Question

A neutrino beam with \(E=143 \mathrm{GeV}\) is passed through a slab of aluminum- 27 (with 27 nucleons in each nucleus). The probability that a neutrino in the beam will scatter off a nucleon in the aluminum slab is \(4.19 \cdot 10^{-12}\) The scattering cross section is given by \(\sigma(E)=\left(0.68 \cdot 10^{-38} \mathrm{~cm}^{2} \mathrm{GeV}^{-1}\right) E,\) and aluminum has a density of \(2.77 \mathrm{~g} / \mathrm{cm}^{3} .\) How thick is the slab?

Short answer

Based on the given information and the step-by-step solution explained above, to find the thickness of the aluminum slab, we should follow these steps: 1. Determine the number of nucleons per unit volume in the aluminum. 2. Determine the scattering cross section per unit volume. 3. Calculate the probability of interaction per unit length. 4. Integrate to find the total probability at a general distance x. 5. Solve the integral to find the total probability at a general distance x. 6. Solve for the thickness of the aluminum slab, x, by dividing the total probability (4.19 × 10^{-12}) by the constant product nσ(E). By substituting the values provided and following the steps, we can calculate the thickness of the aluminum slab.

Step-by-step solution

Determine the number of nucleons per unit volume in the aluminum

To find the number of nucleons per unit volume in the aluminum, first convert the mass of a single aluminum atom to grams. Since there are 27 nucleons in each nucleus, the mass of a single aluminum atom is \(27 \times 1.66 \times 10^{-24} \mathrm{g}\), where \(1.66 \times 10^{-24} \mathrm{g}\) is the mass of a single nucleon. Then, divide the density of aluminum by the mass of a single aluminum atom to find the number of aluminum atoms per unit volume. Finally, multiply by the number of nucleons in each aluminum atom to arrive at the number of nucleons per unit volume.

Determine the scattering cross section per unit volume

Now that we have the number of nucleons per unit volume, we can find the scattering cross section per unit volume by multiplying the scattering cross section formula, \(\sigma(E)\), by the number of nucleons per unit volume. This will give us the scattering cross section per unit volume, which we can then use to find the thickness of the aluminum slab.

Calculate the probability of interaction per unit length

The probability of a neutrino interacting with a nucleon as it traverses a short distance \(\Delta x\) through the aluminum slab is equal to the cross-section per unit volume multiplied by \(\Delta x\). As \(\Delta x \to 0\), this is described as follows: $$ \frac{dP}{dx} = n\sigma(E), $$ where \(n\) is the number of nucleons per unit volume, and \(\sigma(E)\) is the cross-section for interaction.

Integrate to find the total probability at a general distance x

To find the total probability of interaction at a general distance x through the slab, integrate the probability of interaction per unit length: $$ P(x) = \int_0^x \frac{dP}{dx'}dx'. $$

Solve the integral to find the total probability at a general distance x

Using the fact that the cross-section per unit volume is a constant with respect to \(x\), we can rewrite the integral as: $$ P(x) = n\sigma(E) \int_0^x dx' = n\sigma(E)x, $$ where \(n\sigma(E)\) is the constant scattering cross section per unit volume.

Solve for the thickness of the aluminum slab

We know the total probability of interaction is given as \(4.19 \times 10^{-12}\). Using the equation for the total probability at a general distance \(x\), we can solve for the thickness of the aluminum slab, which is equal to the distance \(x\) when the total probability equals \(4.19 \times 10^{-12}\): $$ 4.19 \times 10^{-12} = n\sigma(E) x. $$ Solve for \(x\) by dividing both sides of the equation by the constant product \(n\sigma(E)\): $$ x = \frac{4.19 \times 10^{-12}}{n\sigma(E)}. $$ Using the values calculated in the previous steps, plug in \(n\) and \(\sigma(E)\) to find the thickness of the aluminum slab, \(x\).

Key concepts

Cross-section

In the world of particle physics, one of the key terms that you will encounter is the 'cross-section'. But what exactly does this mean? Think of it like a target area – the cross-section is a measure of the probability of a specific interaction occurring between particles, such as neutrinos and nucleons. When a neutrino passes close enough to a nucleon, if the 'target area' is hit, the neutrino is said to scatter off the nucleon.

Mathematically, the cross-section is often represented in terms of area, despite being a probability. It's measured in strange but standard units like square centimeters (cm²) or even smaller. The greater the cross-section of an interaction, the more likely the event is to occur. In the exercise provided, the cross-section of neutrino scattering off a nucleon was dependent on energy, showcasing a fundamental aspect of particle interactions: as energy changes, so does the likelihood of scattering.

Nucleons

Nucleons make up the core of every atom – they are the protons and neutrons residing within the nucleus. Each element is defined by the number of protons it has, while the number of neutrons can vary, giving rise to different isotopes of the same element. In the context of the given exercise, we refer to the nucleons in an aluminum-27 atom, which has a total of 27 nucleons (13 protons and 14 neutrons).

Understanding nucleons is vital because they serve as the 'scattering centers' for neutrinos, meaning that neutrinos are most likely to interact with these particles. This is due to the weak force, a fundamental force that governs such particle interactions. For the aluminum-27 slab in the exercise, knowing the number of nucleons provides key data for calculating the total probability of a neutrino scattering event.

Neutrino interaction probability

Neutrinos are known for being ghostly particles, often passing through matter without any interaction. Yet, despite their elusive nature, there is a finite probability that a neutrino will interact with matter, governed by what's called the 'neutrino interaction probability.'

This probability is a result of the cross-section, the density of the material it's passing through, and the thickness of the material. For a neutrino to scatter, it must encounter a nucleon, and this likelihood can be mathematically described and predicted. As the exercise describes, this probability is very low. However, by applying precise mathematical methods, such as integration over the distance the neutrino travels, we can determine at what thickness of material a neutrino would have a certain probability of interacting.

Density of materials

The density of a material can be crucial information when considering particle interactions. It merely isn't enough to know how many nucleons there are; we also must consider how tightly these nucleons are packed in a given space – that is the density, typically expressed in grams per cubic centimeter (g/cm³).

Density ties directly into our understanding of neutrino interactions because it dictates the number of potential interaction sites within a given volume of material. The exercise provided uses aluminum's density to calculate the scattering cross section per unit volume. Remember, a higher material density means more nucleons are packed into the same space, thus increasing the probability that a neutrino will interact within that space.