Exploring Alpha-Particle Scattering, Nuclear Decay, & Mass Calculation

[ryan9907]

hello there! your help is really appreciated.

1.For the Ruther alpha-particle scattering experiment, how come we are only concerned with the nucleus of the gold with repulsion on the alpha particle? what about the electrons in gold? Don't they attract the alpha particle?

2. To calculate the Electric potential energy between the alpha particle and the gold nucleus we use the formula Qq/4pi(Eo)R, whereby from my answer scheme Qq= 2(1.6x10^-19) x 79(1.6 x 10^-19).

Is this only concerning the repulsion between the protons? How come the electrons of the Gold is not considered?

Is it different if the question is phrased this way : calculate the e.p.e between the alpha particle and the gold nucleus. how do you determine Qq since Gold has both electrons and protons? which one do i consider?

3. So for EPE, do it apply to any atoms? like between a hydrogen atom and a carbon atom, is there EPE between them? because I'm confused with like the Qq part of the formula,its like hydrogen has electrons outside the nucleus and the nucleus inside has protons, so i don't know if it's attractive of repulsion when you place it beside another atom.

4. For nuclear decay, like alpha decay, the nucleus decay and emit a alpha particle nuclues with 2 protons and 2 neutrons, so what happened to the electrons?

5. how do you calculate the the mass of a uranium 238 atom. My answer says 938u. how come u is used and what does u means? don't protons and neutrons have different mass?

Thanks for helping :)

 

[bcrowell]

Neglecting the electrons is an approximation. When Rutherford and Marsden originally published their results, they did have to carefully discuss the possible effects of electrons. But basically it is not plausible that the electrons could have had any significant effect on their results. An electron is thousands of times less massive than an alpha particle, so even if the alpha particle collides head-on with an electron, it's not possible for the alpha to turn around and go back the way it came. Therefore the alpha can't be scattered at 180 degrees due to a close encounter with a single electron. That doesn't prove that it can't be scattered at 180 degrees by a series of such encounters, but I hope you can see what such a scenario is pretty implausible. (This is all without quantum mechanics, which didn't exist in 1909. With quantum mechanics, you can visualize the electrons as a could spread out over scales of 10^-10 m. In that picture, there is no way for the r between the alpha and the electron cloud to get down to the 10^-14 m scales that would be needed in order to turn the alpha around. When the alpha gets to within ~10^-14 m of the gold nucleus, the electric field it experiences from the electrons is negligible compared to the field from the nucleus.)

 

[Parlyne]

The way to account for the electrons (at least if you're just looking to understand potential energy) is to treat the charge of the atom as a function of r. In effect, what you're doing is summing all the charge within a distance r of the center of the nucleus. (Strictly, thinking quantum mechanically, it's the expectation value of the amount of charge within r of the center; but, that's not a horrible complication.) You should find that on distance scales much larger than 10^-10 m, or so, this function is approximately 0. As you decrease r, it smoothly increases up to the charge of just the nucleus, which it certainly reaches by the time you get down to 10^-14 m.