Lesson Explainer: Electron Diffraction and Microscopy Physics

In this explainer, we will learn how to describe electron beam diffraction, how it is used in electron microscopy, and how other forms of electron microscopy compare to it.

Electrons are often described as particles, whereas diffraction is a wave phenomenon.

It is true that diffraction can occur in media that are made of particles; waves made of water can diffract, for example.

Electron diffraction is not equivalent to the diffraction of water waves, however. Electron diffraction does not refer to the diffraction of waves in a medium made of electrons.

In electron diffraction, electrons move through a material medium. The directions that these electrons travel in after they exit the medium is consistent with the electrons having moved in the medium as waves rather than as particles.

For electrons to move in a medium as waves rather than as particles, it is very helpful to understand how wave properties can be associated with an electron.

All particles are associated with a wavelength, called the de Broglie wavelength.

Formula: De Broglie Wavelength of a Particle

For a particle that has a momentum 𝑝, the de Broglie wavelength 𝜆 of the particle is given by 𝜆=𝑝, where is the Planck constant, which has an approximate value of 6.634×10.Js

Let us now look at an example involving the de Broglie wavelength of an electron.

Example 1: Identifying the Waveform of an Accelerating Electron

A particle accelerator accelerates electrons through the potential difference between 𝑉 and 𝑉, as shown in the diagram. The smallest value of the velocity of the electron is at 𝑉. Which waveform corresponds to that of an electron moving across the accelerator?

Answer

As the electron passes along the accelerator, it increases in velocity. The momentum of the electron therefore also increases.

The effect of an increase in momentum on the de Broglie wavelength of the electron can be seen by considering the formula 𝜆=𝑝, where 𝑝 is the momentum of the electron, 𝜆 is its de Broglie wavelength, and is the Planck constant.

The momentum of the electron is increasing, and therefore, its de Broglie wavelength must be decreasing. The longest de Broglie wavelength must be at 𝑉 and the shortest de Broglie wavelength must be at 𝑉.

The waveform that corresponds to this description is shown within the accelerator in the following figure.

This waveform corresponds to the following answer option:

That electrons do actually have wavelengths has been demonstrated by producing the diffraction of electrons.

We recall that diffraction of waves that pass through an aperture is greatest for an aperture with a width equal to that of the wavelength of the waves.

Let us consider an electron that has a velocity of 1‎ ‎000 m/s. The momentum of the electron is given by 𝑝=𝑚𝑣𝑝9.1×10×1000𝑝9.1×10.kgmskgms

This momentum value results in a de Broglie wavelength of 𝜆6.634×109.1×10𝜆6.634×109.1×10𝜆7.3×10.Jskgkgskgmmsmsms

Electrons with velocities of 1‎ ‎000 m/s passing through an aperture about 730 nanometres wide should be maximally diffracted.

The greater the velocity of electrons passing through an aperture, the less wide the aperture should be to produce maximum diffraction.

Crystalline materials can be used to diffract electrons. The planes of the lattice of a crystal correspond to the slits in screens used in the diffraction of light waves, as shown in the following figure.

Light waves passing through a set of slits produce a light intensity pattern of alternating light and dark fringes, as shown in the following figure.

Light waves

The pattern is formed by interference between the waves passing through the slits.

A beam of electrons diffracted by a crystal lattice produces a similar intensity pattern. The crystal lattice has length in perpendicular directions, unlike a slit that is much longer than it is wide. Rather than fringes, then, an electron beam intensity pattern consists of concentric rings.

If electrons had no wave properties and moved only as particles move, the intensity distribution of electrons exiting a crystal would look something like the following figure.

This intensity distribution shows no interference. If interference occurred, the distribution would look more like the following figure.

Where interference occurs, an intensity distribution contains alternating intensity maxima and minima.

For a crystal to produce alternating intensity maxima and minima for an electron beam exiting the crystal, there would have to be some angles by which electrons were deflected that produce constructive interference and some that produce destructive interference, as shown in the following figure.

Let us now look at an example involving the intensity patterns produced by diffracted electron beams.

