Answer to Question #126171 in Atomic and Nuclear Physics for eli

Question

 The original Rutherford scattering experiment was done using 5 MeV alpha particles incident on a gold target (Z = 79). Suppose, that this experiment had been instead done with 10 MeV alpha particles incident on a silver foil (Z = 47). Discuss the implications of the outcomes of these experiments on the determination of the size of the nucleus

Answer 1

Let's consider conservation energy law:

$\frac{m_\alpha v^2_\alpha}{2} = \frac{1}{4\pi\epsilon_0}\frac{Q_\alpha Q_{nucleus}}{b} \implies b = \frac{1}{4\pi\epsilon_0}\frac{2Q_\alpha Q_{nucleus}}{m_\alpha v^2_\alpha} = \frac{1}{4\pi\epsilon_0}\frac{Q_\alpha Q_{nucleus}}{T_\alpha}$

We've known what the result in Rutherford experiment:

$b_{Au} = \frac{1}{4\pi\epsilon_0}\frac{Q_\alpha Q^{Au}_{nucleus}}{T_\alpha} = 27 *10^{-15 }m$

So:

$b_{Ar} = \frac{1}{4\pi\epsilon_0}\frac{Q_\alpha Q^{Ar}_{nucleus}}{T_\alpha} \implies \frac{b_{Ar}}{b_{Au}} = \frac{5MeV}{10MeV}*\frac{47}{79} = 0.297$

Or

$b_{Ar} = 0.297*27*10^{-15}m = 8.03*10^{-15}m$