Why does mass matter in Rutherford's experiment?

[Jufa]

TL;DR Summary

Why do we neglect defection due to electrons?

Here it goes. I'm reading some notes on the Rutherford (gold foil) experiment and they first state what one should expect if the atom model was like the one described by Thomson (plumb pudding model). In order to somehow predict what the deviation should be when throwing alpha particles towards the foil, the following considerations are commonly made:
-Only electrical force has to be taken into consideration when dealing with alpha particles and atoms interaction.
-The less concentrated is the positive charge the least will be the deviation experimented (that comes out from Coulomb's Law).
-Interaction between electrons can be neglected, since they MASS is known to be negligible.
I don't really understant what roll can mass play here. Indeed, if one assumes that electrons are smaller than the positive charge, it should be its interaction the most strong one, since electric field on its surface would be higher than on the positive charge's.

 

[hutchphd]

My interaction with a hornet may be stronger than with a buffalo but I will bounce backward from the buffalo.

 

[Jufa]

My interaction with a hornet may be stronger than with a buffalo but I will bounce backward from the buffalo.

Then, deflections are caused by impact? What role has electrical force then? I find it quite misleading.

 

[Bandersnatch]

The experiment measures deflection, so a change in momentum after collision. Collision here meaning electric force interaction. Colliding with a particle of negligible mass changes momentum negligibly.

 

[Jufa]

The experiment measures deflection, so a change in momentum after collision. Collision here meaning electric force interaction. Colliding with a particle of negligible mass changes momentum negligibly.

But once electrical force is considered. I don't see how mass is relevant here. If we study it from the point of view of two marbles colliding everything gets as simple as you depict it. Nonetheless, I've read (Wikipedia) that that's not how it works. I see it like the alpha particle never does collide with the atom. It is deflected when it does come near enough to the atom to substantially feel the electric field. Thus, I don't see where mass (of the electron or the positive charge) is important here. I see how the mass of the alpha particle is important but not the others, since interaction (as I understand) is somehow "wireless".

 

[Ibix]

The alpha particle deflects the nucleus and the nucleus deflects the alpha particle. Each one applies an electromagnetic force to the other. Mass comes into it because, given equal forces, the less massive particle will accelerate more.

It's not unlike trying to play snooker with a table tennis ball instead of the white ball. No matter how hard you hit the table tennis ball, the colour balls will barely twitch. The only difference is that the particles are mutually repelled by Coulomb repulsion, not physical contact.

 

[Jufa]

The alpha particle deflects the nucleus and the nucleus deflects the alpha particle. Each one applies an electromagnetic force to the other. Mass comes into it because, given equal forces, the less massive particle will accelerate more.

It's not unlike trying to play snooker with a table tennis ball instead of the white ball. No matter how hard you hit the table tennis ball, the colour balls will barely twitch. The only difference is that the particles are mutually repelled by Coulomb repulsion, not physical contact.

I understand what you say. The less massive is the charge we are dealing, the more it will be accelerated. In this case, if the interaction is with the massive positive charged sphere, the sphere does not deflect and if we consider an electron, it does. But, regarding to the alpha particle, both scenarios are the same. The same momentum is tranferred to it because this momentum exchange only depends on the electrical force and the amount of time the interaction took. I don't see how the electron mass concerns anything else than the electron itself.

 

[Jufa]

I understand what you say. The less massive is the charge we are dealing, the more it will be accelerated. In this case, if the interaction is with the massive positive charged sphere, the sphere does not deflect and if we consider an electron, it does. But, regarding to the alpha particle, both scenarios are the same. The same momentum is tranferred to it because this momentum exchange only depends on the electrical force and the amount of time the interaction took. I don't see how the electron mass concerns anything else than the electron itself.

Indeed, in Thompson's Model the electron was attached to the atom, so its in order to compute its acceleration the whole mass should be taken into consideration (if one considers that it cannot scape due some other kind of stronger forces).

 

[Dale]

But, regarding to the alpha particle, both scenarios are the same. The same momentum is tranferred to it because this momentum exchange only depends on the electrical force and the amount of time the interaction took

Why do you think that the amount of time is the same? (It isn’t) The force is not necessarily the same either.

 

[Jufa]

Why do you think that the amount of time is the same? (It isn’t) The force is not necessarily the same either.

