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In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.

But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.

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    $\begingroup$ The body of your question has nothing in particular to do with neutrons - perhaps you mean nucleus? (Yes, very similar names). As for scattering off of electrons, the alpha particles most certainly were interacting with the electrons, and as a result lost energy as they traversed the material. This so-called electronic stopping is very small compared with a nucleus-nucleus collision event primarily because the mass of the electron is so much smaller than even a proton - you just can't have much energy transfer in such a collision. $\endgroup$
    – Jon Custer
    May 25, 2016 at 17:51

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Protons and neutrons have a mass approximately 1836 times greater than an electron, ignoring relativistic effects and nuclear binding energies. There are 4 of these baryons in an $\mathrm{\alpha}$-particle. There is no way a puny electron could do much to deflect an $\mathrm{\alpha}$-particle that is almost 7500 times more massive. This is like if a 150-lb human tried to block a 747 (the 747-100's maximum takeoff weight is only about 6400 times greater than the person's).

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    $\begingroup$ Your answer is essentially on the right track but you should point out that the "target size" isn't actually measured in mass but in terms of a quantity named the cross-section. $\endgroup$
    – MaxW
    Oct 19, 2016 at 18:25
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    $\begingroup$ @MaxW You're absolutely correct. I was trying to aim more for a more intuitive answer, but yes, a cross section would be the technically correct way to approach this problem. $\endgroup$
    – Zhe
    Oct 19, 2016 at 19:16

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