Deviations from Rutherford Scattering (HL) | HL IB Physics Revision Notes 2025

Deviations from Rutherford Scattering

Rutherford's scattering experiment predicted that
- As the scattering angle increases, the number of alpha particles scattered at that angle decreases
This was found to be correct at low to moderately high energies
However, at very high energies (>27.5 MeV) significant deviations from Rutherford's predictions were observed
Instead of the number decreasing at the expected rate, the number of alpha particles being back-scattered sharply decreases to zero

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The observed back-scattering from alpha particles strongly deviates from the predicted relationship based only on electromagnetic repulsion at 27.5 MeV

Rutherford's alpha scattering experiment originally assumed that the alpha particles only interact through electrostatic repulsion
However, if the energy of the alpha particles exceeds 27.5 MeV, then they will be close enough to interact with the nucleus via the strong nuclear force
Factoring in the interactions due to the strong nuclear force explains the scattering pattern observed in the experimental results
Therefore, deviations from Rutherford scattering provide evidence for the strong nuclear force

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Very high-energy alpha particles can get close enough to the nucleons that the effect of the strong nuclear force becomes significant

Worked example

Alpha particles undergo scattering after being fired at a thin gold ${}_{79}^{197}{Au}$ foil. The gold is then replaced to make a comparison.

Describe the predicted difference in the scattering pattern when the foil is replaced with aluminium ${}_{13}^{30}{Al}$ foil of the same thickness.

Answer:

Step 1: Compare the relative charges of the nuclei

The force between nuclei due to the electric repulsion is

$F = k \frac{q Q}{r^{2}} \Rightarrow r^{2} F = k q Q$

Therefore, the charge of a nucleus is proportional to the square of the distance between it and an alpha particle

$Q \propto r^{2}$

Gold has 79 protons, so $Q_{g o l d} = + 79 e$
Aluminium has 13 protons, so $Q_{a l u m i n i u m} = + 13 e$
Therefore, an alpha particle will get closer to the nucleus with less charge i.e. the aluminium nucleus than the gold nucleus

Step 2: Predict the patterns and deviations from Rutherford scattering

Deviations from Rutherford scattering occur when alpha particles get close enough for the strong nuclear force to begin to become more significant than the electric force
At very small separations (<1.5 fm) the effect of the strong nuclear force becomes significant
Alpha particles will be able to get closer to aluminium nuclei at lower energies than the gold nuclei
Therefore, alpha particles will be less affected by electric repulsion and able to get close enough for interactions with the strong nuclear force
Hence, more deviation will be seen with aluminium foil than with gold foil

Exam Tip

Make sure you can explain when the greatest deviations from Rutherford scattering will occur i.e.

When the alpha particles have high energies
The target nuclei have a low nucleon number

Deviations from Rutherford Scattering (HL) (HL IB Physics)

Revision Note