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when firing a proton (for example) to an atomic nucleus, from a distance $D$, the deflection angle of the proton $\alpha $ to the type of changes atomic nuclei? or always constant?

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2 Answers 2

As with any other scattering reaction, it depends.

It should be obvious that heavier nuclei have a higher charge and thus a stronger repulsive effect. Likewise lower energy projectiles will be deflected through larger angles, but heavier mass projectiles will be less affected.

Assume a elastic interaction and $Q^2$ not more than, say, $0.1\text{ GeV}^2$, you can deduce it from the energy and impact parameter simply by integrating the electric force.

For stronger elastic interactions you'll need to use a more complete treatment including the form factors of the proton and the nucleus. Inelastic interaction are more complicated still.


You can find the first layer of math in an older answer of mine, which still applies to first order despite the differences in the reactions.

Most college quantum mechanics books do a treatment of a simple case. You're looking for the chapter or section in which "scattering" first appears.

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Not an up-to-date formula, but there is Rutherford's impact parameter formula (IIRC experimentally verified). It may be a useful approximation for you.

$$D=\frac{Z_1 Z_2 e^2}{4\pi \epsilon _o mv_0^2} \cot\frac{\alpha}{2}$$

Or,

$$\alpha =2\cot^{-1}\frac{4\pi D\epsilon _0 m v_0^2}{Z_1 Z_2 e^2}$$

where $Z_1,Z_2$ are the atomic numbers of nucleus and projectile, and $m_0$ is the mass of the projectile.

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