Rutherford Scattering - Closest Approach

At the distance of closest approach in a head-on Rutherford collision, the alpha particle's entire initial kinetic energy has been converted into electrostatic potential energy, since its velocity is instantaneously zero. Equating $K=\frac{1}{4\pi {\epsilon }_0}\frac{(2e)(Ze)}{r_0}$ and solving gives $r_0=\frac{2kZ{e}^2}{K}$, where the factor 2 is the alpha charge and $Z=79$. Converting the energy, $K=5\text{MeV}=5\times 10^6\times 1.6\times 10^{-19}=8\times 10^{-13}$ J. Substituting $k=9\times 10^9$, $Z=79$, and $e=1.6\times 10^{-19}$ C: $r_0=\frac{2\times 9\times 10^9\times 79\times (1.6\times 10^{-19}{)}^2}{8\times 10^{-13}}\approx 4.55\times 10^{-14}$ m. The value $2.27\times 10^{-14}$ m omits the factor of 2 for the alpha charge. The value $9.10\times 10^{-14}$ m doubles the result incorrectly. The value $1.60\times 10^{-13}$ m uses the energy in eV without proper conversion. The closest-approach distance scales inversely with the bombarding energy, so faster alpha particles probe deeper toward the nucleus, while a larger nuclear charge $Z$ pushes the turning point farther out. This reproduces the NCERT treatment of Rutherford's gold-foil experiment. A sanity check confirms $r_0$ is far larger than nuclear radii (~$10^{-14}$ m yet well outside ~$7\times 10^{-15}$ m), so the alpha never penetrates the nucleus.