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An alpha particle is at rest at the origin of a Cartesian coordinate system

  • #1

Homework Statement



An alpha particle (the nucleus of a helium atom) is at rest at the origin of a Cartesian coordinate system. A proton is moving with a velocity of v towards the alpha particle in the xˆ direction. If the proton is initially far enough away to have no potential energy, how close does the proton get to the alpha particle? Your answer should be in terms of v, mp (mass of a proton) and e (charge of a proton).



Homework Equations



Wext = [tex]- \Delta K[/tex]

[tex]\Delta U[/tex] = q(Vb-Va)


The Attempt at a Solution


[/B]
I set the change in K equal to q(Vb - Va), with q being just e, and Va going to zero since the starting point has no potential. To find Vb I used the V for a point charge, (V = kq/r). From there I moved the variables around a bit, crossed out 2e on both sides, and got r = (mv^2)/k.

I wasn't sure if this was right because the distance is always changing. So I used the integral equation E dl, setting that equal to Vb, which was equal to the kinetic energy (I used gauss' law to find E as a function of r) and got the same answer. Something just seems wrong though. Also I definitely mixed up negative signs depending on the direction of the work. I am on the right track?

 

Answers and Replies

  • #2
Orodruin
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What about momentum conservation?
 
  • #3
ehild
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Initially, the potential energy is zero, and the speed is v. At the final stage, the speed is zero. The potential energy of the proton is k 2e2/r. Conservation of energy gives 1/2 mv2=2e2k/r. Isolate r. It is definitely NOT r=mv2/k, how did you crossed out 2e2????
 
  • #4
Orodruin
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At the final stage, the speed is zero.
##m_p v + 0 = 0+ 0## then? That does not seem likely.
 
  • #5
ehild
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It is not clear if the alpha particle is hold in rest or free to move, and v is the initial velocity of the proton and changing while approaching the alpha particle, or it is constant. I understood that the alpha particle is stationary at the origin and the speed of the proton decreases as approaching it.


ehild
 
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  • #6
Orodruin
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I thought about that interpretation and decided against it based on that any potential holding the alpha particle stationary also would affect the proton, leading to an additional level of complication and that the alpha and proton were the only things mentioned. Also, assuming the proton mass to be much smaller than the alpha mass seems a bit of an oversimplification.

Of course, the approach is the same in the end with the additional step of having to compute the kinetic energy of the system when the proton and alpha have the same velocity.
 
  • #7
ehild
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I thought about that interpretation and decided against it based on that any potential holding the alpha particle stationary also would affect the proton, leading to an additional level of complication and that the alpha and proton were the only things mentioned. Also, assuming the proton mass to be much smaller than the alpha mass seems a bit of an oversimplification.

Of course, the approach is the same in the end with the additional step of having to compute the kinetic energy of the system when the proton and alpha have the same velocity.
You are right, but it is just a problem for students, the situation need not be real. "It is at rest at the origin " suggests the alpha particle does not move. If it were "initially at rest at the origin" you were certainly right. Anyway, the OP had problems with rearranging an equation.
 
  • #8
Orodruin
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The semantics of the problem definitely are not clear. We are first led to know that there is an alpha particle at rest and that a proton is approaching, nothing about whether or not this affects the alpha particle. If unclear I would tend to go with the more physical interpretation (I would also try to make my problems as physically reasonable as possible so students do not end up with misconceptions).

That being said, I was just trying to make sure that OP had the right equation to rearrange before starting to rearrange it and to make him/her consider whether momentum should be conserved or not.
 
  • #9
ehild
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The semantics of the problem definitely are not clear. We are first led to know that there is an alpha particle at rest and that a proton is approaching, nothing about whether or not this affects the alpha particle. If unclear I would tend to go with the more physical interpretation (I would also try to make my problems as physically reasonable as possible so students do not end up with misconceptions).

That being said, I was just trying to make sure that OP had the right equation to rearrange before starting to rearrange it and to make him/her consider whether momentum should be conserved or not.
I always try to follow the OP's way of thinking if it is reasonable. And assuming the particle in rest is reasonable. He mentioned work-energy theorem (with sign error), potential energy, and potential of a point charge. If he applied conservation of energy, it was acceptable, but he had some error with the maths. Or he did not know what kinetic energy is.

As for physically reasonable set-up, you can build a single layer of 100 nm gold spheres on silicon or even on glass, and implant an alpha particle into one of the gold spheres. It is possible. Then it stays in rest and the gold particle has 2e charge. And you shot the proton from 10 m distance with 100 m/s speed. The distance of approach is great enough so the proton feels only the field of the charged gold particle. (And it is in perfect vacuum... :) )
 
  • #10
Orodruin
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Would the problem then not had said "a gold particle with charge +2e"? ;)
Anyway, my intention is not to quibble over semantics and I think we need some input and feedback from OP on whether or not to even discuss it further than it has already been.
 

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