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# Rutherford scattering

Rutherford scattering is a phenomenon that was observed by Ernest Rutherford in 1911 that led to the development of the orbital theory of the atom. It is now exploited by the materials analytical technique Rutherford backscattering. Rutherford scattering is also sometimes referred to as Coulomb scattering because it relies on static electric (Coulomb) forces.

The discovery was made by Rutherford when he performed a gold foil experiment, in which he fired a beam of alpha particles (helium nuclei) at layers of gold leaf only a few atoms thick. At the time of the experiment, the atom was thought to be analogous to a plum pudding (as proposed by Thomson), with the negative charges (the plums) found throughout a positive sphere (the pudding). If the plum-pudding model were correct, the positive “pudding”, being more spread out than in the current model of a concentrated nucleus, would not be able to exert such large coulombic forces, and the alpha particles should only be deflected by small angles as they pass through.

However, the intriguing results showed that some alpha particles were deflected by very large angles (over 90°). From this, Rutherford concluded that the majority of the mass was concentrated in a minute, positively charged region (the nucleus) surrounded by electrons. When a (positive) alpha particle approached sufficiently close to the nucleus, it was repelled strongly enough to rebound at high angles. The small size of the nucleus explained the small number of alpha particles that were repelled in this way. Rutherford showed, using the method below, that the size of the nucleus was about 10−14 m.

## Details of calculating nuclear size

For head on collisions between alpha particles and nucleus, all the kinetic energy ($\frac{1}{2}mu^2$) of the alpha particle is turned in to potential energy and the particle is at rest. The distance from the centre of the electron to the centre of the nucleus (b) is approximately then the radius.

Applying the inverse-square law between the charges on the electron and nucleus, one can write:

$\frac{1}{2} mu^2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{b}$

Rearranging:

$b = \frac{1}{4\pi \epsilon_0} \cdot \frac{2 q_1 q_2}{mu^2}$

For an alpha particle:

• m (mass) = 6.7×10−27 kg
• q1 = 2×(1.6×10−19) C
• q2 (for gold) = 79×(1.6×10−19) C
• u (initial velocity) = 2×107 m/s

Substituting these in gives the value of about 2.7×10−14 m.

Last updated: 05-07-2005 06:43:25
Last updated: 05-13-2005 07:56:04