Question

A Geiger-Marsden type experiment is done by bombarding a 1.00 - \(\mu\) m thick gold foil with 8.00 - \(\mathrm{MeV}\) alpha rays. Calculate the fraction of particles scattered to an angle a) between \(5.00^{\circ}\) and \(6.00^{\circ}\) and b) between \(30.0^{\circ}\) and \(31.0^{\circ}\). (The atomic mass number of gold is 197 and its density is \(\left.19.3 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

Short answer

Question: For an alpha particle experiment, where alpha particles are scattered from a gold foil with a thickness of 1 µm and the particle kinetic energy given is 8 MeV, calculate the fraction of scattered particles for the following angle ranges: a) From 5.0 to 6.0 degrees b) From 30.0 to 31.0 degrees. Answer: To calculate the fraction of scattered particles for the given angle ranges, follow these steps: Step 1: Compute the area density of gold atoms in the gold foil. Step 2: Use Rutherford's Scattering Formula to calculate the differential cross-section for a given scattering angle. Step 3: Calculate the fraction of scattered particles for the angle ranges a) from 5.0 to 6.0 degrees and b) from 30.0 to 31.0 degrees using the area density and differential cross-section obtained in steps 1 and 2.

Step-by-step solution

Compute the area density of gold atoms

The mass of one gold atom can be calculated from its atomic mass number (197) and the Avogadro number (\(6.022 \times 10^{23}\,\mathrm{atoms/mol}\)). The area density of the gold atoms (atoms per cm^2) can be found by multiplying the gold foil thickness (in cm) by the gold density and dividing by the mass of one gold atom. 1.00 - \(\mu\)m thickness = \(1.00 \times 10^{-4}\,\mathrm{cm}\) Mass of one gold atom: $$ M_{\text{Au}}=\frac{197 \times 1.6605 \times 10^{-24}\,\text{g}}{1\, \text{mol}} $$ Area density (\(n\)) of the gold atoms: $$ n=\frac{1.00 \times 10^{-4}\,\mathrm{cm} \times 19.3\,\frac{\mathrm{g}}{\mathrm{cm}^{3}}}{M_{\text{Au}}} $$

Use Rutherford's Scattering Formula

Rutherford's scattering formula can be used to calculate the differential cross-section (\(d\sigma/d\Omega\)) for a given scattering angle. The formula is: $$ \frac{d\sigma}{d\Omega}= (\frac{Ze^2}{4 \pi \varepsilon_0})^2\frac{1}{4 E^2 \sin^4{(\theta/2)}} $$ Where \(Z\) is the atomic number of the target nuclei (gold: \(Z=79\)), \(e\) is the elementary charge, \(\varepsilon_0\) is the vacuum permittivity, \(4\pi\varepsilon_0\) is the Coulomb's constant (\(8.98755179 \times 10^{9} \, \mathrm{Nm^{2}/C^{2}}\)) and \(E\) is the kinetic energy of the alpha particles in Joules. First, convert the given 8 MeV to Joules: \(8\,\mathrm{MeV} = 1.28 \times 10^{-13}\,\mathrm{J}\). Plug the values into Rutherford's scattering formula to compute the differential cross-section for the given scattering angle range.

Calculate the fraction of scattered particles for the angle ranges

The fraction of scattered particles for a specific angle range can be calculated by integrating the scattering formula with respect to the solid angle. The solid angle \(\Delta \Omega\) can be approximated as \(2\pi (\cos{\theta_1} - \cos{\theta_2})\) for small angle differences. So, the fraction of scattered particles (\(I_{\text{sc}}\)) for the given angle range can be approximated as: $$ I_{\text{sc}} = n \times \Delta \Omega \times \frac{d\sigma}{d \Omega} $$ Compute the scattered particles fraction for the angle ranges: a) \(5.00^{\circ}\) to \(6.00^{\circ}\) b) \(30.0^{\circ}\) to \(31.0^{\circ}.\) Using the given values for the area density, scattering angle ranges, and differential cross-section obtained in steps 1 and 2, calculate the fraction of scattered particles for each angle range. The final answers are the fraction of particles scattered for angle ranges a) and b).

