Closest approach of alpha particle - two separate methods

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Homework Help Overview

The discussion revolves around the closest approach of an alpha particle to a larger nucleus, exploring two different methods for calculating this distance. The subject area includes concepts from nuclear physics and energy conservation principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster presents two methods for determining the distance of closest approach, one assuming the larger nucleus remains stationary and the other allowing for its movement. Participants discuss the validity of these assumptions and the implications for the results.

Discussion Status

Some participants suggest that method 2 should be used unless the nucleus is significantly larger, while others reflect on the challenges of justifying assumptions in different scenarios. There is acknowledgment of the need for practice to develop intuition regarding these assumptions.

Contextual Notes

Participants note that the problem is common in exams and often lacks clarity on whether the larger nucleus can be assumed to be stationary. There is also mention of the significance of the energy gained by the larger mass in relation to the overall energy of the system.

etotheipi
Homework Statement
Find the distance of closest approach of an alpha particle and nucleus (of arbitrary atomic number Z)
Relevant Equations
##E_{k}##, relative velocities
I've found two methods for doing this problem and they give different answers.

Method 1: Assume the larger nucleus does not move, and simply equate energies before the collision and at the point of closest approach:$$\frac{1}{2} m v^{2} = \frac{qQ}{4\pi\epsilon_{0}r}$$

Method 2: Assume the larger nucleus is capable of moving; the distance of closest approach occurs when the relative velocity of the two particles is zero. From this, we can use CLM and then do another energy calculation:

Conservation of Linear Momentum:
$$m_{\alpha}v_{\alpha} = (m_{\alpha} + M_{nucleus})v_{common}$$Energy:
$$\frac{1}{2} m_{\alpha} v_{\alpha}^{2} = \frac{1}{2} (m_{\alpha} + M_{nucleus})v_{common}^{2} + \frac{qQ}{4\pi\epsilon_{0}r}$$It makes sense that the two methods give different answers for ##r##, however this is a very common exam question, and it is never usually stated whether we permit the larger nucleus to move or not. So, I was wondering which approach I should use for this and similar problems. Thank you!
 
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Unless you can assume the nucleus is large enough that its movement is negligible you must use method 2. In the case where the nucleus is very large, methods 1 and 2 should give approximately the same answer.
 
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PeroK said:
Unless you can assume the nucleus is large enough that its movement is negligible you must use method 2. In the case where the nucleus is very large, methods 1 and 2 should give approximately the same answer.

Thank you, this is essentially what I was thinking. It can be quite hard though to determine whether the assumption is justified or not. For instance, in the one below, I'd assume since silver is about 30x as massive as helium we could get away with the simpler method.

I suppose the only way to get a better feel for when certain assumptions are justified is just by doing lots of practice problems.

1576061119118.png
 
Here's an idea. In this collision, the energy gained by the large mass is approx ##\frac{m}{M}## of the energy of the system. Is that significant? In the above problem, this ratio is about ##4/50## = 8%.

So, do you need the answer to within 10%? If not, you can do a quick calculation by method 1.
 
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PeroK said:
Here's an idea. In this collision, the energy gained by the large mass is approx ##\frac{m}{M}## of the energy of the system. Is that significant? In the above problem, this ratio is about ##4/50## = 8%.

So, do you need the answer to within 10%? If not, you can do a quick calculation by method 1.

That's a nice rule of thumb!
 

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