Correct Answer - Option 2 : 47.1fm

__CONCEPT__:

- The distance of the closest approach is the closest distance
**to the nucleus at which all the kinetic energy of the**** projected particle becomes potential **energy and is given by

\(\Rightarrow d = \frac{2Ze^{2}}{4\pi \epsilon_{0}K}\)

Where d = Distance of closest approach, Z = Atomic number, K = Kinetic energy. e = charge

__CALCULATION : __

Given - K = 5 meV = 5 × 1.6 × 10-13J, Z = 82, and e = 1.6 × 10-19 C

- The distance of closest approach is given by

\(\Rightarrow d = \frac{2Ze^{2}}{4\pi \epsilon_{0}K}\)

Substituting the given values the above equation becomes

\(\Rightarrow d = \frac{2\times82\times(1.6\times10^{-19})^{2}}{4\pi \epsilon_{0}\times 5\times1.6\times 10^{-13} }\)

\(\Rightarrow d = \frac{9\times 10^{9}\times 419.84\times 10^{-3 8}}{8\times 10^{-13}} = 47.1 \ \rm fermi\)

- Hence,
** option 2 is the answer**