Suppose that an alpha particle of 4.50 MeV approaches head-on a uranium nucleus (Z = 92). Assuming that the uranium 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 uranium nucleus, the distance of its closest approach is close to:

Correct option is B

​$$\begin{aligned}\text{The potential energy at distance } d \text{ is given by:} \\K &= \frac{1}{4 \pi \varepsilon_0} \frac{(Ze)(2e)}{d} \\\text{Solving for } d\text{:} \\d &= \frac{1}{4 \pi \varepsilon_0} \frac{2Ze^2}{K} \\\text{Substitute the constants:} \\d &= 9 \times 10^9 \times \frac{2 \times 92 \times (1.6 \times 10^{-19})^2}{4.50 \times 10^6 \times 1.6 \times 10^{-19}} \\d &= 59 \times 10^{-15} \, \text{m} \\d &= 59 \, \text{fm}\end{aligned}$$​​