# How can we derive the maximum scattering angle from the experiment of Geiger and Marsden?

Schematic diagram of apparatus used by Geiger and Marsden to observe scattering of $$\alpha$$ particles past 90°. "A small frac­tion of the $$\alpha$$ particles falling upon a metal foil have their di­rections changed to such an ex­ tent that they emerge again at the side of incidence." The scattered $$\alpha$$ particle struck a scintillating screen where the brief flash was observed through the microscope.

From H. Geiger and E. Marsden, Pro­ceedings of Royal Society (London) 82, 495 (1909).  This is about the alpha-particle scattering experiment. The book says that some alpha-particles were scattered with the angle larger than 90 degree, which is impossible if the particles were colliding with electrons of small mass. There's the formula of the maximum scattering angle at the bottom of the second image. I don't get how the formula is derived. What does the angle have to do with the momentum?

• Please don't abuse screenshots and type your question completely and clearly. Also, use Mathjax syntax for equations. Thank you. Oct 18, 2022 at 15:36
• This is classical kinematics. Conserve momentum and energy in the center-of-mass frame. You can’t backscatter if the incident particle is more massive than the other particle. Oct 18, 2022 at 15:43

The equation use this approximation when $$\theta \ll 1$$
$$\theta \approx \tan \theta = \frac{\Delta P_{\alpha}}{P_\alpha}$$