Well, if the electron charge is equally charged (in magnitude) than the positive Sphere, then both forces should be equiparable. Anyway, even if they aren't and if the amount of time is not the same. I would like to know why it is the mass of the electron the reason of this amount of time to be shorter or the magnitude of the force to be lower. I don't find de connexion self-evident nor trivial.

 

[Paul Colby]

As I recall Rutherford worked out an expression for the coulomb differential cross section, two charged massed elastically scattering. Didn't he then compare his data using this? $\alpha$ on electron wouldn't fit the observed differential cross section where as $\alpha$ on gold mass lumped at a point did match the data. Game over.

 

[Dale]

Anyway, even if they aren't and if the amount of time is not the same. I would like to know why it is the mass of the electron the reason of this amount of time to be shorter or the magnitude of the force to be lower. I don't find de connexion self-evident nor trivial.

The best way to see this is to work it out mathematically. For simplicity consider a head on collision so that it is a one dimensional problem, and consider all particles to be point particles. Use the actual charges and masses and just calculate the motion and the forces.

 

[Mister T]

First of all, a collision occurs when objects interact, they do not necessarily need to make contact, which is anyway an illusion.

The issue is that an alpha particle is either going to interact with a gold nucleus or with an electron. The alpha particle is much more massive than the electron, so the electron will undergo a huge deflection compared to the alpha particle. Likewise, a gold nucleus is more massive than an alpha particle, so the alpha particle will undergo the larger deflection.

If you shot a small pellet from a pellet gun would you expect the pellet to undergo just as large a deflection from a collision with a spec of dust as with a bowling ball? You're focusing on the electric charge, which determines the amount of force. But it's the mass which determines the amount of deflection.

 

[Jufa]

The best way to see this is to work it out mathematically. For simplicity consider a head on collision so that it is a one dimensional problem, and consider all particles to be point particles. Use the actual charges and masses and just calculate the motion and the forces.

But if I do so, I only have two equations, momentum conservation and energy conservation (if one considera elàstic rebound).

The best way to see this is to work it out mathematically. For simplicity consider a head on collision so that it is a one dimensional problem, and consider all particles to be point particles. Use the actual charges and masses and just calculate the motion and the forces.

That is What I'm thinking of when I'm trying to depict it Dale. But momentum conservation Will only Tell me the alpha's deflection if I know the momentum transfer. Usually, problems like that are presented in such a form that one knows the deflection of one body (say the electron) and then one can compute the deflection of the other so that they cancel out. I don't see how the electron's mass determines the amount of momentum transfer. To me momentum transfer is fixed parameter fixed by the electron and alpha's particle (electrical forces) and time of interaction. Electron's mass, (to me, and I'm concerned that I'm probably wrong) only tells us how big is the final velocity of the electron. Moreover, in Thompson's model electrons where thought to be attached at the surface of the atom so I guess that if one wanted to compute the electron's deflection, the mass of the whole àtom needed to be considered. The point here is that I'm already convinced that electron interaction can be somehow neglected and I'm missing something, I just don't see it trivial.

 

[Jufa]

First of all, a collision occurs when objects interact, they do not necessarily need to make contact, which is anyway an illusion.

The issue is that an alpha particle is either going to interact with a gold nucleus or with an electron. The alpha particle is much more massive than the electron, so the electron will undergo a huge deflection compared to the alpha particle. Likewise, a gold nucleus is more massive than an alpha particle, so the alpha particle will undergo the larger deflection.

If you shot a small pellet from a pellet gun would you expect the pellet to undergo just as large a deflection from a collision with a spec of dust as with a bowling ball? You're focusing on the electric charge, which determines the amount of force. But it's the mass which determines the amount of deflection.

I'm asking why we can neglect electron's interaction just because they are poorly massive. And by "Why" I mean if It can be clearly shown by looking the equations of some fundamental law. You are just telling me that this is what happens.

 

[Jufa]

I'm asking why we can neglect electron's interaction just because they are poorly massive. And by "Why" I mean if It can be clearly shown by looking the equations of some fundamental law. You are just telling me that this is what happens.

If the electron's acceleration due to interactio was fixed, mass would tell us everything about its momentum gain. But I don't think this is the case.

 

[Drakkith]

I'm asking why we can neglect electron's interaction just because they are poorly massive. And by "Why" I mean if It can be clearly shown by looking the equations of some fundamental law. You are just telling me that this is what happens.

