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de Broglie

de Broglie wavelength in terms of kinetic energy

With kinetic energy K, p = √(2mK) and λ = h / √(2mK). An electron with 120 eV has p = 5.91 × 10⁻²⁴ kg m/s, speed 6.49 × 10⁶ m/s and λ = 0.112 nm.

de Broglie wavelength

Every moving particle has a wave with λ = h / p. Give the speed, the kinetic energy, or the voltage a charged particle was accelerated through. Speeds are taken well below the speed of light.

Particle
Given
Wavelength λ0.112 nm1.12 × 10⁻¹⁰ m
Momentum5.91 × 10⁻²⁴kg m/s
Kinetic energy120 eV1.92 × 10⁻¹⁷ J
Speed6.49 × 10⁶ m/sElectron

Electron: de Broglie wavelength 0.112 nm.

Show the working
  1. Given: electron, m = 9.11 × 10⁻³¹ kg, K = 120 eV.
  2. K = 120 eV = 120 × 1.6 × 10⁻¹⁹ = 1.92 × 10⁻¹⁷ J
  3. p = √(2mK) = √(2 × 9.11 × 10⁻³¹ × 1.92 × 10⁻¹⁷) = 5.91 × 10⁻²⁴ kg m/s
  4. λ = h / p = 6.63 × 10⁻³⁴ / 5.91 × 10⁻²⁴ = 1.12 × 10⁻¹⁰ m = 0.112 nm
Working

Step by step

  1. Given: electron, m = 9.11 × 10⁻³¹ kg, K = 120 eV.
  2. K = 120 eV = 120 × 1.6 × 10⁻¹⁹ = 1.92 × 10⁻¹⁷ J
  3. p = √(2mK) = √(2 × 9.11 × 10⁻³¹ × 1.92 × 10⁻¹⁷) = 5.91 × 10⁻²⁴ kg m/s
  4. λ = h / p = 6.63 × 10⁻³⁴ / 5.91 × 10⁻²⁴ = 1.12 × 10⁻¹⁰ m = 0.112 nm
Same energy

Four particles with 120 eV

ParticleMassMomentumλ
Electron9.11 × 10⁻³¹ kg5.91 × 10⁻²⁴ kg m/s0.112 nm
Proton1.67 × 10⁻²⁷ kg2.53 × 10⁻²² kg m/s2.62 pm
Neutron1.67 × 10⁻²⁷ kg2.54 × 10⁻²² kg m/s2.61 pm
Alpha particle6.64 × 10⁻²⁷ kg5.05 × 10⁻²² kg m/s1.31 pm
Derivation

Why it works

  1. de Broglie proposed that a moving particle has a wave, with λ = h / p, the same relation that holds for a photon.
  2. For a slow particle p = mv, so λ = h / mv.
  3. Kinetic energy K = p² / 2m gives p = √(2mK), so λ = h / √(2mK).
  4. A charge q accelerated through V gains K = qV, so λ = h / √(2mqV).
  5. For an electron, putting in h, m and e gives λ = 1.227 / √V nm.
Questions

de Broglie wavelength from kinetic energy: quick answers

How is the de Broglie wavelength written in terms of kinetic energy?

Since K = p² / 2m, p = √(2mK) and λ = h / √(2mK). Doubling K shortens λ by a factor √2.

What is λ for an electron with 120 eV?

K = 1.92 × 10⁻¹⁷ J, p = √(2mK) = 5.91 × 10⁻²⁴ kg m/s, and λ = h / p = 0.112 nm.

At the same kinetic energy, which particle has the longest wavelength?

The lightest. λ = h / √(2mK), so λ goes as 1 / √m: an electron's is about 43 times a proton's.

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