Particle Accelerator problem with a proton beam hitting a target

Click For Summary
In a particle accelerator problem, a 300μA proton beam with a 4.00-MeV energy and a radius of 1.30 mm is analyzed to determine how many protons hit a target in 3 seconds. The discussion highlights the relationship between current, charge, and time, leading to the calculation of total charge delivered as 900μC over 3 seconds. By dividing this charge by the charge of a single proton (1.60 x 10^-19 C), an estimated total of 5.625 x 10^15 protons is derived. Participants note that the beam's radius does not affect the number of protons hitting the target since the current already accounts for the cross-sectional area. The conversation concludes with agreement on the methodology used to arrive at the solution, despite some initial confusion regarding the parameters.
OmegaFury
Messages
28
Reaction score
0

Homework Statement


In a certain particle accelerator, a current of 300μA is carried by a 4.00-MeV proton beam that has a radius of 1.30 mm. The mass of a proton is 1.67 x 10-27kg. If the beam hits a target, how many protons hit the target in 3s?


Homework Equations


The only equation that even remotely comes to mind regarding this problem is I=e2n(tau)AE/me

I do not know which variable has the electron volt unit, or an equation involving current and time.


The Attempt at a Solution


I don't even know where to begin as I don't know a suitable formula to solve the problem. Perhaps if I knew that, I could get somewhere.
--Also, for personal curiosity-- In the formula for current that I listed, what exactly is tau?
Also, is n the number of particles, or the number density? And is E the electric field? My book didn't exactly break the formula down as well as I'd like.
 
Last edited:
Physics news on Phys.org
If you don't know what the formula means then it's useless. What does an ampere (the A in your formula) mean? I think they gave you a lot of numbers in that problem that don't really matter. Try to understand what an ampere is first.
 
Okay. Well. An ampere is equal to 1C/1s. So I guess that is where the time would come in. 300μA= 300 x 10-6A = 300 x 10-6C/S. So would I just multiply that by 3s to cancel the seconds, leaving only coulombs? Then divide by 1.60 x 10-19C (I suppose that would be the charge of one proton). Which would equal 5.625 x 1015 protons.

I highly doubt this is the case because I'm sure the radius has some influence on how many protons hit the target.
 
Last edited:
OmegaFury said:
Okay. Well. An ampere is equal to 1C/1s. So I guess that is where the time would come in. 300μA= 300 x 10-6A = 300 x 10-6C/S. So would I just multiply that by 3s to cancel the seconds, leaving only coulombs? Then divide by 1.60 x 10-19C (I suppose that would be the charge of one proton). Which would equal 5.625 x 1015 protons.

I highly doubt this is the case because I'm sure the radius has some influence on how many protons hit the target.

But they don't give you the target size, do they? How can you use the beam radius?
 
I'm assuming that the target is two-dimensional and is as big as the cross-sectional area of the beam.
 
OmegaFury said:
I'm assuming that the target is two-dimensional and is as big as the cross-sectional area of the beam.

So am I. So the whole beam hits the target. So how can changing the radius change the number of particles hitting the target?
 
Oh. I think I see what you're getting at. Electric current already incorporates the cross-sectional area as part of the equation. When calculating for the number of protons, you do calculate for the total number that hit that area over time. Looking back at it, the answer does seem to be in the ballpark. It was a multiple choice answer in which all of the answers were wrong. However, all of them are in the 1015 range, which agrees with my answer.
 
Last edited:
OmegaFury said:
Oh. I think I see what you're getting at. Electric current already incorporates the cross-sectional area as part of the equation. When calculating for the number of protons, you do calculate for the total number that hit that area over time. Looking back at it, the answer does seem to be in the ballpark. It was a multiple choice answer in which all of the answers were wrong. However, all of them are in the 1015 range.

I don't see how to get an answer much different from what you got in post 3. Unless they forgot to include part of the problem or something.
 
I'm pretty certain that was the way to go about solving it. Thanks for the help. I appreciate it.
 

Similar threads

Replies
3
Views
4K
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
20
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
1
Views
1K
Replies
3
Views
3K