How to stop a fast moving electron

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To stop a 500 GeV electron using iron, the stopping distance can be calculated with the formula E = E0*e(-x/x0), where x0 depends on the material's density and atomic number. The initial energy E0 is 500 GeV, while the rest energy E is 0.511 MeV, leading to a discussion on whether these values are correctly assigned. For a 100 GeV photon beam passing through a 2cm thick lead absorber, the fraction transmitted is calculated using I = I0*exp(-mu*x), with mu being related to lead's density. The conversation clarifies that the fraction of the beam transmitted is I/I0, while the absorbed fraction is 1 - (I/I0). The importance of energy in calculating intensity is acknowledged, confirming that the ratio of intensities suffices for the analysis.
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How much iron is required to stop a 500 GeV electron if it only deposits energy via ionization?

Is the following approach correct?

E = E0*e(-x/x0)

x0 = 170*A*density/Z^2

x = ? where x is the stopping distance

E0 = 500 GeV

E = 0.511 MeV

Or should E and E0 be flipped?
:confused:
James
 
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I say E0 is what is left after it is stopped and comes to rest..this will be the rest energy. Then E must be the initial energy.

Am I correct?

James
 
Yes,but the question is unclear...Do they mean the mass of the Iron...?How are u going to get that...?

Daniel.
 
How much material...which is x
 
I have one more

What fraction of a 100 GeV photon beam will be transmitted through a 2cm thick lead absorber?

I used

I = I0*exp(-mu*x)

where mu = density of lead * (0.04 cm^2/gm)

The fraction absorbed is 1- (I/I0)

First of all is the fraction correct or should it be just I/I0. Secondly, why do I not have to use the energy when the book (Ferbel) says that mu should typically depend on the enregy.

James
 
"x" is the length of the path the electron takes inside the iron.

Daniel.

Anyway,logrithmate & see whether u get something reasonable.
 
For the electron question I got 13.5 cm.
 
It's okay.It looks good.The energy is very high.

Daniel.
 
Any suggestions about the photon question I posted in this thread. I' ll just repeat it here again

What fraction of a 100 GeV photon beam will be transmitted through a 2cm thick lead absorber?

I used

I = I0*exp(-mu*x)

where mu = density of lead * (0.04 cm^2/gm)

The fraction absorbed is 1- (I/I0)

First of all is the fraction correct or should it be just I/I0. Secondly, why do I not have to use the energy when the book (Ferbel) says that mu should typically depend on the enregy.
 
  • #10
What do you mean you don't get to use the energy...?You do.That I_{0} is proportinal to the enregy of the incoming photons...

Daniel.
 
  • #11
No, I am not using the 100 GeV, so the question could relate to any photon regardless of its energy right?

James
 
  • #12
Wy aren't u using that value...?Isn't intesity:energy in unit time...?

Daniel.
 
  • #13
Ok, so the energy has been taken into account when calculating the intensity. For the question, I am just looking for the ratio of I / I_0 so I mean in evaluating the exopnential, I am not using the energy.
 
  • #14
Yes,okay.Indeed,"what fraction of a beam" requires only the ratio of the 2 intensities...

Daniel.
 
  • #15
Also, is the required fraction I/I_0 or 1 - (I/I_0) ? Why?
 
  • #16
I_{0} is the initial intensity.I(x) is the value of the intensity in the point "x".For the endpoint of the trajectory through the lead,it's the intensity of the ray which gets out.So the ratio is just I(d)/I_{0}...

Daniel.
 
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