How Do You Calculate the Distance of Closest Approach in a Particle Collision?

  • Thread starter Thread starter fayled
  • Start date Start date
  • Tags Tags
    Approach
AI Thread Summary
To calculate the distance of closest approach in a particle collision involving an alpha particle and a stationary Beryllium nucleus, the problem requires considering both kinetic and potential energy. The alpha particle, accelerated through a potential difference of 20 kV, has its speed derived from energy equations, leading to a calculation of potential energy at closest approach. The correct approach involves transforming the problem into the center of mass frame, where the total momentum is conserved, and ensuring that energy conservation is accurately applied. After correcting initial mistakes, the final expression for the distance of closest approach is derived as d = 416 fm. This solution emphasizes the importance of accurately accounting for kinetic and potential energy during particle interactions.
fayled
Messages
176
Reaction score
0

Homework Statement


An alpha particle is accelerated from rest through a potential dierence of 20 kV. It travels directly towards a stationary Beryllium nucleus (4 protons, 5 neutrons). Calculate the distance of closest approach.

The problem definitely wants a solution that takes into account the recoil of the stationary nucleus on approach.

Homework Equations


VCM=m1v1+m2v2/m1+m2
U=Q1Q2/4πε0d
T=mv2/2

The Attempt at a Solution


Let the oncoming alpha particle have speed v. VCM=4v/13. In the CM frame, we have a particle of mass 4m, charge 2e, speed 9v/13 heading towards on of mass 9m, charge 4e, speed -4v/13.

Initially, T=18mv2/3. This all becomes PE which will be 2e2/πε0d for closest approach d. Equating given d=e2/3πε0mv2. T=qV for a p.d V so v2=2qV/m. Then d=e/3πε0V. Plugging in the numbers given d=96fm.

I don't have a solution so would appreciate somebody having a check. I'm unsure about my approach because it assumes that the initial scenario (i.e as soon as the alpha particle has finished being accelerated through the p.d) has the PE as zero, which isn't realistic and the question doesn't indicate this is valid. Thanks.
 
Physics news on Phys.org
You went wrong when you said all the energy becomes PE. That's not the case. At closest approach both particles are moving at speed VCM
 
dauto said:
You went wrong when you said all the energy becomes PE. That's not the case. At closest approach both particles are moving at speed VCM

But I'm working in the CM frame?
 
Last edited:
dauto said:
You went wrong when you said all the energy becomes PE. That's not the case. At closest approach both particles are moving at speed VCM
Fayled has transformed the problem into the CM frame, where VCM is zero. So the approach is correct.

But I did spot two mistakes or typos.
fayled said:

The Attempt at a Solution


Let the oncoming alpha particle have speed v. VCM=4v/13. In the CM frame, we have a particle of mass 4m, charge 2e, speed 9v/13 heading towards on of mass 9m, charge 4e, speed -4v/13.
Looks good so far.

Initially, T=18mv2/3.
It should be "/13" at the end, not "/3". Do you agree?

This all becomes PE which will be 2e2/πε0d for closest approach d. Equating given d=e2/3πε0mv2.
Agree, once you fix the earlier mistake.

T=qV for a p.d V so v2=2qV/m.
I agree.

Then d=e/3πε0V.
I don't think that follows. Can you show the steps leading up to this expression (following the v^2=\frac{2qV}{m} result)? What are you using for q?

Plugging in the numbers given d=96fm.

I don't have a solution so would appreciate somebody having a check. I'm unsure about my approach because it assumes that the initial scenario (i.e as soon as the alpha particle has finished being accelerated through the p.d) has the PE as zero, which isn't realistic and the question doesn't indicate this is valid. Thanks.
It is safe to assume this is valid. The alpha should be a considerable distance from the berylium when it exits the accelerator, so there would be negligible PE at that time.
 
Last edited:
Redbelly98 said:
Fayled has transformed the problem into the CM frame, where VCM is zero. So the approach is correct.

But I did spot two mistakes or typos.

Looks good so far.It should be "/13" at the end, not "/3". Do you agree?Agree, once you fix the earlier mistake.I don't think that follows. Can you show the steps leading up to this expression? What are you using for q?It is safe to assume this is valid. The alpha should be a considerable distance from the berylium when it exits the accelerator, so there would be negligible PE at that time.

Right:

Yes I agree! In fact a follow up question of mine was going to be why doesn't TLAB=TTCM+T', with T' the KE of a particle of the total mass of the system moving at speed VCM? That error solves that (should have known better than thinking physics had broken and I hadn't made an error).

So T=18mv2/13

U=2e2/πε0d

d=13e2/9πε0mv2

Now v2=eV/m. Before I had v2=2qV/m. But I would then need q=2e,m=4m (i.e I should have used M say - I think this confused you).

d=13e/9πε0V=416fm (simply substituting the above expression into the one above it).

Hopefully that's right. Thanks for your help!
 
Last edited:
Looks good. You're welcome! :smile:

p.s. You're right, I was confused about m vs. 4m.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top