Multiple Scattering in Rutherford Experiments

In Rutherford scattering experiments, a metal foil of finite thickness is bombarded by alpha particles. At what thickness of foil do you have to worry about multiple scattering? Some simple calculations will be explored here to suggest that with the gold foil thicknesses around 1 micrometer in the historical experiments, there was very little multiple scattering. Using the concept of cross section, the fraction of the particles scattered can be taken to be the effective area of the scatterer compared to the total area.

在卢瑟福散射实验中，有限厚度的金属箔被α粒子轰击。在什么厚度的箔上需要考虑多次散射？在这里，将通过一些简单的计算来探讨这个问题，表明在历史上使用的金箔厚度约为1微米时，多次散射的影响非常小。利用截面的概念，可以将被散射的粒子分数视为散射体的有效面积与总面积的比值。

Gold, for example, has a nuclear radius about 7 fermis and an atomic radius of about 0.13 nm. The geometrical cross section of the nucleus is about 154 fm² or 1.54 barns. The cross sectional area of the atom is then about 3.45 x 10⁸ times the area of the nucleus. You would then expect that if you fired a neutron into a gold atom, the probability of hitting the nucleus would be about one in 345 million!

例如，金的核半径约为7费米，原子半径约为0.13纳米。核的几何截面积约为154平方费米或1.54拜耳。原子的截面积则约为核截面积的3.45×10⁸倍。因此，如果你向金原子发射一个中子，击中核的概率应约为1 in 345 million！

One simple approach to estimating the probability for multiple scattering it to suggest that for a given scattered particle, the remaining targets would have to cover 1% of the area for you to have a 1% probability of a second scattering event. Using the area ratio, you would project 3.45 x 10⁶ layers of atoms to cover 1% of the area with nuclei, assuming a random distribution. At a nominal distance between layers equal to the atomic diameter, this projects to about 900 micrometers in thickness.

一种估算多重散射概率的简单方法是，建议对于给定的散射粒子，剩余的目标必须覆盖1%的面积，才能有1%的第二次散射事件的概率。使用面积比，你将投影出3.45×10⁶层原子，以覆盖1%的面积中的核，假设随机分布。在层间名义距离等于原子直径的情况下，这将投影出约900微米的厚度。

For alpha particle scattering, the cross section is much larger than the geometric cross section because of the Coulomb force. For 6 MeV alpha particles, the cross section calculation for scattering above 30° gives 15730 fm² = 157.3 barns. The ratio of atomic area to this area is about 3.38 x 10⁶. Using the same simple approach, it would take about 3.38 x 10⁴ atomic layers to scatter 1%, which projects to a thickness of about 9 micrometers.

对于α粒子散射，由于库仑力的作用，其截面远大于几何截面。对于6 MeV的α粒子，散射角超过30°的截面计算结果为15730 fm² = 157.3 barns。这个面积与原子面积的比值约为3.38 x 10⁶。采用同样的简单方法，要使1%的粒子发生散射，需要约3.38 x 10⁴个原子层，这对应于约9微米的厚度。

This simple approach suggests that the 1 micrometer thickness of the gold foils used in Rutherford experiments was thin enough to minimize concern about multiple scattering.

这种简单的方法表明，用于卢瑟福实验的金箔厚度为1微米，足够薄，可以最小化对多次散射的担忧。