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

[fayled]

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

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

The Attempt at a Solution

Let the oncoming alpha particle have speed v. V_CM=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.

 

[dauto]

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 V_CM

 

[fayled]

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 V_CM

But I'm working in the CM frame?

 

[Redbelly98]

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 V_CM

Fayled has transformed the problem into the CM frame, where V_CM 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. V_CM=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.

 

[fayled]

Fayled has transformed the problem into the CM frame, where V_CM 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 T_LAB=T_TCM+T', with T' the KE of a particle of the total mass of the system moving at speed V_CM? 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!

 

[Redbelly98]

Looks good. You're welcome!

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