View Full Version : How to stop a fast moving electron
JamesJames
Mar5-05, 09:21 AM
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
JamesJames
Mar5-05, 09:39 AM
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
dextercioby
Mar5-05, 10:02 AM
Yes,but the question is unclear...Do they mean the mass of the Iron...?How are u gonna get that...?
Daniel.
JamesJames
Mar5-05, 10:07 AM
How much material.....which is x
JamesJames
Mar5-05, 10:10 AM
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
dextercioby
Mar5-05, 10:11 AM
"x" is the length of the path the electron takes inside the iron.
Daniel.
Anyway,logrithmate & see whether u get something reasonable.
JamesJames
Mar5-05, 10:13 AM
For the electron question I got 13.5 cm.
dextercioby
Mar5-05, 10:15 AM
It's okay.It looks good.The energy is very high.
Daniel.
JamesJames
Mar5-05, 10:16 AM
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.
dextercioby
Mar5-05, 10:20 AM
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.
JamesJames
Mar5-05, 10:29 AM
No, I am not using the 100 GeV, so the question could relate to any photon regardless of its energy right?
James
dextercioby
Mar5-05, 10:33 AM
Wy aren't u using that value...?Isn't intesity:energy in unit time...?
Daniel.
JamesJames
Mar5-05, 10:39 AM
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.
dextercioby
Mar5-05, 10:41 AM
Yes,okay.Indeed,"what fraction of a beam" requires only the ratio of the 2 intensities...
Daniel.
JamesJames
Mar5-05, 10:42 AM
Also, is the required fraction I/I_0 or 1 - (I/I_0) ? Why?
dextercioby
Mar5-05, 10:47 AM
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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