All Problems

Scattering of Particles. Rutherford-Bohr Atom

Problem 6.10

A narrow beam of alpha particles with kinetic energy TT =0.50MeV=0.50 \mathrm{MeV} and intensity I=5.0105I=5.0 \cdot 10^{5} particles per second falls normally on a golden foil. Find the thickness of the foil if at a distance r=15 cmr=15 \mathrm{~cm} from a scattering section of that foil the flux density of scattered particles at the angle θ=60\theta=60^{\circ} to the incident beam is equal to J=40J=40 particles /(cm2s)/\left(\mathrm{cm}^{2} \cdot \mathrm{s}\right)

Reveal Answer
d=(4Jr2T2/nIZ2e4)sin4(θ/2)=1.5μm, where n is the concentration of nuclei.d=\left(4 J r^{2} T^{2} / n I Z^{2} e^{4}\right) \sin ^{4}(\theta / 2)=1.5 \mu \mathrm{m}, \text { where } n \text { is the concentration of nuclei.}