An $$\alpha $$-particle of energy $$5\,MeV$$  is scattered through $${180^ \circ }$$ by a fixed uranium nucleus. The distance of closest approach is of the order of

Solution :
KEY CONCEPT:
Distance of closest approach
$$\eqalign{ & {r_0} = \frac{{Ze\left( {2e} \right)}}{{4\pi {\varepsilon _0}E}} \cr & {\text{Energy, }}E = 5 \times {10^6} \times 1.6 \times {10^{ - 19}}J \cr & \therefore {r_0} = \frac{{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)}}{{5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} \cr & \Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm \cr} $$