I understood the result of this experiment that the nucleus is nearly empty and things like that. But what I have on mind is that when an alpha particle goes nearer to the thin gold foil why couldn't it capture electrons from gold and convert to very stable helium gas? It could also happen right? All the textbooks only mention the deviation of alpha particles. Why couldn't the above happen or is there anything wrong about my question?
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asked Dec 11, 2016 at 21:58
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2 Answers
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Your assumption is correct. For alpha particles, the main contribution to the total stopping power can be attributed to the electronic stopping power, i.e. inelastic collisions with electrons. Only a small contribution comes from the nuclear stopping power, i.e. elastic Coulomb collisions in which recoil energy is imparted to atoms.
The stopping power of a material is defined as the average energy loss per path length that the alpha particle suffers when travelling through the material.
According to the International Commission on Radiation Units and Measurements (ICRU) Report 49 Stopping Powers and Ranges for Protons and Alpha Particles (1993), the contributions to the total stopping power for alpha particles in gold are as follows.
Typical low-energy alpha particles with $E=1\ \mathrm{MeV}$:
- Electronic stopping power: $3.887\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
- Nuclear stopping power: $8.394\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$
Typical high-energy alpha particles with $E=10\ \mathrm{MeV}$:
- Electronic stopping power: $1.650\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
- Nuclear stopping power: $1.315\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$
Since the energy that is required for excitation or ionization is only a few $\mathrm{eV}$, an alpha particle with an initial energy of a few $\mathrm{MeV}$ can liberate many electrons on its path. When the alpha particle has sufficiently slowed down due to the stopping interactions, it can finally catch two electrons to form a neutral helium atom. Hence, you can release and detect helium when you dissolve old minerals of uranium or thorium.
In the Rutherford experiment, however, the used gold foil was very thin ($<1\ \mathrm{\mu m}$; note that gold is very malleable and can be beaten thin enough to become semi-transparent). Even for low-energy alpha particles of only $E=1\ \mathrm{MeV}$, the range in gold is still $r/\rho=3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}$, which can be calculated using the continuous-slowing-down approximation (CSDA), i.e. by integrating the reciprocal of the total stopping power with respect to energy. Considering a density for gold of $\rho=19.3\ \mathrm{g\ cm^{-3}}$, the range $r$ can be calculated as
$$\begin{align} r&=\frac{3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}}{19.3\ \mathrm{g\ cm^{-3}}}\\[6pt] &=2.06\times10^{-4}\ \mathrm{cm}\\[6pt] &=2.06\ \mathrm{\mu m} \end{align}$$
I.e., the experiment was deliberately designed so that the alpha particles were not completely stopped and converted to neutral helium atoms within the gold foil.
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The atom, not the nucleus, is mostly empty. That's a pretty big conceptual difference.
Now, there's no reason to believe that the alpha particles haven't picked up an electron on their way to the gold foil before they even strike the foil. The detector on the other side just detects high energy particles, so we can't actually tell if how they're charged. The deviation of the alpha particles (very slight at best) would still be the roughly the same.
