In Rutherford scattering experiment, the number of $$\alpha $$-particles scattered at $${60^ \circ }$$ is $$5 \times {10^6}.$$  The number of $$\alpha $$-particles scattered at $${120^ \circ }$$ will be

Solution :
$$\eqalign{ & N \propto \frac{1}{{\frac{{{{\sin }^4}\theta }}{2}}};\frac{{{N_2}}}{{{N_1}}} = \frac{{{{\sin }^4}\left( {\frac{{{\theta _1}}}{2}} \right)}}{{{{\sin }^4}\left( {\frac{{{\theta _2}}}{2}} \right)}} \cr & {\text{or}}\,\,{N_2} = 5 \times {10^6} \times {\left( {\frac{1}{2}} \right)^4}{\left( {\frac{2}{{\sqrt 3 }}} \right)^4} = \frac{5}{9} \times {10^6} \cr} $$