Calculate Coulomb Barrier Height & Distance with 7.68MeV Alpha Particle

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To calculate the Coulomb barrier height and distance for a 7.68 MeV alpha particle colliding with gold nuclei, the key is to equate the alpha particle's kinetic energy to the electric potential energy at the point of closest approach. The barrier height is at least 7.68 MeV, as this is the energy required for the alpha particle to reach the nucleus. The distance can be calculated using the formula d = (keqα*qAu)/(7.68*10^6*1.6*10^-19), resulting in approximately 2.96 x 10^-14 m. This approach assumes the alpha particle starts from infinity, simplifying the calculations. Understanding the Coulomb barrier is crucial for analyzing nuclear interactions in scattering experiments.
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Its something that I should be able to figure out easily but my brain is refusing to work this term. So any help will be appreciated.


In a scattering experiment, gold nuclei (Au Z=79 A=197) are bombarded by alpha particles (He Z=2 A=4)

If the Kinetic Energy of the apha particles is 7.68MeV calculate the height of the Coulomb Barrier in MeV, and the distance between the alpha particle and the barrier when the particle collides with the barrier.



So can anyone help. I should really be able to do this but my brain is goo, my fault for giving myself an easy summer.
 
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Uh to be more specific I am sure that's the answer is worked out using conservation of Energy, however I'm not sure what formulae to use as everyone I try requires prior knowledge of the distance between the particles, which I don't have.
 
Have you tried assuming the alpha particles are coming from infinity. The nature of the fields is such that you can start arbitrarily far away and arrive at the same result.

Njorl
 
I just started studying electricity a few weeks ago so I haven't seen the term Coulomb barrier before, but is that just the point where the repulsion between the alpha particle and the protons in the nucleus becomes strong enough to stop the alpha particle?

If so, wouldn't you solve this by finding the distance from the center of the gold radius at which the alpha particle's electric potential energy is equal to its initial kinetic energy?
 
Looks a bit like a trick question to me. I'd say, if we know that the alpha collides, then the barrier must be at least 7.68MeV high, or the alpha would go over it. Well, what is the distance of two objects colliding?

(I've been bashed before for answers like this...)
 
We don't know we have to figure it out in the next question, using the answer to the 1st (I assume)
 
I don't know it this is any help to you but I had a homework problem like this last week. It didn't refer to the "Coulomb barrier". It just asked how close an alpha particle would get to a gold nucleus if the particle started out at a certain initial velocity (so we had to compute the initial kinetic energy from the given velocity).

In your case, you want
(1/2)mAuvi2 = (keqαqAu)/d

and you are given that (1/2)mAuvi2 = 7.68 MeV

d = (8.99*10^9*2*1.6*10^-19*79*1.6*10^-19)/(7.68*10^6*1.6*10^-19)

d = 2.96 * 10^-14 m

Now, if the "height" of the barrier is defined as the amount of potential energy (and I don't know that this is the definition) you can compute the potential energy at this distance using
U = (keqαqAu)/d
 
Thanks I'll give that a try
 
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