de Broglie hypothesis and matter waves
If light waves can act as particles, moving particles can act as waves: λ = h/p = h/mv. For electrons accelerated through V volts, λ ≈ 12.3/√V Å.
de Broglie's idea
Light shows both wave and particle nature. In 1924, the French physicist Louis de Broglie argued that nature loves symmetry. If light can behave as particles, then moving particles such as electrons should also behave as waves. These are called matter waves. Their wavelength is λ = h/p, where p = mv is the momentum. So a wave property (λ) is linked to a particle property (p). His examiners at first doubted the idea, but experiments soon proved him right.
Derivation: λ for an electron accelerated through V volts
- 1de Broglie relation: λ = h/p, with p = mv.
- 2An electron (charge e) accelerated from rest through V volts gains kinetic energy: ½mv² = eV.
- 3So v = √(2eV/m), and p = mv = √(2meV).
- 4Hence λ = h/√(2meV).
- 5Put in h = 6.63 × 10⁻³⁴ J s, m = 9.1 × 10⁻³¹ kg and e = 1.6 × 10⁻¹⁹ C: λ = 1.23 × 10⁻⁹/√V m.
- 6★So, λ ≈ 12.3/√V Å, and λ ∝ 1/√V. At 100 V, λ ≈ 1.23 Å.
★ key step
Worked example: a cricket ball and an electron
Find the de Broglie wavelength of (a) a cricket ball of mass 0.16 kg moving at 25 m/s, (b) an electron moving at 2.2 × 10⁶ m/s. (h = 6.63 × 10⁻³⁴ J s, m = 9.1 × 10⁻³¹ kg)
- (a) p = mv = 0.16 kg × 25 m/s = 4.0 kg m/s.
- λ = h/p = 6.63 × 10⁻³⁴ ÷ 4.0 = 1.66 × 10⁻³⁴ m.
- (b) p = 9.1 × 10⁻³¹ × 2.2 × 10⁶ = 2.0 × 10⁻²⁴ kg m/s.
- λ = 6.63 × 10⁻³⁴ ÷ 2.0 × 10⁻²⁴ = 3.3 × 10⁻¹⁰ m = 3.3 Å.
(a) about 1.66 × 10⁻³⁴ m, far too small to detect; (b) about 3.3 Å, about the spacing of atoms in a crystal
Worked example: electron accelerated through 400 V
An electron starts from rest and is accelerated through 400 V. Find its de Broglie wavelength. What happens to the wavelength if the voltage is made 100 V?
- Use λ = 12.3/√V Å.
- At 400 V: λ = 12.3/√400 = 12.3/20 = 0.615 Å ≈ 0.62 Å.
- At 100 V: λ = 12.3/√100 = 12.3/10 = 1.23 Å.
- Making V one quarter doubles λ, since λ ∝ 1/√V.
0.62 Å at 400 V; 1.23 Å at 100 V, twice as long
Key points about matter waves
- Wave nature is noticed only when λ is comparable to the size of the slit or obstacle.
- λ ∝ 1/p: the heavier or faster the particle, the shorter its wavelength.
- Matter waves are not electromagnetic waves. Any moving particle has them, charged or neutral, such as a neutron.
- The same relation holds for photons: p = h/λ.
- For a 100 eV electron, λ is about 1 Å, close to X-ray wavelengths and to the spacing of atoms in a crystal. So a crystal can diffract electrons.
Dual nature of radiation vs dual nature of matter
A cricket ball's wavelength is about 10⁻³⁴ m, so it never spreads like a wave. An electron's wavelength is about the size of an atom, which is why electron beams can reveal the arrangement of atoms in a crystal.
5 marks: “State de Broglie's hypothesis. Show that the wavelength of an electron accelerated through V volts is λ = 12.3/√V Å. Why is the wave nature of a cricket ball not seen?” Moving particles have waves of λ = h/p. ½mv² = eV gives p = √(2meV), so λ = h/√(2meV); with h, m and e put in, λ ≈ 12.3/√V Å. A ball's λ (about 10⁻³⁴ m) is far smaller than any opening, so no wave effect is seen.
2 marks: “Find the de Broglie wavelength of an electron accelerated through 225 V.” λ = 12.3/√225 Å = 12.3/15 Å = 0.82 Å.
1 marks: “Why do we never see a moving ball or bus behave like a wave?” Everyday objects have a large momentum, so their wavelength h/p is far too small to notice.
Heavy or fast, the wave is tiny: λ = h/mv; for electrons λ ≈ 12.3/√V Å.