Rutherford Scattering - Closest Approach

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An alpha particle carrying kinetic energy 5 MeV is fired head-on toward a stationary gold nucleus of atomic number 79. Estimate the distance of closest approach at which the particle momentarily halts before reversing.

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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 $K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_0}$ and solving gives $r_0 = \dfrac{2kZe^2}{K}$, where the factor 2 is the alpha charge and $Z = 79$. Converting the energy, $K = 5\,	ext{MeV} = 5	imes10^6 	imes 1.6	imes10^{-19} = 8	imes10^{-13}$ J. Substituting $k = 9	imes10^9$, $Z = 79$, and $e = 1.6	imes10^{-19}$ C: $r_0 = \dfrac{2	imes 9	imes10^9 	imes 79 	imes (1.6	imes10^{-19})^2}{8	imes10^{-13}} \approx 4.55	imes10^{-14}$ m. The value $2.27	imes10^{-14}$ m omits the factor of 2 for the alpha charge. The value $9.10	imes10^{-14}$ m doubles the result incorrectly. The value $1.60	imes10^{-13}$ 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 ~$7	imes10^{-15}$ m), so the alpha never penetrates the nucleus.

Rutherford Scattering - Closest Approach

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