Key concepts

Rutherford's Scattering Formula

The key to understanding the Geiger-Marsden experiment lies in Rutherford's scattering formula. This powerful equation allowed scientists to derive information about the structure of atoms by analyzing how alpha particles scatter when they strike a target, like a thin gold foil.

At its core, Rutherford's scattering formula estimates the differential cross-section \( \frac{d\sigma}{d\Omega} \) of alpha particles when deflected by the Coulomb forces from the gold nuclei. It is expressed as:
\[ \frac{d\sigma}{d\Omega}= \left(\frac{Ze^2}{4 \pi \varepsilon_0}\right)^2\frac{1}{4 E^2 \sin^4{(\theta/2)}} \]
This equation takes into account the atomic number \( Z \) of the gold, the elementary charge \( e \) of the particles, the permittivity of free space \( \varepsilon_0 \) and the kinetic energy \( E \) of the alpha particles. The angle \( \theta \) here is the angle of scattering.

Understanding Rutherford's formula is crucial because it reveals that the scattering of particles is inversely proportional to the fourth power of the sine of half the scattering angle. This implies that very few particles are deflected at large angles, which was an unexpected outcome at the time, leading to the conclusion that the positive charge within an atom is concentrated in a very small nucleus.

Area Density of Gold Atoms

When we talk about the area density of gold atoms, we are essentially looking at how many gold atoms are packed into a specific area of the gold foil. This concept is vital for predicting the fraction of alpha particles that will scatter at certain angles.

To calculate this, we use the thickness of the gold foil, its density, and the atomic mass number of gold. Here's a brief overview:
\[ n = \frac{t \times d}{M_{\text{Au}}} \]
where \( n \) is the area density (\( \text{atoms/cm}^2 \) ), \( t \) is the thickness of the foil, \( d \) is the density of gold, and \( M_{\text{Au}} \) is the mass of one gold atom. By multiplying the gold's density by the foil's thickness and dividing by the atomic mass, we establish how densely the gold atoms are arranged across the foil. This data, combined with the scattering formula, lets us predict the scattering pattern of the particles hitting the foil.

Differential Cross-Section Calculation

The differential cross-section calculation is a key part of determining the scattering behavior of alpha particles. To calculate the probability of alpha particles being deflected by a certain angle, we need to find the differential cross-section for our particular setup.

To compute this value, we apply Rutherford's scattering formula and adjust it for the energy of the alpha particles and the angle of scattering. Once we have the differential cross-section \( \frac{d\sigma}{d\Omega} \) from the formula, we can then find the solid angle \( \Delta \Omega \) for the specific scattering angle range. For small differences in angles, this is approximately \( 2\pi (\cos{\theta_1} - \cos{\theta_2}) \).

Here, \( \theta_1 \) and \( \theta_2 \) denote the angle range we are considering, such as between \(5.00^\circ\) and \(6.00^\circ\). By integrating the differential cross-section across this solid angle, we can estimate the fraction of alpha particles that will scatter within these boundaries, giving us a quantitative understanding of the atom's structure.

Atomic Mass Number

The atomic mass number plays an integral role in the calculations involving the Geiger-Marsden experiment. It is essentially the total number of protons and neutrons in the nucleus of an atom, which for gold is 197. This number helps us to determine the mass of a single gold atom, a building block in calculating the area density of a gold foil.

In our calculations, we used the atomic mass number in conjunction with Avogadro's number \( (6.022 \times 10^{23} \, \text{atoms/mol}) \) to find the mass of a single gold atom:
\[ M_{\text{Au}} = \frac{197 \times 1.6605 \times 10^{-24} \, \text{g}}{1 \, \text{mol}} \]
With this information, we can then compute the area density and subsequently use it to find out the fraction of alpha particles deflected at different angles. It reveals how molecular and atomic constants interplay to provide a window into the microscopic world, allowing us to explore the fundamental principles governing matter.