If scattering particles are 56 for 90^o angle then find the number of particles at 60^o angle.

According to scattering formula

$N\propto \frac{1}{{\sin }^4(\theta /2)}\Rightarrow \frac{N_2}{N_1}={\left[\frac{\sin \left({\theta }_1/2\right)}{\sin \left({\theta }_2/2\right)}\right]}^4$

$$\begin{gathered} \Rightarrow \frac{N_2}{N_1}={\left[\frac{\sin \frac{{90}^{\circ }}{2}}{\sin \frac{{60}^{\circ }}{2}}\right]}^4={\left[\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}\right]}^4 \\ \Rightarrow N_2=(\sqrt{2}{)}^4\times N_1=4\times 56=224 \end{gathered}$$