Example 2: Comparing the Intensity Distributions of Different Diffracted Electron Beams

A beam of electrons passes through a crystal. A diffraction pattern of concentric rings is formed on a screen behind the crystal that records the positions of electrons that arrive at it, as shown in the diagram. The intensity of the rings is plotted against the radial distance from the center of the pattern. The resulting intensity distribution is shown three times, each time compared to another intensity distribution that is shown below it.

  1. Which of the intensity distributions would result from decreasing the velocity of the electrons in the beam?
  2. Which of the intensity distributions would result from decreasing the charge density of the electron beam while not changing the velocity of the electrons in the beam?

Answer

Part 1

Only changing the velocity of electrons in the beam would not affect the number of electrons in the beam. The magnitude of the peaks of the intensity distribution would not be affected by a change in electron velocity. We can therefore eliminate distribution I, as in this distribution, we see that the intensity has decreased across the distribution.

Decreasing the velocity of electrons decreases the momentum of the electrons. The effect of this on the de Broglie wavelengths of the electrons can be seen by considering the formula 𝜆=𝑝, where 𝑝 is the momentum of an electron, 𝜆 is its de Broglie wavelength, and is the Planck constant.

Decreasing electron momentum would increase the electron de Broglie wavelength. The question is therefore asking whether intensity distribution II or III could correspond to an increased electron de Broglie wavelength.

Increasing the electron de Broglie wavelength would result in greater separations of the concentric rings produced by the interfering diffracted electrons.

Careful inspection of distribution II shows that it corresponds to the ring separation decreasing slightly, while distribution III shows a slight increase in the ring separation. We see then that distribution III is the correct option.

Part 2

The velocity of the electrons is kept constant, and so, therefore, is the de Broglie wavelength of the electrons. This tells us that the separations of the concentric rings in the intensity distribution will not change. This is only shown in distribution I.

Reducing the charge density in the electron beam corresponds to using a beam that consists of fewer electrons. The intensities in a pattern produced by a smaller number of electrons would be decreased. In distribution I, the intensity has decreased across the distribution. Distribution I is then the correct option.

Let us now look at an example in which the diffraction of electrons in a beam by a crystal is considered.

Example 3: Comparing the Paths of Electrons in a Diffracted Electron Beam

The diagram shows some parts of an electron beam that is passing through a crystal lattice. The lattice has parallel planes separated by a perpendicular distance 𝑑. Some of the electrons in the beam are scattered by the atoms of the lattice. The electrons all have a wavelength 𝜆. Each of the blue dotted lines in the diagram corresponds to a separate wave. The waves at points 𝐴 and 𝐵 are in phase with each other, and the waves at points 𝐵 and 𝐶 are in phase with each other. Lines 𝐿 and 𝐿 are parallel.

  1. Which of the following correctly describes the length of the path traveled by electrons between point 𝐴 and point 𝐶?
    1. The length is equal to 𝑛𝜆, where 𝑛 is an integer.
    2. The length is equal to 𝑛𝜆2, where 𝑛 is an integer.
    3. The length is equal to 𝑑.
    4. The length is equal to 𝑛𝜆𝑑, where 𝑛 is an integer.
    5. The length is equal to 𝑛𝜆𝑑, where 𝑛 is an integer.
  2. Which of the following correctly describes the relationship between angles 𝜃 and 𝜃?
    1. 𝜃=𝜃
    2. 𝜃>𝜃
    3. 𝜃<𝜃

Answer

Part 1

The question states that the diagram shows some parts of an electron beam. This tells us that not all the electrons in the beam are shown. To be specific, the diagram shows us only two parts of the beam:

  • electrons that are transmitted through the crystal without deflection, which contribute to the central maximum of the intensity distribution,
  • electrons that are deflected by one particular angle.

All the other electrons in the beam that would be deflected by other angles than these are not shown. These other electrons are omitted to make it possible to clearly see what happens to the electrons that are shown.

The question asks about the length of the path that electrons travel from the point 𝐴 to the point 𝐶. This is shown in the following figure.

