- #1
ognik
- 643
- 2
Homework Statement
A proton of mass m, charge +e and (asymptotic) momentum mv is incident on a Nucleus of charge +Ze at an impact parameter b. Consider only coulomb repulsion and classical mechanics, what is distance of closest approach - d?
Homework Equations
Haven't encountered impact parameter before so please check the following:
I found ## b = \frac{kq_1q_2}{mv^2} \sqrt{ \frac{1+Cos \theta}{1-Cos \theta}} ##
I gather if b = 0, the proton would be aimed straight at the center of the nucleus, let's say that approach is a vector ##\vec{r_0}##
Then b is the distance between ##\vec{r_0}## and the actual path ##\vec{r_i}##, parallel to ##\vec{r_0}##.
And after closest approach, the proton moves off in direction ##\vec{r_f}##, at an angle ##\theta## to ##\vec{r_i}##?
Having said all that, I am unsure of how to apply b...
The Attempt at a Solution
I couldn't see using conservation of energy would help, the proton doesn't stop...
Conservation of momentum looks promising, especially as we don't have the initial distance the proton was when we measured v. Aligning the x-axis with ##\vec{r_i}## we have ##\vec{p_i} = \hat{x}.mv_i## and ##\vec{p_f} = \hat{x}.mv_f Sin \theta + \hat{y}.mv_f Cos \theta##
I'm stuck here, can't see how to involve b - and anyway b is not d, d will be greater than b because the proton starts diverging from ##\vec{r_i}## at some distance along ##\vec{r_i}## before the nucleus - where the coulomb repulsion starts affecting the protons path.
I did wonder if I could make an approximation as to the points where it starts (and symmetrically stops) diverging, I could then approximate d by taking a straight line between those 2 points and minimizing the distance between that line and the nucleus?