energy Eₖ
- Speed v
- Momentum p
- Wavelength λ
- Ring 1 (d₁): 2θ, R
- Ring 2 (d₂): 2θ, R
eV = Eₖ = p²/2m
p = √(2meV)
λ = h/p
2d sin θ = λ
ring radius R = L tan 2θ
p = √(2meV)
λ = h/p
2d sin θ = λ
ring radius R = L tan 2θ
Try this
- Set V = 2.0 kV and note λ, then raise V to 8.0 kV. Check that λ (and roughly each ring radius) has halved, because λ ∝ 1/√V.
- At V = 3.0 kV, read the inner ring radius R and L, find θ from tan 2θ = R/L, then use 2d sin θ = λ to recover d₁.
- Switch from graphite to aluminium at a fixed V: which ring belongs to the larger plane spacing, and why does a larger d give a smaller ring?
Common error: the electron's energy is eV in joules, not V itself — convert to momentum with p = √(2meV) before using λ = h/p, and remember that doubling V shrinks λ only by a factor of √2.