Set up a spreadsheet to track the position, velocity, momentum, and force on each of two particles in a 1d collision. I think you'll quickly see for yourself what's happening.

 

[Jufa]

Set up a spreadsheet to track the position, velocity, momentum, and force on each of two particles in a 1d collision. I think you'll quickly see for yourself what's happening.

I try. But I encounter a problem: I have a huge coin (alpha particle) directed towards a tiny one (electron) now I have that if, when colliding, the tiny coin gains an ordinary amount of velocity in one direction, the huge one will gain a negligible amount in the opposite sense. I think we do agree until this moment. But my question is: why the velocity gain has to be ordinary?
Remember this refers to Thompson's model, so the electrons were believed to be attached to the atom. I don't understand why we don't just assume that the moment transferred to the electrons has to be somehow (shared) by the whole atom.

 

[Dale]

But if I do so, I only have two equations, momentum conservation and energy conservation (if one considera elàstic rebound).

That is definitely the easiest approach. That will also get you the right answer, but you won’t see any details about the duration of the interaction or the force.

Instead, you can use Coulomb’s law and Newton’s 2nd law. You will get the same overall result, but you will see the details of the interaction.

You might want to do it both ways to see that they both agree and bolster your confidence in using the simple approach in the future.

But momentum conservation Will only Tell me the alpha's deflection if I know the momentum transfer.

Write down the momentum conservation and the energy conservation equations and solve them simultaneously for the two final velocities. You will get two solutions, post them and let’s discuss their meaning.

And by "Why" I mean if It can be clearly shown by looking the equations of some fundamental law.

Yes. You can show the same overall result using either conservation of momentum and energy or using Coulomb’s law and Newton’s 2nd law. The conservation laws are easier (algebra instead of calculus) but less detailed (no forces or interaction time).

 

[Jufa]

That is definitely the easiest approach. That will also get you the right answer, but you won’t see any details about the duration of the interaction or the force.

Instead, you can use Coulomb’s law and Newton’s 2nd law. You will get the same overall result, but you will see the details of the interaction.

You might want to do it both ways to see that they both agree and bolster your confidence in using the simple approach in the future.

I think I've got somewhere Dale. Energy conservation sets a limit on how big can the velocity gain be. Indeed, the collision might not be elastic, since heat losses, setting an even lower upper bound for velocity gain. Therefore, non-massive objects cannot transfer (effectively) momentum. Just as if you have a huge coin and make it collide towards a little one. Momentum conservation does not forbid a huge deflection but energy conservation does. Am I right? (Thank you all for being that patient

 

[jbriggs444]

Momentum conservation does not forbid a huge deflection but energy conservation does.

Yes. Imagine shooting a massive ball at a tiny one. But coat the tiny ball with explosives. Now the massive ball can rebound at an arbitrarily large speed and at an arbitrarily great deflection angle.

 

[Dale]

Energy conservation sets a limit on how big can the velocity gain be.

Yes. It also limits how small it can be (assuming an elastic collision). There are only two velocities that are consistent with both conservation of momentum and energy.

 

[Jufa]

Nice. Thank you guys for helping, can sleep well tonight now :).

 

[Drakkith]

Remember this refers to Thompson's model, so the electrons were believed to be attached to the atom. I don't understand why we don't just assume that the moment transferred to the electrons has to be somehow (shared) by the whole atom.

Imagine a baseball that's attached to a 30 ton dump truck with a very strong bundle of bungee cords. Now hit the ball with a baseball bat moving at mach 58,000 (2x10⁷ m/s, or 6.6% the speed of light, the velocity of an incoming alpha particle in Rutherford's experiment). The baseball (assuming it remains intact) is going to go flying off at an enormous speed. The bungee cords holding it to the truck are going to break, and the truck isn't going to move hardly at all since it is 200,000 times more massive than the baseball and the bungees simply can't exert enough force before breaking. Even setting the truck on a frictionless surface won't change things much. A negligible amount of momentum will be transferred to the truck.

Of course, our bat should have a mass of about 1 ton in this case (7,340 times as massive as the baseball, same as an alpha particle vs an electron). So it isn't deflected much from the collision either. Perhaps a better example would be a large motorcycle hitting the baseball at mach 58,000.

I wonder if the insurance would cover the damage to the bike...

Edit: Just realized I was confusing pounds and kg. My motorcycle was close to 1000 pounds, but that's only half a ton. Oh well, I think I've gotten my point across.