# Motion of an electron near a proton [closed]

Statement of the problem:

Consider an electron and a proton that are initially at rest separated $$a$$ meters. Do not take into account the movement of the proton, because its mass is much greater than the electron's.
1. What is the minimum kinetic energy at which the electron must be "launched" so that the electron gets to be $$b$$ meters away from the proton?
2. What is the corresponding minimum speed for that situation?
3. What distance away from the proton will the electron reach when it has double the initial kinetic energy?

(The original problem says 2.00 nm instead of $$a$$ meters, and 12.0 nm instead of $$b$$ meters)

My attempt:

Question 1 and 2
We say $$q$$ represents the elementary charge, i.e. the charge of the proton (positive) and of the electron (negative). Then I can assume that when the electron is $$b$$ meters away, the KE is zero. Then the difference in potential energy would be the same as the lost KE. $$\Delta U=\frac{1}{4\pi \varepsilon_o} \left ( \frac{q^2}{a} - \frac{q^2}{b} \right ) =K= mc^2 \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right ) \approx \frac{1}{2}m v^2$$ Doing some algebra we can get $$v$$, which is what they are asking for.

My questions:

4. Is it possible to have an electron and a proton at rest?
I imagine we would need another force that is "pulling" the electron away from the proton, and that that force cancels out with the electrostatic force from the proton. But, what would be a realistic example of a source of that force? And, how would that source alter the entire system?

5. How can we launch an electron that is near a proton?
Question 3 suggests that the speed of the electron suddenly goes from zero to some positive value. But, doesn't that ("launching the electron") imply infinite acceleration? If that is indeed possible, how can we realistically do it?

EDIT: The original problem was written in a confusing way. Thanks to the help of @user7777777 I was able to interpret it the right way and fix the wording.

## closed as off-topic by Kyle Kanos, stafusa, user191954, John Rennie, glSSep 24 '18 at 18:45

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You are absolutely correct for Questions 1 and 2. For Question 3, we can use the result from Question 2, but replacing $$\frac{1}{2} m {v_1}^2$$ with $$2 (\frac{1}{2} m {v_1}^2)$$, and comparing the two results in terms of $$b$$.