Lesson Explainer: Matter Waves Physics • 9th Grade

In this explainer, we will learn how to calculate the de Broglie wavelength of massive particles that have a given momentum or velocity.

Recall that light can be described using both a wave model and a particle model. Phenomena such as refraction and diffraction can be explained using the wave model of light. The particle model of light is useful for describing phenomena such as the photoelectric effect. Recall that particles of light are massless and known as photons.

In the 20th century, physicist Louis de Broglie proposed that it was not just light that could exhibit both wave-like and particle-like characteristics; he suggested that particles with mass, such as electrons and protons, could also behave like waves. He further proposed that some relations describing the dual nature of light also applied to matter.

Recall that the momentum of a photon, ๐‘, is given by ๐‘=โ„Ž๐œ†, where โ„Ž is the Planckโ€™s constant and ๐œ† is the wavelength of the photon. De Broglie proposed that the same relation is true for particles of matter. Rearranging the above equation to solve for wavelength, ๐œ†=โ„Ž๐‘, gives the de Broglie wavelength of a particle, given its momentum.

Definition: De Broglie Wavelength

The de Broglie wavelength, ๐œ†, of a particle with momentum ๐‘ is given by ๐œ†=โ„Ž๐‘, where โ„Ž is Planckโ€™s constant.

Recall that, for a particle moving at a speed much slower than the speed of light, the momentum is equal to the product of the particleโ€™s mass, ๐‘š, and its speed, ๐‘ฃ. Thus, the de Broglie wavelength can also be found using ๐œ†=โ„Ž๐‘š๐‘ฃ.

This concept also applies to collections of particles or objectsโ€”even very large objects, such as the items we interact with in everyday life. Thus, any massive object with momentum has a de Broglie wavelength. It should be noted that โ€œmassiveโ€ refers to anything that has mass, not just those that are very large.

The concept of a massive object behaving as a wave is often puzzling because we do not observe wave effects such as diffraction with the everyday items we interact with. This is because the de Broglie wavelength is very small for large objects.

For example, one might wonder why people, who move and have mass, do not experience diffraction when they walk through a doorway. To understand why, we can calculate the de Broglie wavelength for the average human and recall that diffraction is best observed when a wave encounters an obstacle the size of its wavelength. Assuming a mass of 62 kg and walking speed of 1.5 m/s, the de Broglie wavelength of a human is ๐œ†=โ„Ž๐‘š๐‘ฃ6.63ร—10โ‹…/(62)(1.5/)=7.13ร—10.๏Šฑ๏Šฉ๏Šช๏Šจ๏Šฑ๏Šฉ๏Šฌkgmskgmsm

Even though the de Broglie wavelength for a human exists in theory, its value is much, much smaller than anything we can physically measure. Thus, we do not observe wave effects for the everyday objects we are familiar with. This is due to the fact that the de Broglie wavelength of an object is inversely proportional to its momentum.

We can explore this proportionality with a couple of examples.

Example 1: Graphically Relating Momentum to the de Broglie Wavelength

The graph shows a number of curves. Which curve shows the relation between the momentum of a particle and its de Broglie wavelength?

Answer

Let us begin by recalling the equation for the de Broglie wavelength of a particle, ๐œ†=โ„Ž๐‘.

Because โ„Ž represents Planckโ€™s constant, which is an unchanging value, the proportionality relating the two variables in this equation is ๐œ†โˆ1๐‘.

Thus, we can say that the de Broglie wavelength is inversely proportional to momentum. This inverse relationship means that a greater wavelength corresponds to a smaller momentum, so we can expect that the graph of wavelength as a function of momentum should only decrease as ๐‘ gets larger. Thus, we know that the purple, blue, and green curves are incorrect.

This leads us to compare the red and orange curves. Notice that the orange curve intersects the ๐‘ฆ-axis, while the red curve has a vertical asymptote. In order to determine which is correct, let us examine the behavior of the de Broglie wavelength equation near ๐‘=0 (the ๐‘ฆ-axis).

Notice that ๐‘ appears in the denominator of the equation, and we know that it is not possible to meaningfully divide by zero. Thus, as ๐‘ goes to 0, the de Broglie wavelength function goes to infinity. Therefore, the graph of the de Broglie wavelength versus momentum cannot have a defined value at ๐‘=0. Therefore, the red curve correctly illustrates the relationship between the momentum of a particle and its de Broglie wavelength.

Example 2: Relating Momentum to the de Broglie Wavelength

If an electron and a muon have the same speed, which particle has the greater de Broglie wavelength?

Answer

Let us begin by recalling the equation for the de Broglie wavelength of a particle, ๐œ†=โ„Ž๐‘.

Furthermore, recall that, for a particle moving at a speed much slower than the speed of light, momentum is equal to the product of mass, ๐‘š, and speed, ๐‘ฃ. Thus, the de Broglie wavelength can be found using ๐œ†=โ„Ž๐‘š๐‘ฃ.

The de Broglie wavelength is inversely proportional to momentum. So, because we are told that the particles are moving at the same speed, we can compare their masses to learn about their respective momentum values. A muon has a mass of 1.89ร—10๏Šฑ๏Šจ๏Šฎ kg, while an electron has a mass of 9.11ร—10๏Šฑ๏Šฉ๏Šง kg. The muon has a larger mass, and therefore a larger momentum, than the electron moving at the same speed. Because the de Broglie wavelength is inversely proportional to momentum, a greater momentum indicates a smaller de Broglie wavelength. Thus, we know that the muon has a smaller de Broglie wavelength.

