What fraction of the \(\alpha\) particles in Rutherford's gold foil experiment are scattered at large angles? Assume the gold foil is two layers thick, as shown in Figure \(2.9,\) and that the approximate diameters of a gold atom and its nucleus are 270 \(\mathrm{pm}\) and \(1.0 \times 10^{-2} \mathrm{pm}\), respectively. Assume that the gold nuclei in each layer are offset from each other.

To calculate the fraction of alpha particles scattered at large angles, we need to know the number of gold nuclei in the foil and the number of alpha particles incident on the foil. The gold foil is two layers thick, and the nuclei are offset from each other. Let's assume N represents the number of gold nuclei in the foil and M represents the number of alpha particles incident on the foil.

We will now compute the probability of an alpha particle interacting with a gold nucleus. The probability of interaction can be determined by considering the cross-sectional area of a gold nucleus and comparing it to the area occupied by a gold atom. The cross-sectional area of a gold nucleus can be approximated as a circle with radius \(r_n = 1.0 \times 10^{-2} \mathrm{pm}\), and the area of a gold atom as a circle with radius \(r_a = 270 \mathrm{pm}\). The probability of interaction can be determined as the ratio of the cross-sectional area of the nucleus to the area of the atom: \( P_{interaction} = \frac{\pi {r_n}^2}{\pi {r_a}^2} \) Notice that the pi term cancels out: \( P_{interaction} = \frac{{r_n}^2}{{r_a}^2} \)

Now, plug in the values for the radii of the gold atom and its nucleus: \( P_{interaction} = \frac{(1.0 \times 10^{-2} \mathrm{pm})^2}{(270 \mathrm{pm})^2} \) Calculate the probability of interaction: \( P_{interaction} = 1.37 \times 10^{-9} \)

Now we need to take into account that the alpha particles have the possibility to interact with two layers of gold nuclei. The probability that an alpha particle does not interact with either layer can be calculated as: \( P_{no\ interaction} = (1 - P_{interaction})^2 \) Calculate the probability of no interaction: \( P_{no\ interaction} = (1 - 1.37 \times 10^{-9})^2 \approx 0.9999999973 \) Now, the fraction of alpha particles scattered at large angles is equal to the complement of the probability of no interaction: \( P_{large\ angles} = 1 - P_{no\ interaction} \) Calculate the fraction of alpha particles scattered at large angles: \( P_{large\ angles} = 1 - 0.9999999973 \approx 2.7 \times 10^{-9} \) The fraction of alpha particles scattered at large angles in Rutherford's gold foil experiment is approximately \(2.7 \times 10^{-9}\).