# The use of cross section in a particle collision exercise.

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
Homework Helper
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.

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
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