How Close Will an Alpha Particle Get to a Gold Nucleus?

[DMac]

[SOLVED] Electrostatics Problem

I've been trying to solve this problem for the longest time, but it doesn't seem to give enough information.

Homework Statement

If an alpha particle is fired straight at a nucleus of a gold atom, (q = 79 e) at a velocity of 3.3 x 10^4 m/s, how close will it get to the gold nucleus?

mass of alpha particle = 6.64 x 10^27 kg
e = 1.6 x 10^-19 C
(I looked this up on the internet, it's not given in the question, but: charge on alpha particle = +2e)

Homework Equations

I have v1 and v2 (I'm assuming it's zero), but that's pretty much it. I can't calculate the force between these charges because I don't know their distance.

I also tried: W = Delta Ek
W = (0.5)(m)(v)^2
W = (0.5)(6.64x10^-27)(3.3x10^4)^2
W = 3.62x10^-18 J
F * d = 3.62x10^-18

But, as I said, I can't calculate the force, and the "d" in that formula gives me the distance it traveled, not how close it got to the nucleus.

The Attempt at a Solution

I'd really like to post an attempt, to show that I truly tried this problem, but I'm really stumped. I have thought of using the electrostatic force equation, but that didn't work out. I'm really not sure where to start.

Could someone please tell me the general direction I should go? I don't need an actual answer, I just want some guidance. Thanks.

 

[marcusl]

You should look up the charge of a gold nucleus, then go back to Coulomb's law--specifically, the energy between two charged particles. Think about how the energy changes as you bring the two particles towards each other from infinity.

 

[DMac]

Well I know the charge on the gold nucleus is 79e. (1.264 x 10^-14 C)

The alpha particle is repelled by the nucleus, so the electric potential energy increases as it is brought towards the nucleus.

E = k q1 q2 / r
E = 3.65 x 10^-26 / r

For the force,

F = 3.65 x 10^-26 / r^2.

But I still don't see how this can get an answer.

 

[DMac]

Is it because all of the kinetic energy is converted into electric potential energy? If that is the case, I guess I could isolate for r. =P

 

[Redbelly98]

Is it because all of the kinetic energy is converted into electric potential energy?

Yes.

 

[DMac]

Thanks!