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Chapter 2: Problem 17

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 2.7 A and \(1.0 \times 10^{-4} \hat{\mathrm{A}}\) , respectively. Hint: Calculate the cross sectional area occupied by the nucleus as a fraction of that occupied by the atom. Assume that the gold nuclei in each layer are offset from each other.

Short Answer

Expert verified
Approximately \(2.7394 \times 10^{-9}\) (0.0000000027394) of the α-particles are scattered at large angles in Rutherford's gold foil experiment, assuming a two-layer thick gold foil with offset nuclei.

Step by step solution

01

Calculate the area occupied by the gold atom

The diameter of the gold atom is given as \(2.7 \, \text{Å}\) = \(2.7 \times 10^{-10} \, \text{m}\). We can calculate the area occupied by the gold atom using: Area of gold atom = \(\frac{\pi (2.7 \times 10^{-10})^{2}}{4}\) = \(5.7325 \times 10^{-20} \, m^2\)
02

Calculate the area occupied by the gold nucleus

The diameter of the gold nucleus is given as \(1.0 \times 10^{-4} \, \text{Å}\) = \(1.0 \times 10^{-14}\, \text{m}\). Area of gold nucleus = \(\frac{\pi (1.0 \times 10^{-14})^{2}}{4}\) = \(7.8539 \times 10^{-29} \, m^2\) #Step 2: Calculate the cross-sectional area fraction# To calculate the fraction of the total area occupied by the gold nuclei as a fraction of that occupied by the gold atoms, we will divide the area occupied by the gold nucleus by the area occupied by the gold atom. Fraction of cross-sectional area = Area of gold nucleus / Area of gold atom
03

Calculate the fraction of the cross-sectional area occupied by the gold nucleus

Fraction of cross-sectional area = \(\frac{7.8539 \times 10^{-29} \, m^2}{5.7325 \times 10^{-20} \, m^2}= 1.3697\times 10^{-9}\) This fraction indicates that only approximately 1.3697 x \(10^{-9}\) (0.0000000013697) of the cross-sectional area is occupied by the gold nuclei as a fraction of the total area occupied by the gold atoms. #Step 3: Double the calculated fraction of nuclear cross-sectional area# Since the gold foil is two-layer thick, and the nuclei in each layer are offset from each other, we need to double the value obtained in Step 2 in order to account for the increased likelihood of α-particles scattering at large angles.
04

Double the fraction to account for two layers of gold nuclei

Probability of scattering α - particle at large angles = \(2 \times 1.3697 \times 10^{-9}\) = \(2.7394 \times 10^{-9}\) Thus, approximately \(2.7394 \times 10^{-9}\) (0.0000000027394) of the α-particles are scattered at large angles in Rutherford's gold foil experiment, assuming a two-layer thick gold foil with offset nuclei.

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