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Old Jun 12th 2011, 12:29 AM   #1
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Find Electron Velocity

How does the velocity of a falling electron change as it falls toward a proton?

As the distance between a stationary proton and a moving electron toward the proton decreases by factors of 10, what factor is the increase in velocity?

Assume the distance between the proton and electron starts at 1,000 meters, then what is the velocity of the electron at 1,000 meters, 100 meters, 10 meters, 1 m, O.1 m, 0.01 m etc.

I'm wondering as the distance decreases by a factor of 10, does the velocity increase by a factor of a 100 or what? That is my interest in the solution!

My attempt at a solution is as follows:

a0 = F / m = 253.27 / x0^2 [m / s^2]

a1 = 253.27 / x1^2

a01 = (a0 + a1) / 2

(x0 - x1) = 0.00001 = x01

(t1 - t0) = ((x0 - x1) / a01))^0.5

v0? = (x0 - x1) / (t1 - t0)

v1? = ((v0^2 + 2 * a01 * (x0 - x1))^0.5

The above requires a computer solution as (x0 - x1) = 0.00001 changes by amounts of 0.00001 to (x1 - x2) = 0.00001, etc.

I have no requirement to write any software or to solve this problem. Perhaps such software exists?

My interest is theoretical in what the electrical field around the proton is?

Could this electric field be made of ultra small particles of mass coming off the proton? This positron mass would decrease by a factor of 100 as the radius increases by a factor of 10.

What is your opinion and hopefully an answer to my velocity interest?
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I solved the problem using coulombs energy equation replacing charge q with the number of charged particles n as follows:

Fr = 2.3071 E-28 (n1) (n2) / r = m (v ^2) / 2

2.3071 E-28 / 9.1094 E-31 = 253.27

v = (506.54 / r) ^0.5

r0 = 1 E+0 m then v = 22.506 m / s
r1 = 1 E-1 m then v = 71.172 m / s

r9 = 1 E-9 m then v = 7.1172 E+5 m / s
r10 = 1 E-10 m then v = 2.2506 E+6 m / s

The factor velocity increase when r decreases by a factor of 10 is:
r1 / r0 = 71.172 / 22.506 = 3.1622
r10 / r9 = 2.2506 E+6 / 7.1172 E+5 = 3.1622

The factor velocity decrease when r increases by a factor of 10 is:
r0 / r1 = 22.506 / 71.172 = 0.31622
r9 / r10 = 7.1172 E+5 / 2.2506 E+6 = 0.31622

Solved!
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The general solution to find the velocity of the falling electron to a nucleus with n protons is as follows:

v = 22.506 (n / r) ^0.5

v1 = v2 (r2 / r1) ^0.5

v2 = v1 (r1 / r2) ^0.5

Last edited by DKseed; Jun 13th 2011 at 09:40 PM. Reason: Below I solved for the general solution of a falling electron to a nucleus with n protons
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Old Jun 13th 2011, 12:59 AM   #2
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I'm not sure where you're getting your values... anyway, this is how I would have done it:

I will first assume that the velocity of the electron is 0 when it is 1000 m away since yo said it 'starts' there. I will not go around using the basic equations of motion since the acceleration is not constant. The force acting on the electron varies as the distance varies and this will be a real pain to find the acceleration at each and every point in between the points you want to get the velocity. Instead, it's easier to go around energy equations.

Potential energy 1000 m away from the proton:



Using the values that we have:



Now, at 100 m;



And so on. The velocity at 100 m is given by the kinetic energy formula and the change in potential energy.

Lost in PE = Gain in KE



You solve for v.

To get the velocity at 10 m, you use:

http://latex.codecogs.com/gif.latex?...\frac12%20mv^2

(the site wouldn't let me post a 5th image... -.-" )

And so on.

I hope it helped!

PS: Eo [ Epsilon nought is the permittivity of free space and is a very small value. There are some schools which teack V = kq/r though, where k = 1/(4pi Eo) ]
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