The use of cross section in a particle collision exercise.

[Coffee_]

1. Situation: I have no specific exercise in mind but just the general form. ''A beam of protons is colliding with a plate of a given particle density $\rho$ and thickness $l$. The cross section is $\sigma$. What should the thickness of the plate be such that the intensity of the outgoing particles on the other side is 0.1 of the incoming intensity'' The exercises we made in class were literally worded like this with no more info.2. Equations: $W=\rho l \phi \sigma$ where $W$ is the amount of elastic collisions per second and $\phi$ the amount of incoming particles per second. 3. My general reasoning is something like this. If I look at a 1 second interval I have an amount $\phi$ of incoming particles, I'd like that the collision rate would be $0.9 \phi$ so that only 0.1 of the particles have not collided. In that case the relevant equation becomes $0.9= \rho l \sigma$.

CONFUSION:

The point where I'm confused is the given cross section as I underlined. I make an implicit assumption here that this cross section is just the cross section of collision, and that after collision the particles ''dissapear'' so to speak or move very slow relative to the ingoing beam. If this isn't the case then there will be collisions after which particles will keep moving forward without losing all too much momentum and my reasoning would be wrong.

QUESTION:

Is my reasoning of a solution correct? If so, how can I lift my confusion about the meaning of the cross section in this situation?

 

[BvU]

Your assumption is completely correct. No confusion necessary in this simple case.

For scattering at small angles, there is no simple sigma and an angular dependence comes in.

 

[Coffee_]

Your assumption is completely correct. No confusion necessary in this simple case.

For scattering at small angles, there is no simple sigma and an angular dependence comes in.

Then what about the particles that keep on having a quite large velocity component in the direction after colliding? Those particles will still contribute to the outgoing beam of particles.

 

[BvU]

Absolutely. In fact for protons with a reasonable kinetic energy that is what normally happens: they lose energy mainly through collisions with bound electrons (which they ionize) through Coulomb interaction. Google Bethe Bloch, browse through http://www.kip.uni-heidelberg.de/~coulon/Lectures/Detectors/Free_PDFs/Lecture2.pdf , or get it all here, but realize that you are by now way beyond the scope and level of the original exercise. Welcome to the wonderful world of elementary particle physics !

 

[HalfLostAndSearching]

RESPONSE:

Your reasoning for finding the thickness of the plate is correct. However, your confusion about the meaning of the cross section is valid and needs to be addressed. In this situation, the cross section represents the effective area of the plate that the incoming particles will collide with. It is a measure of the probability of a collision occurring between the incoming particles and the plate.

In other words, the cross section is not just the physical area of the plate, but it also takes into account the interaction between the incoming particles and the plate. So, after a collision occurs, the particles will either be absorbed or scattered, but they will not continue moving forward with the same momentum as before. This is because the collision will transfer some of the particle's energy and momentum to the plate.

To lift your confusion, it may be helpful to think of the cross section as a measure of the target's "effectiveness" in causing a collision. In this exercise, the thickness of the plate is directly related to the cross section, as a thicker plate will have a larger effective area and therefore a higher collision rate. So, in order to achieve a 0.9 collision rate, the thickness of the plate must be adjusted according to the given cross section value.

Overall, your reasoning is correct and your confusion about the cross section is understandable. It is important to remember that the cross section is not just a measure of the physical area, but it also takes into account the effectiveness of the target in causing a collision.