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