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sorry if the question is too elementary. From: The Britannica Guide to Particle Physics:

The sizes of atoms, nuclei, and nucleons are measured by firing a beam of electrons at an appropriate target. The higher the energy of the electrons, the farther they penetrate before being deflected by the electric charges within the atom. For example, a beam with an energy of a few hundred electron volts (eV) scatters from the electrons in a target atom. The way in which the beam is scattered (electron scattering) can then be studied to determine the general distribution of the atomic electrons.
At energies of a few hundred megaelectron volts (MeV; 106 eV), electrons in the beam are little affected by atomic electrons. Instead, they penetrate the atom and are scattered by the positive nucleus. Therefore, if such a beam is fired at liquid hydrogen, whose atoms contain only single protons in their nuclei, the pattern of scattered electrons reveals the size of the proton. At energies greater than a gigaelectron volt (GeV; 109 eV), the electrons penetrate within the protons and neutrons, and their scattering patterns reveal an inner structure. Thus, protons and neutrons are no more indivisible than atoms are. Indeed, they contain still smaller particles, which are called quarks.

I'm confused by this. I understand that the increasing velocity of the electron particles in the beam can overcome repulsive force of the electron cloud in an atom. Thus detecting size of the atom and even particular electrons.

But what happens to beam particles in MeV range? At higher velocities, when they do reach nucleus, doesn't the positive charge then attract beam particles? How does it repel them? Is it that they have so much speed that they on near miss of a proton deflect or is something else happening here?

What happens on GeV range? Do electron beam particles pass through proton to allow detection of quarks?

Or is there happening something completely differentTM, like total annihilation of collided particles and recreation of particles from energy.

Thanks for your time...

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up vote 4 down vote accepted

Good question!

To understand what happens consider the analogy of firing tennis balls at a planet. The tennis balls are attracted by the planet's gravity, so if you fire the tennis balls slowly (specifically at lower than the escape velocity) they'll hit the planet or possibly end up orbiting around it. However suppose you're firing the tennis balls much faster than the escape velocity. If the tennis ball misses the planet by a wide margin then nothing much happens. If the tennis ball grazes the planet then it's direction will be changed by the planet's gravity, but it will still carry on past the planet. If the tennis ball is aimed directly at the planet it will hit the surface, but because it has far more than the escape velocity it will bounce off i.e. it will scatter by a large angle. So by looking at the scattering angles of the tennis balls you can tell how big the planet is.

The same thing happens when you fire electrons at protons. The binding energy of a hydrogen atom is only about 20eV, so even low energy electron beams won't form hydrogen atoms. Instead the electrons bounce off and by looking at the distribution of scattering angles you can get an effective size for the proton. However things are more complicated than the tennis ball/planet analogy because at high energies the electrons can react to produce other particles. Nevertheless you can still extract a proton size subject to some mathematical modelling.

Once the de Broglie wavelength of the electron is much smaller than the size of the proton the electron no longer interacts with the proton as a whole. Instead it starts scattering off the quarks (real and virtual) inside the proton. Again, you look at the scattering angles to get some idea of the quark size, but so far at the highest energies we can generate quarks still appear to be pointlike.

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Thank you very much for a clear answer! :o) –  Hidden Scholar Dec 5 '12 at 9:01
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To expand a little on John's nice answer in the intermediate (few GeV) range you can map the momentum distributions of the nuclear components and relate it to the spacial distribution. –  dmckee Dec 5 '12 at 14:14
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