If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from

Question

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is $\_\_\_\_$ m. (Atomic number of gold $=79$ and $\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 	imes 10^9$ in SI units)

Professor PC Thomas Classes · Answer

At the distance of closest approach $r_0$, the entire kinetic energy of the alpha particle is converted into electrostatic potential energy. $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0} \Rightarrow r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$. Substituting values: $r_0 = \frac{9 	imes 10^9 	imes 2 	imes 79 	imes (1.6 	imes 10^{-19})^2}{7.7 	imes 10^6 	imes 1.6 	imes 10^{-19}} = \frac{9 	imes 10^9 	imes 158 	imes 1.6 	imes 10^{-19}}{7.7 	imes 10^6} = 2.95 	imes 10^{-14} 	ext{ m}$.