- #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.
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.
Last edited: