Problem 17 - Chemistry The Central Science In Si Units

Question

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. 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.

Step-by-Step Solution

Answer

The fraction is approximately \(2.74 \times 10^{-9}\), indicating a very small scattering probability at large angles.

Step 1: Understand the problem

We need to compute the fraction of alpha particles scattered at large angles in Rutherford's gold foil experiment, which involves understanding the relation between the size of the nucleus and the whole atom. We need to calculate the cross-sectional area covered by the nucleus and compare it with the area covered by the entire atom.

Step 2: Calculate the area of a gold atom

The diameter of a gold atom is 270 pm. The radius is half of the diameter, which is 135 pm. We can use the formula for the area of a circle, which is \( A = \pi r^2 \), where \( r \) is the radius, to calculate the cross-sectional area of the atom.\[ A_{\text{atom}} = \pi (135 \text{ pm})^2 \approx 57253.36 \text{ pm}^2 \]

Step 3: Calculate the area of a gold nucleus

The diameter of a gold nucleus is given as \(1.0 \times 10^{-2} \) pm. Thus, the radius is \(0.5 \times 10^{-2} \) pm. Again, using the area of a circle formula:\[ A_{\text{nucleus}} = \pi (0.5 \times 10^{-2} \text{ pm})^2 \approx 7.85 \times 10^{-5} \text{ pm}^2 \]

Step 4: Compute the fraction of the area occupied by the nucleus in one layer

To find the fraction of the area that the nucleus occupies compared to the entire atom, divide the area of the nucleus by the area of the atom:\[ \text{Fraction in one layer} = \frac{A_{\text{nucleus}}}{A_{\text{atom}}} = \frac{7.85 \times 10^{-5} \text{ pm}^2}{57253.36 \text{ pm}^2} \approx 1.37 \times 10^{-9} \]

Step 5: Consider the two-layer foil

Since the gold foil is two layers thick and the nuclei are offset, the chances of an alpha particle hitting a nucleus increase. For two layers offset, the fraction is doubled:\[ \text{Fraction for two layers} = 2 \times 1.37 \times 10^{-9} = 2.74 \times 10^{-9} \]

Step 6: Interpret the result

The probability or fraction of alpha particles being scattered at large angles corresponds to this small fraction. Thus, very few alpha particles are scattered at large angles due to the small size of the nucleus compared to the entire atom.

Key Concepts

Alpha Particle Scattering

In Rutherford's gold foil experiment, alpha particles were directed at a thin sheet of gold. These particles are essentially helium nuclei, being composed of two protons and two neutrons. The key observation was that most alpha particles passed through the foil with little to no deflection. However, a tiny fraction was deflected at significant angles, and an even smaller number rebounded almost directly back. This scattering phenomenon was critical in shaping our understanding of the atomic structure.

• When alpha particles encounter the dense, positively charged nucleus of a gold atom, they experience strong repulsive electrostatic forces.
• This causes large-angle deflections, contrasting drastically with the expected behavior if the atom had been a homogenous spread of positive charge.
• Such encounters are rare because the nucleus occupies only a tiny fraction of the atom's volume.

Nuclear Dimensions

The size of the nucleus compared to the atom is pivotal in understanding the results of the alpha particle scattering. In the case of gold, the diameter of an atom is about 270 picometers (pm), while the nucleus is much smaller, roughly 0.01 pm. This disparity helps explain why only a small fraction of alpha particles are deflected at significant angles.

• The nucleus is incredibly dense and holds almost all the atomic mass, despite its small size.
• This density means that when an alpha particle meets the nucleus, a considerable force acts on it, leading to significant deflection.
• The vast empty space surrounding the nucleus, where electrons are located, offers little obstruction, allowing most alpha particles to pass through undisturbed.

Atomic Structure

Rutherford's experiment reshaped the model of the atom. Before this, the atom was thought to be a "plum pudding" model, a homogenous mix of positively charged material with electrons embedded within. The experiment's outcomes necessitated a new model, introducing the concept of a central nucleus.

• The nucleus was posited as a small, dense center comprising protons and neutrons.
• The electrons orbit this nucleus in the much larger area of the atom, maintaining overall electrical neutrality.
• This nuclear model of the atom laid the foundation for future discoveries about atomic structure and behavior.

Cross-Sectional Area Calculation

Calculating the cross-sectional area is a crucial step in understanding why so few alpha particles are deflected by the nuclei in Rutherford's experiment. The cross-section is visualized by imagining a slice through the atom, capturing the relative space occupied by the nucleus.
The area of the nucleus and the entire atom can be estimated using the formula for the area of a circle: \( A = \pi r^2 \). This helps determine the likelihood of an alpha particle encountering a nucleus.

• For a gold atom with a diameter of 270 pm, its radius is 135 pm, leading to a cross-sectional area of approximately 57253.36 pm².
• Comparatively, a gold nucleus with a diameter of 0.01 pm has a radius of 0.005 pm, and its area is approximately 7.85 x 10⁻⁵ pm².
• When calculated, the nucleus occupies an immeasurably small fraction of the total area, showing why only a few alpha particles are deflected at larger angles.