Distance of closest approach

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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.
 

Answers and Replies

  • #2
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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
 
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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?
 
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  • #4
Redbelly98
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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.

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 [itex]v^2=\frac{2qV}{m}[/itex] 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.
 
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  • #5
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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!
 
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  • #6
Redbelly98
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Looks good. You're welcome! :smile:

p.s. You're right, I was confused about m vs. 4m.
 

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