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Wize High School Grade 12 Physics Textbook > Modern Physics

Matter Waves

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de Broglie Wavelength (Particles as Waves)


If light can behave like a particle, then particles can also behave like waves!

If we design an experiment that highlights the wave nature of things (e.g. interference, diffraction), but do it with particles instead of waves (e.g. with electrons instead of light), then those particles are forced to exhibit their wave-like properties.



Assuming that massive particles can behave like waves, we can equate the momentum of a particle with the momentum of a wave and solve for the wavelength. What we get is the de Broglie wavelength of a massive particle:


 λ=hmv \boxed{\ λ = \dfrac{h}{mv} \ }

  • λ\lambda is the de Broglie wavelength of the particle
  • mm is the mass of the particle
  • vv is the speed of the particle
  • hh is Planck's constant







Exam Tip
As the mass or the velocity of the particle increases, the de Broglie wavelength decreases.


Wize Concept
  • We can only detect wavelike properties of matter when the wavelength is comparable to the size of the particle.
  • We don't observe the objects around as as waves because their very large mass means their wavelength is undetectably small.


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Example: De Broglie Wavelength


a) What is the de Broglie wavelength of an electron moving at 1.00×1061.00\times10^6 m/s ?

b) What is the de Broglie wavelength of a baseball of mass 100100 g traveling at 9090 km/hr ?

c) At what speed does a 4545 g golf ball need to move in order to achieve a de Broglie wavelength of 100100 nm?


Part a)


The de Broglie wavelength is:

λ=hmv\lambda=\dfrac{h}{mv}

=6.626×10−34(9.11×10−31)(106)=\dfrac{6.626\times10^{-34}}{(9.11\times10^{-31})(10^6)}

=7.27×10−10=7.27\times10^{-10} (m)

=0.727=0.727 (nm)


NOTE: The atomic diameter is in the range of 10−1010^{-10} m. Therefore this wavelength of an electron can generate measurable diffraction patterns, e.g. when passed through crystals. In other words, this wavelength is significant in the atomic world.


Part b)


The de Broglie wavelength is:

λ=hmv\lambda=\dfrac{h}{mv}

=6.626×10−34(0.1)(90×10003600)=\dfrac{6.626\times10^{-34}}{(0.1)(90\times\dfrac{1000}{3600})}

=2.65×10−34=2.65\times10^{-34} (m)


NOTE: This wavelength is too short to be significant, even in the microscopic world!


Part c)


The speed is given by:

λ=hmv   →   v=hmλ\lambda=\dfrac{h}{mv} \ \ \ \to \ \ \ v=\dfrac{h}{m\lambda}

=6.626×10−34(0.045)(100×10−9)=\dfrac{6.626\times10^{-34}}{(0.045)(100\times10^{-9})}

=1.47×10−25=1.47\times10^{-25} (m/s)

Practice: Light vs. Particle

a) What is the kinetic energy of alpha particle whose wavelength is the same as a 3 MeV3 \ MeV photon? (mass of alpha particle = 6.645 x 10-27 kg)

b) For the same problem, if the energy of the photon doubles, by what factor does the corresponding kinetic energy of the alpha change?

c) Using the results from part a) and b), predict the kinetic energy of the alpha if it has the same wavelength as that of a 24 MeV24\ MeV photon.