$7.9 \mathrm{MeV} \alpha$-particle scatters from a target material of atomic number 79 . From the gi

Question

$7.9 \mathrm{MeV} \alpha$-particle scatters from a target material of atomic number 79 . From the given data the estimated diameter of nuclei of the target material is (approximately) $\_\_\_\_$ m. $\left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 	imes 10^9 \mathrm{Nm}^2 / \mathrm{C}^2ight.$ and electron charge $\left.=1.6 	imes 10^{-19} \mathrm{C}ight]$

Professor PC Thomas Classes · Answer

The distance of closest approach $r_0$ is given by equating the kinetic energy to the electrostatic potential energy: $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. Solving for $r_0 = \frac{9 	imes 10^9 	imes 2 	imes 79 	imes (1.6 	imes 10^{-19})^2}{7.9 	imes 10^6 	imes 1.6 	imes 10^{-19}} = 2.88 	imes 10^{-14} 	ext{ m}$. The estimated diameter of the nucleus is approximately twice the distance of closest approach, $D \approx 2r_0 = 2 	imes 2.88 	imes 10^{-14} = 5.76 	imes 10^{-14} 	ext{ m}$.