Exercises for Feyman Lectures on Physics

AI Thread Summary
The discussion focuses on the concept of effective target area, or cross-section, in scattering theory, particularly in relation to a gas layer's impact on particle beams. A 1mm layer of argon is discussed as the thickness through which a beam passes, affecting particle intensity and velocity. It is noted that if particles are scattered, they are considered lost from the beam, with 3% of particles not passing through the argon layer. The importance of understanding the number of argon molecules per volume is emphasized for accurate calculations. A related problem about light transmission through a window with dots illustrates the application of these concepts in determining area coverage.
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Homework Statement
1.10 the intensity of a collimated,parrallel beam of potassium atoms is reduced 3.0% bya layer of argon gas 1.0mm thick at a pressure of 6.0x10e4(e=exponent)mmHg.
Calculate the effective target area A per argon atom
Whats a effective target area?
Relevant Equations
no equations
Whats a effective target area?
 
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Also known as cross-section in scattering theory. Look it up and ask questions as needed.
 
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thanks
 
another question, the 1mm layer is the depth or the diameter I am really confuse in that
 
That is the thickness of the layer of gas through which the beam passes. Do you think it matters? Why (or not)?
 
I think is important because if the layer of argon doesn't exist the intensity the beam throws the particles will be greater
 
and if the layer is thicker the velocity of the particle will be less than in the original problem
 
The assumption here is that any particle in the beam that is scattered is lost from the beam. The ones that get through do not hit anything.
Are there more questions that follow this set up (I am assuming yes}?
 
so,with the layer of argon, 3% of the particle don't get through the beam.Im correct?
(Sorry from the bad grammar I am good at it)
 
  • #10
Yes. So we assume they are sparse enough that they don't "shadow" each other (3% is a small number so that's OK). We need the number/volume of Argon in the chamber. Do you see why?
 
  • #11
number =argon ifthis is corect i see why number of molecules/by the volume
because do that you find the volume of 1 molecule of argon atom
Im correct?
 
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  • #12
Here's a similar problem. I have a window 1mx1m. I festoon it with 100 round black dots and notice that the total light streaming through diminishes by 5%. How big (area) is a black dot?
 
  • #13
the total dots should cover 5%of the window so if 100% =100m2 that means that the total area of the dots are 5m2 and is one dot = 0.05m2?
 
  • #14
i finish the problem,THANK YOU SOOOO MUCH
 
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