# Deflection of Alpha Particles in Rutherford's model of atom

Rutherford stated that deflection of alpha particles is due to repulsive positive charge of nucleus. Why can't alpha particles be deflected by attraction of so many electrons revolving around nucleus?

• Mar 12, 2017 at 21:52
• Mar 12, 2017 at 21:55

See, what the Geiger-Marsden-Rutherford experiment achieved was the following: by bombarding (with alpha particles) a one-atom thick gold sheet and counting how many alpha particles passed through, they were able to relate the already known atomic radius with the actual area that could get collided by alpha particles. Figure 1 shows a sketch of the apparatus. Figure 1: depiction of the apparatus (courtesy of Wikipedia).

Now follows a simplified rationalisation.

Assume you have an atom-thick sheet of some material. By knowing the element's molar mass and the mass of the sheet you can calculate how many atoms $$n$$ are there, and knowing the sheet's surface area $$A$$ allows you to calculate the atomic surface density $$\sigma = n/A$$.

Now if the atoms have radius $$\rho$$, an area $$A_\rho = n \pi \rho^2$$ of the sheet should be covered by atoms. This means a ratio

$$r = \frac{A_\rho}{A} = \pi \rho^2 \sigma$$

is actually covered by atoms and $$f = 1 - r$$ is the ratio of "free space".

Rutherford actually tested if this made sense by throwing particles at the sheet: those scattered simply hit something, the rest passed through. Imagine $$M$$ particles were thrown and $$m$$ of those passed through. If $$M$$ is big enough we can set $$f_R = m/M$$ as Rutherford's "free space" estimate. Going backwards we may set the coverage ratio

$$r_R = 1 - f_R = 1 - \frac{m}{M} = \frac{A_R}{A} = \pi \rho_R^2 \sigma$$

were $$\rho_R$$ above is the measured radius of the collision cross section of an atom. The experiment has shown that $$r_R \ll r$$ and thus $$\rho_R \ll \rho$$, which implies that most of the atomic mass is buried inside the atom. In fact (Wikipedia),

The diameter of the nucleus is in the range of $$1.75$$ fm ($$1.75 \times 10^{-15}$$ m) for hydrogen (the diameter of a single proton) to about $$15$$ fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about $$23,000$$ (uranium) to about $$145,000$$ (hydrogen).

It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge.

--- Ernest Rutherford

Thus, nuclei are simply very small, but very dense, as stated by Enigma and Ilmari Karonen.

• Don't you mean "simply very small" (but dense)? A big low-density nucleus would match the Thomson plum pudding model, which is precisely what Rutherford's experiment disproved. Mar 12, 2017 at 22:15
• You are right! I'll change the wording, thanks! Mar 12, 2017 at 22:42

The simple reason for this being the ratio of masses of alpha particles and electrons.....Both experience equal electrostatic forces which leaves no option for the electron but to get displaced from its current position ( it gains potential energy to do so) whereas in the case of alpha-nucleus interaction,the size of a nucleus is comparable (in the case of the gold foil he used...way more than that of an alpha particle) to that of the projected alpha particle hence causes a measureable deflection on the alpha particle....Hopefully you got your answer

• For comparison, if a 747 runs into a person, which one is deflected more? I actually ran this analysis, and the ratio of rest masses is comparable.
– Zhe
Mar 12, 2017 at 16:07

I was also wondering upon the fact that it could be the electrons which caused the sway, but you see if a certain force were acting on two particles, one lighter than the other, the lighter one will be the one feeling the "force". For more clarity refer to Newtons second law in acceleration format. So the only left out thing was a nucleus, which is way more massive than the alpha particle, again apply our first logic here(stated above) and you see, this time its the He(2+) that has to do the displacement. Hope it resolves your issue.

Well gentlemen we can't deny that fact that there might be some forces of attraction between the alpha particles as well as the electrons,because the alpha particles passed through the atom by small distance margin from nucleus and stroked the zinc sulphide coating but it wasn't clearly stated by Ernest Rutherford that whether the deflection was also caused due to alpha particle attraction and election attraction. Also, attraction (interaction)is between comparable things like a heavy nucleus and heavy alpha particle but it's silly to think about alpha particle and an electron , the alpha particle is approximately 4000 times heavier than electron and it is travelling with a high speed (energy) so we can say that the interaction time might be very much less in order to show its effect and we can underestimate it . though most of the physics practicals are based on assumptions and we have to underestimate those assumptions that's way because the reason for performing the experiments was to propose a model of an atom (accidentally) , to find where is the mass of the atom is concentrated. that's why we got the so we don't have anything to do with the alpha particle and electron attraction.