Suppose that an alpha particle of 3.20 MeV approaches head-on a lead nucleus (Z = 82). Assuming that the lead nucleus remains at rest and the alpha particle momentarily comes to rest and reverses its direction at a distance much more than the radius of the lead nucleus, the distance of its closest approach is:

Correct option is C

​$$\text{To calculate the \textbf{distance of closest approach} } r, \text{ we equate the initial kinetic energy of the alpha particle to the electrostatic potential energy at the point of closest approach:} \\\frac{1}{4\pi \varepsilon_0} \cdot \frac{Z_1 Z_2 e^2}{r} = K.E. \\\\\text{Where:} \\\quad Z_1 = 2 \ (\text{alpha particle}), \\\quad Z_2 = 82 \ (\text{lead nucleus}), \\\quad e = 1.6 \times 10^{-19}\ \text{C}, \\\quad K.E. = 3.20\ \text{MeV} = 3.20 \times 10^6 \times 1.6 \times 10^{-19}\ \text{J}, \\\quad \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\ \text{Nm}^2/\text{C}^2$$

$$r = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z_1 Z_2 e^2}{K.E.}$$

$$r = \frac{9 \times 10^9 \times 2 \times 82 \times (1.6 \times 10^{-19})^2}{3.20 \times 10^6 \times 1.6 \times 10^{-19}} \\r = \frac{9 \times 10^9 \times 2 \times 82 \times 2.56 \times 10^{-38}}{5.12 \times 10^{-13}} \\r = \frac{3.78 \times 10^{-26}}{5.12 \times 10^{-13}} = 7.38 \times 10^{-14}\, \text{m} \\\\\text{Convert to femtometers (fm):} \\r = 73.8\, \text{fm}$$​​​​