How Close Does the Alpha Particle Get to the Platinum Nucleus?

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BubbaJonze
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Homework Statement



An Alpha particle of a Rutherford-type scattering experiment is shot directly toward a platinum nucleus. One of the particles was apparently aimed directly at the centre of the nucleus cause it shot 180 degrees back along the same path. How close does it come to the platinum nucleus?

Homework Equations



Kinetic energy of Alpha particle: 4.0 MeV

The Attempt at a Solution


Doubt this is right, but I figured energy conservation had something to do with it.
[tex]\sum[/tex]E[tex]_{}initial[/tex]=[tex]\sum[/tex]E[tex]_{}final[/tex]

After some re-arranging, this is what I came up with:
[tex]\frac{Kq_{1}q_{2}}{r}[/tex]=E

So I figured I was solving for r, in which case i got this formula [tex]\frac{Kq_{1}q_{2}}{E}[/tex]
K= Planck's constant
q= elementary charge
E= 4.0MeV (?)

Can someone tell me if I'm anywhere near the right track? We didn't really learn about it, but my instructor stated it was a "challenger" question.
 
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You are on the right track.

K is not Planck's constant though. [itex]K =9X10^9[/itex].

Planck's constant, on the other hand, is usually denoted by h and is [itex]6.626X10^-34[/itex]

Just remember that while q_1=electronic charge, q_2 will be the charge of the other body in the problem, the platinum nucleus.
 
G01 said:
You are on the right track.

K is not Planck's constant though. [itex]K =9X10^9[/itex].

Planck's constant, on the other hand, is usually denoted by h and is [itex]6.626X10^-34[/itex]

Just remember that while q_1=electronic charge, q_2 will be the charge of the other body in the problem, the platinum nucleus.
Oh, that's right. I guess I got a little confused since Coulomb's Law was in my last unit and now my constants are all messed up. Thanks.