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Scattering of Particles. Rutherford-Bohr Atom

Problem 6.4

An alpha particle with kinetic energy T=0.50MeVT=0.50 \mathrm{MeV} is deflected through an angle of θ=90∘\theta=90^{\circ} by the Coulomb field of a stationary Hg nucleus. Find: (a) the least curvature radius of its trajectory; (b) the minimum approach distance between the particle and the nucleus.

Reveal Answer
 6.4. (a) ρmin⁡=(Ze2/T)cot⁡2(θ/2)=0.23 pm; (b) rmin⁡==[1+cosec⁡(θ/2)]Ze2/T=0.56pm\begin{aligned} &\text { 6.4. (a) } \rho_{\min }=\left(Z e^{2} / T\right) \cot ^{2}(\theta / 2)=0.23 \text { pm; (b) } r_{\min }=\\ &=[1+\operatorname{cosec}(\theta / 2)] Z e^{2} / T=0.56 \mathrm{pm} \end{aligned}
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