The question describes the electrons as waves and states that they all have the same wavelength. Electrons have a de Broglie wavelength, so this is a valid description of the electrons.

The question states that the de Broglie waves of the electrons at 𝐴 are in phase with de Broglie waves of the electrons at 𝐵.

The question also states that the de Broglie waves of the electrons at 𝐶 are in phase with de Broglie waves of the electrons at 𝐵.

For the de Broglie waves of the electrons at both 𝐴 and at 𝐶 to be in phase with the de Broglie waves of the electrons at 𝐵, the path followed by electrons between 𝐴 and 𝐶 must have a length that is exactly divisible by the de Broglie wavelength of the electrons. This is represented in the following figure.

In the figure, the distance between 𝐴 and 𝐶 is 𝜆. The waves at 𝐴 and 𝐶 would also be in phase if this distance was 2𝜆, 3𝜆, or any integer multiple of 𝜆.

We see then that the waves are in phase if the distance between 𝐴 and 𝐶 is equal to 𝑛𝜆, where 𝑛 is an integer.

If the distance between 𝐴 and 𝐶 was equal to 𝑛𝜆2, however, the waves would be out of phase.

Let us now consider how 𝑑, the distance between the planes of the crystal lattice, relates to the distance between 𝐴 and 𝐶. This occurs in the options where the distance equals either 𝑑, 𝑛𝜆𝑑, or 𝑛𝜆𝑑.

Neither the values of 𝑑 nor those of 𝜆 are known, so it is not obvious whether any particular value of d will result in the waves at 𝐴 and 𝐶 being in phase with each other. For these options to be correct, they must be correct for any value of 𝑑.

We can suppose that 𝑑 is much shorter than the de Broglie wavelengths of the electrons. If we assume this, then the distance between 𝐴 and 𝐶 cannot equal 𝑑.

We can just as well suppose that this is true for the case of 𝑛𝜆𝑑 and for 𝑛𝜆𝑑. We can always assume a value of 𝑑 for which the waves at 𝐴 and 𝐶 are not in phase with each other.

The only distance between 𝐴 and 𝐶 for which the waves at those points must be in phase with each other is 𝑛𝜆.

Part 2

The angles 𝜃 and 𝜃 are shown in the following figure.

The angle 𝜃 relates to the distance that electrons deflected by adjacent atoms in a crystal lattice must travel to be in phase with each other.

The angle 𝜃 relates to the deflection of electrons exiting the crystal that interfere constructively.

We know that constructive interference is the result of adding waves that are in phase with each other, so we might expect that 𝜃 is equal to 𝜃. This is correct. The equality of the angles is demonstrated in the following sequence of figures.

We see from the final figure that 𝜃2=𝜃2.

The angles 𝜃 and 𝜃 depend on the de Broglie wavelength of diffracted electrons and on 𝑑, the distance between the planes of the crystal lattice.

As these angles are equal, it is possible to use the angles that electrons are deflected by to arrive at the maxima and minima of an intensity distribution to determine the distance between the planes of a crystal lattice used to diffract electrons.

Suppose that a substance has a structure that will diffract an electron beam passed through an object made of that substance. Suppose also that the details of the structure are not known.

We know that there is a relationship between the distances between lattice planes of a regular structure and the de Broglie wavelength of electrons diffracted when passing through the structure. Assuming that the de Broglie wavelength is known, the distances can be determined.

An object made of a substance with an unknown structure can be used to diffract electrons that are incident on the object at different angles and that have different de Broglie wavelengths.

By comparing the intensity distributions of the electrons at different angles and with different wavelengths, it is possible to construct a detailed model of the structure of the substance. This technique is known as transmission electron microscopy.

The following figure shows the components of a transmission electron microscope.

A transmission electron microscope uses either magnetic or electrostatic lenses to deflect electron beams. A magnetic lens consists of multiple curved magnets. An electrostatic lens consists of charged parallel plates.

Let us now look at an example concerning using electrons to produce images.

Example 4: Identifying the Advantages of Using Electrons to Produce Images

Which of the following correctly states the advantage of using electrons to produce images of very small objects compared to using electromagnetic waves?

