# Homework Help: How close can two protons get if...

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1. Feb 21, 2016

### PAK108

Mentor note: Thread got moved to the homework section

How close can two protons get if one is at rest and the other has a kinetic energy equal to the average energy at T =107 K?

I know that the kinetic energy of the moving proton is 3/2kT, but what is the kinetic energy of the proton
at rest? This question is from my book in astronomy. The answer are supposed to be r= r = 1.1 × 10−12m

Any help appreciated

Last edited by a moderator: Feb 21, 2016
2. Feb 21, 2016

### Staff: Mentor

Zero. What else?

At the moment of closest approach, both protons will move and have non-zero kinetic energy.

Is this homework?

3. Feb 21, 2016

### drvrm

In your opinion what should be kinetic energy of a body which is not in motion?
the other protons kinetic energy is given so by virtue of its energy it can approach the other proton -in spite of repulsion from it so you must calculate the net work done by it in getting closer.........

4. Feb 21, 2016

### PAK108

Aha, I think I understand now, that means that all the kinetic energy becomes potential energy when the electron
are at the closest approach...thank you :)

Yes, this is homework (I realized that this should be posted at the homework section after posting. I am sorry for posting here)

5. Feb 21, 2016

### Staff: Mentor

No it does not.
For that to occur, both would have to stop at the same time, which violates momentum conservation.

I moved the thread to the homework section.

6. Feb 21, 2016

### Staff: Mentor

It's not clear from the problem statement that the "at rest" proton is meant to be able to move, or if it is to remain at rest (fixed in place). Is the problem statement complete and exactly as it was given?

7. Feb 22, 2016

### PAK108

Yes this is the complete problem statement

I tried to solve the problem like this:

3/2kT + mc2=k(e2/r2), solving for r gave r=1.2*10-9, which is like 1000 order bigger than the correct answer :(

Last edited: Feb 22, 2016
8. Feb 22, 2016

### Staff: Mentor

That does not even have matching units, and I don't understand where you got that equation from.

9. Feb 22, 2016

### drvrm

what this energy term ( equivalent to mass) is doing there -is the problem talks about velocities comparable to velocity of light

10. Feb 22, 2016

### PAK108

The equation was just put together from the equation for "thermal energy and rest energy" = "potential energy"

the k's on the left hand side is not the same on the right hand side
LHS k=1,38065*10-23 J/K
RHS k=8,99*109 Nm2/C2

11. Feb 22, 2016

### PAK108

No, the problem does not talk about velocities comparable to light...basically I was just trying something since the units matched...the chapter mentions the rest energy of a proton (E=mc2) so I just gave it a shot

12. Feb 22, 2016

### Staff: Mentor

Why should that be true? Do you annihilate a particle to extract its rest energy? Even if that would be true, why one and not both?

That is clear and no problem. The potential energy still does not follow a 1/r2 law.

13. Feb 22, 2016

### drvrm

well you can not do these things as physics theory does not mixes newtonian and sTR -it will be a mess.

14. Feb 22, 2016

### PAK108

Ah of course, it is supposed to be 1/r, that means that the calculations are completely off