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## Homework Statement

In an experiment, an alpha particle of mass 6.7e-27 kg; charge 3.2e-19C travels in a straight line with a speed 2.0e7m/s towards the centre of a gold nucleus.

Find:

1. The kinetic energy([tex]KE[/tex]) of the alpha particle;

2. The potential(V) at a distance '[tex]x[/tex]' from the centre of the gold nucleus(of charge 1.3e-17C)

3. The closest distance of approach, and prove that it is 2.8e-14m

## Homework Equations

[tex]KE=\frac{mv^{2}}{2}[/tex]

[tex]E=\frac{F}{Q}[/tex]

[tex]E=\frac{V}{x}[/tex]

## The Attempt at a Solution

1. [tex]KE=\frac{mv^{2}}{2} = \frac{1}{2}·{(6.7*10^{-27}) · (2*10^{7})^{2}}[/tex] = [tex]_1.34*10^{-12}J[/tex]

This part I'm pretty sure of, and I know I should be doing these in my sleep by now, but it was 3:00am when I did this and haven't checked any of these much.

2. E=F/Q and E=V/x

F/Q=V/x -> V=(Fx)/Q

I really had no idea what I was being asked here; yes I realize it tells me clear-as-day, but I just can't seem to get anywhere with this question.

2. I really didn't have a clue. I have tried so many approaches with each being as tame as the last: I first tried initial KE = final PE, to no avail; but that may just be since every time I go over the question I end up answering it in the same WRONG way.

Sorry, I'm not asking anyone to

**do**my homework for me, but I just feel like I'm missing that little piece of info I happened to overlook.

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