  1. Electrons can easily be accelerated to velocities at which they have wavelengths much shorter than those of electromagnetic waves with wavelengths that are useful in forming images.
  2. Electrons can penetrate deeper into objects than electromagnetic waves.
  3. Electrons are reflected from objects more strongly than electromagnetic waves.
  4. A beam of electrons will in no way affect an object that it produces an image of, so it produces a more valid image than the ones that can be produced by electromagnetic waves.

Answer

Let us first see which options can be eliminated.

We recall that some frequencies of electromagnetic waves, such as X-ray and gamma-ray frequencies, can pass through even very dense solid objects. Gamma rays of high energy can penetrate several metres of lead. From this, we see that the ability to penetrate objects is not the main reason for using electrons to produce images.

We can recall that some frequencies of electromagnetic waves are almost completely reflected from some substances, such as mirrored or white surfaces. From this, we see that the ability to reflect from objects is not the main reason for using electrons to produce images.

When a beam of electrons passes through an object, electrons in the beam can transfer energy to the object. An electron beam is a type of electric current. It is not sensible to suppose that passing an electric current through an object has no effect on the object.

Let us consider the de Broglie wavelength of an electron. This is given by the formula 𝜆=𝑝, where 𝑝 is the momentum of the electron, 𝜆 is its de Broglie wavelength, and is the Planck constant.

Let us consider what momentum an electron would have and thereby have the same wavelength as the shortest wavelength of visible light, 400 nanometres. We rearrange the formula to make 𝑝 the subject, setting 𝜆 to equal 4×10 m: 𝑝=6.634×104×10𝑝=6.634×104×10𝑝=1.685×10/.Jsmkgsmkgmsms

The momentum of the electron is given by the formula 𝑝=𝑚𝑣.

We can make 𝑣 the subject of the formula and use 9.1×10 kg for the electron mass: 𝑣=𝑝𝑚=1.685×109.1×10𝑣1823/.kgkgmsms

An electron can be accelerated to this velocity using a potential difference. A potential difference, 𝑉, will transfer do work, 𝑊, on an electron given by 𝑊=𝑉𝑒, where 𝑒 is the charge of the electron, 1.6×10 C.

The kinetic energy, 𝐸, of an electron that has a velocity of 1‎ ‎823 m/s is given by the formula 𝐸=12𝑚𝑣𝐸=12×9.1×10×(1823/)𝐸1.5×10.kgmsJ

The value of 𝐸 equals the work done on an initially stationary electron to increase its velocity to 1‎ ‎823 m/s.

We have then that 𝑉×1.6×10=1.5×10𝑉=1.5×101.6×10=9.375×10.CJJCV

We see that only a very small potential difference is needed to accelerate an electron sufficiently to reduce its wavelength to 400 nanometres.

Suppose a greater voltage was used, say 125 V. We have then that 125=𝐸1.6×10𝐸=125×1.6×10=2×10.VCVCJ

The velocity of the accelerated electron is then given by 2×10=12×9.1×10×(𝑣/).Jkgms

Making 𝑣 the subject, we find that 𝑣=2×10×9.1×106.6×10/.Jkgms

The momentum of the electron is given by 𝑝=9.1×10×6.6×10=6.006×10/.kgmskgms

The wavelength of the electron is given by 𝜆=6.634×106.006×101.1×10.kgskgmmsms

The distance between atoms in a solid object is typically approximately 10 m.

We see then that a not very great potential difference, only 125 V, is sufficient to reduce the wavelength of electrons to the distances between atoms in a solid object. This is the main reason for using electrons rather than electromagnetic waves to image objects.

Let us now summarize what has been learned in this explainer.

Key Points

  • Electrons can be diffracted by the planes of the lattice of a crystal.
  • The intensity distributions of electron beams with different wavelength electrons, incident on an object at different angles, can be used to determine the internal structure of the object. This is called transmission electron microscopy.
  • Electron microscopy can determine the structure of an object at scales comparable to the distances between atoms in an object.
  • Electron beams are focused using magnetic and electrostatic lenses.

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