As far as I know Rutherford measured only the $$I\propto\frac{1}{\sin(\theta/2)^4}$$ intensity dependence, which we would expect for isolated gold atoms. Why didn't he measure Bragg scattering at the gold crystal?
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$\begingroup$ Because the alpha particles were not diffracting off the gold. $\endgroup$– Jon CusterAug 22, 2020 at 14:15
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$\begingroup$ Did they have too high energy and only interacted with the core or why is that? @JonCuster $\endgroup$– user224659Aug 22, 2020 at 14:17
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$\begingroup$ Indeed, they were just scattering from nuclei. You can calculate the de Broglie wavelength and see just how small it is. $\endgroup$– Jon CusterAug 22, 2020 at 14:20
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$\begingroup$ @JonCuster Of course.. Should have thought about that from the beginning! Thank you. $\endgroup$– user224659Aug 22, 2020 at 14:23
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1 Answer
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Rutherford used $^{222}$Ra as an alpha particle source. The resulting $\sim 5.6$ MeV alpha particles have a de Broglie wavelength on the scale of femtometers. On the other hand, the Bragg condition for constructive interference is $n\lambda = d\sin(\theta)$, with $d\sim 1$ nm the spacing of the atomic planes. Since $\lambda/d \sim 10^{-6}$, the scattering maxima are far too close together to resolve.
As a general rule, the energy/momentum of a particular probe (in this case, the alpha particle) defines a length scale which corresponds to the size of features you can expect to observe. If you want to resolve the atomic structure of a solid, you need to use a probe which is sensitive to nanometer-scale phenomena. A $5.6$ MeV alpha is, in a sense, too small; from its point of view, the atoms which comprise a gold crystal may as well be isolated.
answered Aug 22, 2020 at 14:20
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$\begingroup$ Now it makes perfect sense. Thank you. When reading your answer and Jon's comment I remembered that I thought about the very same thing a year ago and got the answer right back then, but totally forgot about it. $\endgroup$– user224659Aug 22, 2020 at 14:25