Let  a $5MeV \alpha$-particle is scattered by $74°$ when it approaches a gold nucleus $(Z=79)$. Find the impact parameter.

Since, $Z=79$

     $K_{\alpha }=5MeV=5\times 1.6\times {10}^{-13}J=8\times {10}^{-13}J$\[

      $\theta =74°$

$$\begin{gathered} \Rightarrow \frac{\theta }{2}={37}^{\circ } \\ \Rightarrow cot\left(\frac{\theta }{2}\right)=\frac{4}{3} \end{gathered}$$

Using the relation

$$\begin{gathered} \tan \left(\frac{\theta }{2}\right)=\frac{1}{b}\left(\frac{Z{e}^2}{4\pi {\epsilon }_0{K}_{\alpha }}\right) \\ \Rightarrow b=\left(\frac{Z{e}^2}{4\pi {\epsilon }_0{K}_{\alpha }}\right)cot\left(\frac{\theta }{2}\right) \\ \Rightarrow b=\frac{79\times {\left(1.6\times {10}^{-19}\right)}^2\times \left(9\times {10}^9\right)}{8\times {10}^{-13}}\times \frac{4}{3} \\ \Rightarrow b=3.03\times {10}^{-16} m \end{gathered}$$