How do alpha particles interact with electrons in gold foil experiments?

[tbn032]

My current understanding:

The mass of alpha particle is approximately 7340.6 times higher than the mass of electron.in the gold foil scattering experiment the deflection of alpha particle due to electron is approximately zero.
The reason that is given is that since the electron is much lighter than the alpha particle, it is unable to deflect the alpha particle and the electron itself get scatted instead of deflecting the alpha particle.

My questions are:

1)How can the electron be pushed or scattered by the alpha particle when the electron is revolving in its orbit around the nucleus, and when the electron is bound to the nucleus?

2)I am unable to understand how is the mass of electron instead of charge (alpha particle has 3.2x10^-19C charge and gold foil has 79 electrons each having -1.6x10^19C charge)relevant in deciding whether it will deflect the alpha particle. The reason around this is given that alpha particle will push through the electron and hence it will not get deflected. How does it exactly matter that the mass of electron is significantly less than electron, the electrostatic forces between the electrons and alpha particle would be the same regardless of mass.

 

[PeroK]

If you fire a cannonball at a pea then the pea gets scattered and the cannonball will hardly change direction.

You can see the effect of mass qualitatively by studying particle collisions and using energy and momentum conservation. Mass affects both these quantities. Momentum is $m\vec v$ and all that.

My current understanding
The mass of alpha particle is approximately 7340.6 times higher than the mass of electron.in the gold foil scattering experiment the deflection of alpha particle due to electron is approximately zero.
The reason that is given is that since the electron is much lighter than the alpha particle, it is unable to deflect the alpha particle and the electron itself get scatted instead of deflecting the alpha particle.
My questions are.
1)How can the electron be pushed or scattered by the alpha particle when the electron is revolving in its orbit around the nucleus, and when the electron is bound to the nucleus?

The answer to this is more subtle. There is a probability that the alpha particle interacts with the electron. You need QM to get the fully story. The energy of the alpha particle must be sufficient to unbind the electron from the nucleus. A cannonball could take your head off!

How does it exactly matter that the mass of electron is significantly less than electron, the electrostatic forces between the electrons and alpha particle would be the same regardless of mass.

Newton's second law $F = ma$. Same force for the same duration accelerates a small particle more than a large one.

 

[Nugatory]

How does it exactly matter that the mass of electron is significantly less than electron, the electrostatic forces between the electrons and alpha particle would be the same regardless of mass.

If you’re thinking classically, there is an electrostatic force on the electron from the alpha because they are both charged, and there is an equal and opposite force on the alpha from the electron (by Newton’s third law). But the alpha is far more massive, so the same force moves it much less - basically the electron gets shoved out of the way without appreciably changing the path of the alpha.

Be aware though that there is only so far you can push that classical model. We now know from quantum mechanics that electrons aren’t small solid objects orbiting the nucleus. It’s better (still not right, for that we need quantum electrodynamics with its fairly daunting mathematical price of admission) to think of the electrons as a sort of cloud around the nucleus that no more deflects the alpha than a cloud in the sky deflects a bullet.

 

[sophiecentaur]

think of the electrons as a sort of cloud around the nucleus

Also, the total charge (79⁺) of the gold nucleus could be thought of as a replulsive force which would perhaps yield a smaller measured diameter of that nucleus.