Flux Calculation for Sphere of Radius R and Detector Acceptance D

In summary, the conversation discussed the calculation of particle flux over a single side of a detector of area A inside a sphere with isotropic particle density n(r,E) and radius R. The first approach suggested calculating the flux as n*V/4pi where V is the mean velocity of particles and 4pi is the total solid angle. It was also noted that the flux over one side of the detector would be half of the total flux. The second approach considered the acceptance of the detector, with a suggestion to multiply the total flux by the acceptance to get the flux over the detector's surface. A suggestion was made to integrate over the solid angle from the surface of the detector to a random point on an arbitrary concentric surface to
  • #1
hokutose
2
0
Hi everyone,
I have a simple problem for you. If we have an isotropic particle density n(r,E) where r is the radius and E the energy in a sphere of radius R, What would be the particle flux over a single side of a detector of area A inside the sphere?
What would it be also if the detector's acceptance would be D?
The first approach i have tried is that the flux should be around n*V/4pi where n is the particle density and v the mean velocity of the particles and 4pi the total solid angle.The flux over one side of the detector should be 1/2 of the total flux as it is isotropic.
In the case of the acceptance of the detector, simply multiplying the total flux n*v/4pi by the acceptance would be the flux over the detector´s surface.
Thanx and regards.
 
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  • #2
I have tried the following course; if the surface detector is squared shape with Area equal to A ,then, integrating over the solid angle from the surface of the detector to a random point of an arbitrary surface concentric and coaxial with the detector(lets say a cube), we have that the acceptance will be pi*A.(The result of integrating the product of the differential area of the surface and the solid angle ).
With this, the flux in the detector would be the total flux,ie n*v/4pi multiplied by pi*A.
I'm not sure if this approach is correct, so any suggestion will be welcome.
Thanks in advanced
 
  • #3


Hello, thank you for presenting this problem. To calculate the particle flux over a single side of a detector with area A inside a sphere of radius R, we can use the following formula:

Flux = (n * V * A) / (4 * pi * R^2)

Where n is the particle density, V is the mean velocity of the particles, and R is the radius of the sphere. This formula takes into account the total solid angle of the sphere and the area of the detector.

If we want to consider the detector's acceptance, we can simply multiply the above formula by the acceptance factor D. This would give us the flux over the detector's surface as:

Flux = (n * V * A * D) / (4 * pi * R^2)

I would also like to note that this formula assumes that the particles are uniformly distributed within the sphere and that the detector is located at the center of the sphere. If these assumptions do not hold, the formula may need to be adjusted accordingly.

I hope this helps. Best regards.
 

1. What is flux and why is it important?

Flux is a measure of the amount of energy or particles passing through a surface. In the context of the sphere and detector, it represents the number of particles emitted from the sphere that are detected by the detector. Flux is important because it allows us to understand the behavior and interactions of particles in a given system.

2. How do you calculate flux for a sphere of radius R and detector acceptance D?

To calculate flux, you will need to use the formula: Flux = (Number of particles detected / Detector area) x (1 / Sphere surface area). The number of particles detected can be measured experimentally. The detector area is the area of the detector that is exposed to the particles. And the sphere surface area is calculated using the formula: 4πR^2.

3. What is the relationship between the radius of the sphere and the flux?

The radius of the sphere has a direct impact on the flux. As the radius increases, the surface area of the sphere also increases, resulting in a larger area for particles to be emitted from. This means that the flux will also increase. Conversely, a smaller radius will result in a lower flux.

4. How does the detector acceptance affect the flux?

The detector acceptance, which refers to the portion of the detector that is able to detect particles, also plays a role in the calculation of flux. A larger detector acceptance means that more particles will be able to be detected, resulting in a higher flux value. A smaller detector acceptance will lead to a lower flux value.

5. Can this calculation be applied to other shapes besides a sphere?

While this specific formula is designed for a sphere and detector system, similar calculations can be applied to other shapes as well. The key is to determine the surface area of the emitting object and the detector acceptance, and then use those values in the formula to calculate the flux.

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