Question

The differential cross section that will cause particles to scatter at an angle \(55^{\circ}\) off a target is \(4.0 \cdot 10^{-18} \mathrm{~m}^{2} / \mathrm{sr}\). A detector with an area of \(1.0 \mathrm{~cm}^{2}\) is placed \(1.0 \mathrm{~m}\) away from the target in order to detect particles that have been scattered at \(55^{\circ} .\) If \(3.0 \cdot 10^{17}\) particles hit the \(1.0-\mathrm{mm}^{2}\) -area target every second, how many will strike the detector every second?

Short answer

Answer: Approximately 120 particles strike the detector every second at an angle of 55 degrees.

Step-by-step solution

Determine the number of particles scattered at \(55^{\circ}\)

Using the given differential cross-section, we can find the number of particles scattered in a solid angle \(d\Omega\) at \(55^{\circ}\) using the equation: $$ dN = \frac{d\sigma}{d\Omega} \cdot \frac{N}{A}. $$ Here, \(\frac{d\sigma}{d\Omega} = 4.0 \cdot 10^{-18} \mathrm{~m}^2/ \mathrm{sr}\), \(N = 3.0 \cdot 10^{17}\) particles/s, and \(A = 1.0 \mathrm{~mm}^2 = 1.0 \cdot 10^{-6} \mathrm{~m}^2\). Plugging these values into the equation: $$ dN = 4.0 \cdot 10^{-18} \mathrm{~m}^2/ \mathrm{sr} \cdot \frac{3.0 \cdot 10^{17} \mathrm{particles/s}}{1.0 \cdot 10^{-6} \mathrm{~m}^2}. $$

Calculate dN

Now, calculate the value of \(dN\): $$ dN = 4.0 \cdot 10^{-18} \mathrm{~m}^2/ \mathrm{sr} \cdot 3.0 \cdot 10^{23} \mathrm{particles/ (\mathrm{m}^{2} \cdot \mathrm{s})} = 1.2 \cdot 10^6 \mathrm{particles/(\mathrm{sr}\cdot\mathrm{s})}. $$ This means that \(1.2 \cdot 10^6\) particles are scattered per steradian per second at an angle of \(55^{\circ}\).

Calculate the solid angle

Now, we need to calculate the solid angle \(d\Omega\) subtended by the detector. This can be found using the detector's area and the square of its distance from the target: $$ d\Omega = \frac{A_\text{detector}}{r^2}, $$ where \(A_\text{detector} = 1.0 \mathrm{~cm}^2 = 1.0 \cdot 10^{-4} \mathrm{~m}^2\) and \(r = 1.0 \mathrm{~m}\). Plugging in the values: $$ d\Omega = \frac{1.0 \cdot 10^{-4} \mathrm{~m}^2}{(1.0 \mathrm{~m})^2} = 1.0 \cdot 10^{-4} \mathrm{sr}. $$

Find the number of particles striking the detector

Finally, we can determine the number of particles that strike the detector every second using \(dN\) and \(d\Omega\). Multiply these two values to find the answer: $$ N_\text{detector} = dN \cdot d\Omega = 1.2 \cdot 10^6 \mathrm{particles/(\mathrm{sr}\cdot\mathrm{s})} \cdot 1.0 \cdot 10^{-4} \mathrm{sr} = 1.2 \cdot 10^2 \mathrm{particles/s}. $$ Thus, the detector is hit by approximately \(120\) particles every second.

Key concepts

Particle Scattering

Particle scattering is a fundamental process in physics where particles, such as photons, protons, or electrons, collide with a target and deviate from their original path. This phenomenon is crucial for understanding the microscopic structure and properties of materials. When a particle collides with a target, the likelihood of it scattering in a particular direction is described by the differential cross section.

In the practice problem, students calculate the number of particles scattering at a certain angle from the target. Observing particle scattering allows scientists to extract information about the forces and interactions at play. It's the bread and butter of experiments in nuclear and particle physics.

• It's imperative to know the incident particle flux (the number of particles that hit the target area per second).
• Equations connecting the differential cross section with the number of scattered particles are critical in these calculations.

Through particle scattering experiments, students not only learn about the interactions at a quantum level but also enhance their skills in applying mathematical concepts to physical scenarios.

Solid Angle

The solid angle is a measure of the 'spread' of an object as seen from a certain point, analogous to the way we consider angles in two dimensions for flat figures. Just as a regular angle is measured in degrees, a solid angle is measured in steradians, which represents the area on a sphere's surface enclosed by the angular spread of the object.

For instance, in our sample problem, the solid angle is crucial because it quantifies the area over which particles are scattered and thus detected. Calculating the solid angle requires knowledge of the area of the detector and its distance from the point of scattering.

• Remember, as the distance from the target increases, the subtended solid angle decreases for a fixed area of the detector, leading to fewer particles being detected.

Understanding solid angles aids in many fields, from astronomy to medical imaging, because it helps in analyzing systems where phenomena spread out in three dimensions.

Steradian

The steradian is the SI unit of solid angle. It helps us describe how large an area appears when looking from a certain point in three-dimensional space, essentially the 3D counterpart to the plane angle's radian. One steradian corresponds to the solid angle that, having its vertex at the center of a sphere, cuts off an area on the surface equal to the square of the sphere's radius.

Understanding the steradian is fundamental when working through problems involving particle scattering, as seen in the exercise where particles scatter at an angle and are detected over a solid angle.

• The differential cross section in the exercise is given in units of square meters per steradian, highlighting how many particles are expected to scatter per unit solid angle.
• Knowing this, students can connect the number of particles detected with the solid angle subtended by the detector.

Understanding the steradian enriches students' comprehension of geometrical concepts in a three-dimensional context.