Episode 538: Electron scattering

This could follow work on Rutherford scattering, where alpha particles are used to identify the nucleus as the region of the atoms containing all the positive charge and most of the mass. Calculation of closest approach of alpha particles gives an estimate of nuclear size, but more direct experimental evidence is given by scattering of electrons.

Lesson Summary

• Student experiment: An optical analogue (10 minutes)
• Discussion: Diffraction graphs (10 minutes)
• Worked example: Calculating nuclear diameter (10 minutes)
• Student questions: Further examples (20 minutes)

Student experiment: An optical analogue

When waves pass through a collection of spheres, which look like circles to the oncoming waves, they diffract around them in the same way as through holes of the same size. This means that they produce a diffraction pattern with a minimum at roughly the angle given by the single-slit diffraction equation sin( θ ) =  λ d (it should really be sin( θ ) = 1.22 λ d for a circle; you could use that form, but the familiar equation is close enough here).

Look at a point source of light (a small circuitry lamp is fine) through a pinhole made in aluminium foil – the smaller the hole the better, so use a fine sewing needle or else pull thin copper wire until it snaps, and push the work-hardened tip through the aluminium foil. A diffraction pattern consisting of a circle with a surrounding ring (more than one if you’re lucky) should be visible. If you look at the point source through a microscope slide dusted with lycopodium powder (cornflour will work, but is not as good, because the particle sizes vary more) you see a similar pattern. You may like to refer to the halo seen around the Moon in the winter, which is due to similar scattering by ice crystals in the upper atmosphere.

The point here is that waves encountering round (spherical) objects will give a diffraction pattern, involving a central maximum, and then a minimum before the secondary maximum of the surrounding ring.

Discussion: Diffraction graphs

The graph below compares a typical electron diffraction pattern with the Rutherford alpha -scattering pattern and the single slit pattern. Show this graph, and identify the differences.

Episode 538-1: Rutherford alpha-scattering pattern & single slit pattern (Word, 26 KB)

The electron scattering clearly shows features of both Rutherford scattering (due to electrostatic interaction: the fact that the force is attractive rather than repulsive is not important) and diffraction. Concentrating on the minimum, it suggests that the nucleus is behaving as a circular object of diameter d.

Worked examples: Calculating nuclear diameter

The nuclear diameter d can be calculated from the electron scattering data. If your students can cope with the maths, you can work through a sample calculation.

(The virtue of the approach is not that students should be able to reproduce the mathematics, but that they should be able to see that a systematic application of the mathematics related to diffraction and the wave nature of the electron can give the size of the nucleus. Note: The calculation assumes that the rest energy of the electron is negligble.)

energy of electron = 100 MeV

energy of electron = 100 × 10⁶ V × 1.6 × 10^-19 C

energy of electron = 1.6 × 10^-11 J

electron momentum, p

p = Ec

p = 1.6 × 10^-11 J3 × 10⁸ m s^-1

momentum = 5.3 × 10^-20 kg m s^-1

de Broglie relationship, λ  = hp

λ  = 6.6 × 10^-34 J s^-15.3 × 10^-20 kg m s^-1

wavelength = 1.2 × 10^-14 m

The first diffraction minimum occurs at about 22 ° , so using the single slit diffraction equation

sin( θ ) =  λ d

rearrange this to get

d =  λ sin( θ )

d = 1.2 × 10^-14 msin(22 ° )

nuclear diameter = 3.3 × 10^-14 m

Student questions: Further examples

A structured question will allow your students to follow through this logic for themselves. It should be covered here if you did not do so in Episode 522.

Episode 522-6: Electrons measure the size of nuclei (Word, 42 KB)