Talk:Geiger–Marsden experiments

Maths

@Mikhail Ryazanov: Are you good at maths? What do you think of this modification to the equation? The lateral force F exerted on the alpha particle by the gold atom is not constant as the alpha particle crosses the space 2r, so I added something to the start to get the average force exerted. It's unnecessary for this article's needs, but is it correct? Kurzon (talk) 17:08, 31 March 2023 (UTC)

$\theta =\arctan {\frac {\Delta p}{p}}={\frac {2}{\pi }}\cdot \left(\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\cos(x)\mathrm {d} x\right)\cdot k{\frac {Q_{\alpha }Q_{g}}{r^{2}}}\cdot {\frac {2r}{v}}\cdot {\frac {1}{mv}}$

The actual correct expression can be seen in the article Impact parameter. But regarding Rutherford's motivation, it's enough to show that the expected scattering angle is very small, and thus the extremely simplified expression to estimate just the order of magnitude would be perfectly sufficient and more appropriate than anything with integrals and trigonometry (which is useless in the small-angle limit). — Mikhail Ryazanov (talk) 02:00, 1 April 2023 (UTC)

@Mikhail Ryazanov: Hah. That's the story of my life. Just when I think I have gotten the hang of this math stuff, a real math swot comes along and shows me how utterly hopeless I am. It's the Dunning-Kruger effect. Kurzon (talk) 07:50, 1 April 2023 (UTC)

@Chetvorno: Now I'd like to do some maths for an alpha particle that goes right through the center of the gold atom. Have I got this right?

The amount of exerted by the gold atom on the alpha particle as the alpha particle approaches is given by

$\int _{r}^{\infty }k{\frac {Q_{\alpha }Q_{g}}{x^{2}}}dx=\left[-k{\frac {Q_{\alpha }Q_{g}}{x}}\right]_{r}^{\infty }=2.5315\times 10^{-16}~{\text{Joules}}$

where x is the distance between the alpha particle and the center of the atom.

The amount of work exerted on the alpha particle as it passes through the atom is given by

$\int _{0}^{r}k{\frac {Q_{\alpha }Q_{g}x}{r^{3}}}dx=\left[{\frac {1}{2}}k{\frac {Q_{\alpha }Q_{g}x^{2}}{r^{3}}}\right]_{0}^{r}=1.266\times 10^{-16}~{\text{Joules}}$

The total amount of work is 3.797 x 10^-16 J.

The initial kinetic energy of the alpha particle is given by

$E_{\alpha }={\frac {1}{2}}mv^{2}={\frac {1}{2}}\times 6.645\times 10^{-27}\times (1.53\times 10^{7})^{2}=7.7776\times 10^{-13}~{\text{Joules}}$

So the Thomson gold atom does not transfer enough energy into the alpha particle to bring it to a halt, never mind send it flying back to the emitter. Kurzon (talk) 15:20, 10 June 2023 (UTC)

@Leyo: You have a PhD in chemistry. Did I get the above stuff right? Kurzon (talk) 15:54, 27 June 2023 (UTC)

Ref list

1st reference in auto-generated references list doesn’t actually link to the reference in the bibliography. Anyone with a deeper knowledge of wiki templates - is this something specific to this page or is it a wider issue with the template itself? 3nt0 (talk) 00:30, 21 March 2024 (UTC)