Because the electron has the smaller momentum, we can conclude that the electron has the greater de Broglie wavelength.

Example 3: Calculating the de Broglie Wavelength of a Particle

What is the de Broglie wavelength of an electron that has a momentum of 4.56ร—10๏Šฑ๏Šจ๏Šญ kgโ‹…m/s? Use a value of 6.63ร—10๏Šฑ๏Šฉ๏Šช Jโ‹…s for the Planck constant. Give your answer in scientific notation to two decimal places.

Answer

We can begin by recalling the de Broglie wavelength equation, ๐œ†=โ„Ž๐‘.

Here, we are given values for the Planck constant, โ„Ž, and the momentum, ๐‘, of an electron. Thus, we have all we need to substitute into the equation: 6.63ร—10โ‹…4.56ร—10โ‹…/=1.454ร—10.๏Šฑ๏Šฉ๏Šช๏Šฑ๏Šจ๏Šญ๏Šฑ๏ŠญJskgmsm

Rounding to two decimal places, we have found that the de Broglie wavelength of this electron is 1.45ร—10๏Šฑ๏Šญ m.

If we are not directly given a value for momentum, we may need to calculate it ourselves, as shown in the next two examples.

Example 4: Calculating the de Broglie Wavelength of a Particle

A muon has a rest mass of 1.89ร—10๏Šฑ๏Šจ๏Šฎ kg. If the muon is moving at a speed of 20 m/s, what is its de Broglie wavelength? Use a value of 6.63ร—10๏Šฑ๏Šฉ๏Šช Jโ‹…s for the Planck constant. Give your answer in scientific notation to two decimal places.

Answer

Recall the equation for the de Broglie wavelength, ๐œ†=โ„Ž๐‘, where โ„Ž is the Planck constant and ๐‘ is the momentum. We do not yet know the momentum of the muon, but we do know that, for a particle of mass ๐‘š moving at a relatively low speed, ๐‘ฃ, momentum can be found using ๐‘=๐‘š๐‘ฃ. Since we have values for โ„Ž, ๐‘š, and ๐‘ฃ, we can substitute the momentum equation in and solve for ๐œ†: ๐œ†=โ„Ž๐‘=โ„Ž๐‘š๐‘ฃ6.63ร—10โ‹…(1.89ร—10)(20/)=1.754ร—10.๏Šฑ๏Šฉ๏Šช๏Šฑ๏Šจ๏Šฎ๏Šฑ๏ŠญJskgmsm

Rounding to two decimal places, we have found that the de Broglie wavelength for this muon is 1.75ร—10๏Šฑ๏Šญ m.

Example 5: Calculating the de Broglie Wavelength of a Particle

An electron has a rest mass of 9.11ร—10๏Šฑ๏Šฉ๏Šง kg. If the electron has a kinetic energy of 1.14ร—10๏Šฑ๏Šจ๏ŠญJ, what is its de Broglie wavelength? Use a value of 6.63ร—10๏Šฑ๏Šฉ๏Šช Jโ‹…s for the Planck constant. Give your answer in scientific notation to two decimal places.

Answer

We want to find the de Broglie wavelength, which is given by ๐œ†=โ„Ž๐‘=โ„Ž๐‘š๐‘ฃ, where โ„Ž is the Planck constant and momentum, ๐‘, is equal to the product of mass, ๐‘š, and speed, ๐‘ฃ. Since we already know the values of โ„Ž and ๐‘š, we only need to find a value for ๐‘ฃ to solve for the de Broglie wavelength. We are given the kinetic energy of an electron, so we can use the equation ๐ธ=12๐‘š๐‘ฃ๏Šจ to find the speed. First, let us rewrite the kinetic energy equation to solve for ๐‘ฃ, and then we can substitute in our known values for ๐ธ and ๐‘š: ๐‘ฃ=๏„ž2๐ธ๐‘š๏„Ÿ2(1.14ร—10)9.11ร—10=50.027/.๏Šฑ๏Šจ๏Šญ๏Šฑ๏Šฉ๏ŠงJkgms

Now, we are ready to solve for the de Broglie wavelength of this electron: ๐œ†=โ„Ž๐‘š๐‘ฃ6.63ร—10โ‹…(9.11ร—10)(50.027/)=1.4548ร—10.๏Šฑ๏Šฉ๏Šช๏Šฑ๏Šฉ๏Šง๏Šฑ๏ŠซJskgmsm

Rounding to two decimal places, we have found that the de Broglie wavelength of this electron is 1.45ร—10๏Šฑ๏Šซ m.

Let us finish by summarizing some important concepts.

Key Points

  • Particles with mass such as electrons and protons exhibit wave-like properties.
  • The wavelength of a massive particle is known as its de Broglie wavelength.
  • The de Broglie wavelength can be found using ๐œ†=โ„Ž๐‘, given an objectโ€™s momentum, ๐‘, and the Planck constant, โ„Ž.
  • The de Broglie wavelengths of the everyday objects we interact with are extremely small, so we do not notice their wave-like properties.

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