1
$\begingroup$

Can an alpha particle (or any charged particle) can penetrate through nucleus of gold (or any other) atom ?

Today I was watching a lecture on the "Estimation of Size of Nucleus" which was a subtopic of Rutherford's Atomic Model, where a respected teacher elaborated the calculations made by Rutherford to estimate the size of nucleus of gold atom. In short, Rutherford considered that the size of nucleus would be less than the closest distance that an alpha particle can reach near the nucleus and made calculations for it taking the maximum Kinetic Energy of a naturally occurring alpha particle, charge of alpha particle, charge of nucleus etc. There he (the teacher) said this statement "The alpha particle cannot get inside the nucleus."

Then I began wondering about it. Why an alpha particle cannot get inside a nucleus ?

What would have happened if Rutherford had accelerated the alpha particle to a very high Kinetic Energy and then bombard it on the gold nucleus ? Will it penetrate through it or will it undergo some nuclear reaction destroying Sir Rutherford's laboratory ?

At first I thought since the alpha particles have high Kinetic energy so they will penetrate through gold nucleus but slowly my thought process shifted to a more reasonable argument that since the nucleus and alpha particles are both positively charged so when they will reach closer to each other the repulsion among them will tend to infinity and whatever energy the alpha particle have it cannot overcome such repulsion thus it cannot penetrate through or more precisely cannot even go much closer to the nucleus.

Are there any other possible arguments that can be made about this or I hit the bulls' eye ?

$\endgroup$
1
  • $\begingroup$ There are two, rather different questions in your title alone. Please ask one question at a time. $\endgroup$ – my2cts Oct 26 '20 at 10:37
2
$\begingroup$

Your first thoughts were correct. $\alpha$ particles can and do penetrate the nucleus if they have enough energy.

These deviations from the Rutherford Scattering formula are actually one of the standard ways that the $~\sim 10^{-15} m$ size of the nucleus has been established. See for example http://www.personal.soton.ac.uk/ab1u06/teaching/phys3002/course/03_diffraction.pdf

$\endgroup$
0
1
$\begingroup$

Are there any other possible arguments that can be made about this or I hit the bulls' eye

For low energies yes. For higher energies, the electric repulsion is overcome by the extra energy. For example two alphas colliding are proposed for a fusion reactor.

For the high energies of the LHC ion ion collisions, a lot of things happen, and are due to the destruction of the nuclei.

cmsleadlead

Lead ions collide in the CMS detector

It seems gold is not too expensive to put up for destruction in experiments !

$\endgroup$
2
  • 1
    $\begingroup$ FWIW, gold ions have been collided with each other, in quark-gluon plasma experiments. There's a nice image at en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider $\endgroup$ – PM 2Ring Oct 26 '20 at 8:52
  • $\begingroup$ @PM2Ring thanks for the info, I am out of date. I will use the link .from your link $\endgroup$ – anna v Oct 26 '20 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.