de Broglie relation: wavelength of accelerated particles
For a particle with kinetic energy K, λ = h/√(2mK). A charge q accelerated from rest through V volts has λ = h/√(2mqV).
From λ = h/p to λ = h/√(2mqV)
- 1de Broglie relation: λ = h/p, where p = mv.
- 2Kinetic energy K = ½mv² = p²/(2m), so p = √(2mK).
- 3So λ = h/√(2mK).
- 4★A charge q accelerated from rest through a potential difference V gains K = qV. So λ = h/√(2mqV).
- 5For an electron, put h = 6.63 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg and q = e = 1.6 × 10⁻¹⁹ C.
- 6Extra reading, not in your textbook: these values give the shortcut λ ≈ 1.227/√V nm (V in volts). Use it only to check; in the exam, work from λ = h/√(2meV).
★ key step
Comparing de Broglie wavelengths
| What is the same | Relation | Example |
|---|---|---|
| Momentum | λ is the same | A photon and an electron with equal momentum have equal λ |
| Speed | λ ∝ 1/m | An electron's λ is about 1836 times a proton's |
| Kinetic energy | λ ∝ 1/√m | λₑ/λₚ = √(mₚ/mₑ) = √1836 ≈ 43 |
| Accelerating potential | λ ∝ 1/√(mq) | Proton and alpha particle: λₚ/λ_α = √(4 × 2) = 2√2 |
Worked example: electron accelerated through a potential difference
An electron starts from rest and is accelerated through a potential difference of 100 V. Find (a) its de Broglie wavelength, (b) its wavelength if the potential difference is made 400 V. (h = 6.63 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
- For an electron: λ = h/p, with p = √(2meV).
- (a) p = √(2 × 9.11 × 10⁻³¹ kg × 1.6 × 10⁻¹⁹ C × 100 V) = √(2.92 × 10⁻⁴⁷) kg m/s ≈ 5.40 × 10⁻²⁴ kg m/s.
- λ = h/p = (6.63 × 10⁻³⁴ J s) ÷ (5.40 × 10⁻²⁴ kg m/s) ≈ 1.23 × 10⁻¹⁰ m = 0.123 nm.
- (b) λ ∝ 1/√V. V becomes 4 times, so λ becomes 1/√4 = 1/2 of its value.
- λ = 0.123 nm ÷ 2 ≈ 0.061 nm.
(a) about 0.123 nm; (b) about 0.061 nm
Worked example: proton and alpha particle
A proton and an alpha particle start from rest and are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths. (Mass of alpha particle = 4 × mass of proton; charge of alpha particle = 2e)
- λ = h/√(2mqV), so at the same V, λ ∝ 1/√(mq).
- λₚ/λ_α = √(m_α q_α) ÷ √(mₚ qₚ).
- λₚ/λ_α = √((4mₚ × 2e) ÷ (mₚ × e)) = √8 = 2√2.
- So λₚ : λ_α = 2√2 : 1, or about 2.83 : 1.
λₚ : λ_α = 2√2 : 1
Worked example: electron and photon of the same wavelength
An electron and a photon each have a wavelength of 1 nm. Find the energy of the photon and the kinetic energy of the electron, in eV. (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, mₑ = 9.11 × 10⁻³¹ kg)
- Photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) ÷ (1 × 10⁻⁹) J = 1.989 × 10⁻¹⁶ J.
- In eV: 1.989 × 10⁻¹⁶ ÷ 1.6 × 10⁻¹⁹ ≈ 1243 eV, about 1.24 keV.
- Electron: p = h/λ = 6.63 × 10⁻³⁴ ÷ 10⁻⁹ = 6.63 × 10⁻²⁵ kg m/s, the same as the photon.
- K = p²/(2m) = (6.63 × 10⁻²⁵)² ÷ (2 × 9.11 × 10⁻³¹) J = (43.96 × 10⁻⁵⁰) ÷ (18.22 × 10⁻³¹) J ≈ 2.41 × 10⁻¹⁹ J.
- In eV: 2.41 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ ≈ 1.51 eV.
Photon ≈ 1243 eV (1.24 keV); electron ≈ 1.51 eV. The same λ and momentum, but very different energies.
How to draw the de Broglie graphs
- λ against p: a curve that falls as p rises and never touches the axes (a rectangular hyperbola, since λp = h).
- λ against 1/p: a straight line through the origin with slope h.
- λ against 1/√V for electrons: a straight line through the origin, because λ = h/√(2meV) and h/√(2me) is a constant.
An electron microscope accelerates electrons through tens of thousands of volts. By λ = h/√(2meV), their wavelength is a few thousandths of a nanometre, far shorter than visible light (400 to 700 nm). That is why it can show viruses.
2 marks: “An electron is accelerated from rest through a potential difference of 64 V. Find its de Broglie wavelength.” λ = h/√(2meV) = (6.63 × 10⁻³⁴) ÷ √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 64) m = (6.63 × 10⁻³⁴) ÷ (4.32 × 10⁻²⁴) m ≈ 1.53 × 10⁻¹⁰ m ≈ 0.153 nm.
3 marks: “A proton and an alpha particle have the same kinetic energy. Which has the shorter de Broglie wavelength? Find the ratio of their wavelengths.” λ = h/√(2mK); at equal K, λ ∝ 1/√m. λₚ/λ_α = √(m_α/mₚ) = √4 = 2. The alpha particle has the shorter wavelength; λₚ : λ_α = 2 : 1.
2 marks: “Draw graphs of the de Broglie wavelength λ of a particle against (a) its momentum p, (b) 1/p.” (a) A curve falling as p increases (rectangular hyperbola, λp = h). (b) A straight line through the origin with slope h.
λ = h/p, with p = √(2mK) = √(2mqV). A bigger m, q or V gives a shorter λ.