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By the end of this section, you will be able to:
  • Describe de Broglie’s hypothesis of matter waves
  • Explain how the de Broglie’s hypothesis gives the rationale for the quantization of angular momentum in Bohr’s quantum theory of the hydrogen atom
  • Describe the Davisson–Germer experiment
  • Interpret de Broglie’s idea of matter waves and how they account for electron diffraction phenomena

Compton’s formula established that an electromagnetic wave can behave like a particle of light when interacting with matter. In 1924, Louis de Broglie proposed a new speculative hypothesis that electrons and other particles of matter can behave like waves. Today, this idea is known as de Broglie’s hypothesis of matter waves    . In 1926, De Broglie’s hypothesis, together with Bohr’s early quantum theory, led to the development of a new theory of wave quantum mechanics    to describe the physics of atoms and subatomic particles. Quantum mechanics has paved the way for new engineering inventions and technologies, such as the laser and magnetic resonance imaging (MRI). These new technologies drive discoveries in other sciences such as biology and chemistry.

According to de Broglie’s hypothesis, massless photons as well as massive particles must satisfy one common set of relations that connect the energy E with the frequency f , and the linear momentum p with the wavelength λ . We have discussed these relations for photons in the context of Compton’s effect. We are recalling them now in a more general context. Any particle that has energy and momentum is a de Broglie wave    of frequency f and wavelength λ :

E = h f
λ = h p .

Here, E and p are, respectively, the relativistic energy and the momentum of a particle. De Broglie’s relations are usually expressed in terms of the wave vector k , k = 2 π / λ , and the wave frequency ω = 2 π f , as we usually do for waves:

E = ω
p = k .

Wave theory tells us that a wave carries its energy with the group velocity    . For matter waves, this group velocity is the velocity u of the particle. Identifying the energy E and momentum p of a particle with its relativistic energy m c 2 and its relativistic momentum mu , respectively, it follows from de Broglie relations that matter waves satisfy the following relation:

λ f = ω k = E / p / = E p = m c 2 m u = c 2 u = c β

where β = u / c . When a particle is massless we have u = c and [link] becomes λ f = c .

How long are de broglie matter waves?

Calculate the de Broglie wavelength of: (a) a 0.65-kg basketball thrown at a speed of 10 m/s, (b) a nonrelativistic electron with a kinetic energy of 1.0 eV, and (c) a relativistic electron with a kinetic energy of 108 keV .

Strategy

We use [link] to find the de Broglie wavelength. When the problem involves a nonrelativistic object moving with a nonrelativistic speed u , such as in (a) when β = u / c 1 , we use nonrelativistic momentum p . When the nonrelativistic approximation cannot be used, such as in (c), we must use the relativistic momentum p = m u = m 0 γ u = E 0 γ β , where the rest mass energy of a particle is E 0 = m 0 c 2 and γ is the Lorentz factor γ = 1 / 1 β 2 . The total energy E of a particle is given by [link] and the kinetic energy is K = E E 0 = ( γ 1 ) E 0 . When the kinetic energy is known, we can invert [link] to find the momentum p = ( E 2 E 0 2 ) / c 2 = K ( K + 2 E 0 ) / c and substitute in [link] to obtain

Questions & Answers

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Practice Key Terms 5

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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