Time it takes for an electron and proton to collide

In summary, the problem is to find the time it takes for an electron and positron, with masses me and separation R, to collide when initially at rest. The force is not constant, so integration is necessary. One idea is to divide by mass and integrate twice with respect to time, but this method may not work due to the changing force with respect to distance. Another method is to use the concept of reduced mass to make both sides of the equation integrable.
  • #1
gsingh2011
115
1

Homework Statement


An electron and positron have mass me and separation R and are initially at rest. Find the time T it takes for them to collide.


Homework Equations


F=kq1q2/R2


The Attempt at a Solution


My problem is that the force is not constant. Do I integrate the force then? And from what bounds?
 
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  • #2
What do you get when you use F=ma?
 
  • #3
vela said:
What do you get when you use F=ma?

I don't know, acceleration is changing. Let say we have F=kq1q2/r2=ma. Since this problem is symmetrical, I was thinking the two particles would meet in the middle, so each would travel a distance of R/2. So I could divide by m and integrate twice to find the position function, and set that equal to R/2. The only problem with that idea is the force is changing with respect to r, so I can't integrate with respect to t. So I guess that method won't work.
 
  • #4
gsingh2011 said:
I don't know, acceleration is changing. Let say we have F=kq1q2/r2=ma. Since this problem is symmetrical, I was thinking the two particles would meet in the middle, so each would travel a distance of R/2.
The difference in masses breaks the symmetry. The proton won't move as far as the electron because it's so massive in comparison.
So I could divide by m and integrate twice to find the position function, and set that equal to R/2. The only problem with that idea is the force is changing with respect to r, so I can't integrate with respect to t. So I guess that method won't work.
The standard trick is to multiply by [itex]\dot{r}[/itex]. You'll find it makes both sides of the equation integrable, but first you need to find the correct equation.

Are you familiar with the concept of the reduced mass?
 
Last edited:
  • #5
The proton won't move as far as the electron because it's so massive in comparison.
It's a positron not a proton, so the mass is the same.
 
  • #6
Oops, not sure why I thought it was a proton. Yes, you're right, they'll meet in the middle.
 

1. How long does it take for an electron and proton to collide?

The time it takes for an electron and proton to collide depends on the energy of the particles and the distance between them. In most cases, it happens very quickly, within a fraction of a second.

2. What factors affect the time it takes for an electron and proton to collide?

The main factors that affect the time it takes for an electron and proton to collide are the speed and direction of the particles, as well as any external forces or interactions that may influence their movement.

3. Can the time it takes for an electron and proton to collide be calculated?

Yes, the time it takes for an electron and proton to collide can be calculated using equations from quantum mechanics and particle physics. However, the calculations can be complex and require knowledge of the particles' energies, velocities, and positions.

4. How does the time it takes for an electron and proton to collide affect chemical reactions?

The time it takes for an electron and proton to collide can greatly influence the rate of chemical reactions. In some cases, a faster collision can lead to a faster reaction, while in others, a slower collision may be necessary for the reaction to occur.

5. Is there a minimum or maximum time for an electron and proton to collide?

There is no minimum or maximum time for an electron and proton to collide. The time can vary depending on the conditions and interactions between the particles. In some cases, they may not collide at all if their paths do not intersect or if there are other forces at play.

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