Answer to Question #141258 in Mechanics | Relativity for Desmond

Question

Calculate the distance of the closest approach for a head-on collision between an alpha particle having initial energy of 0.500 MeV and a gold nucleus at rest. Assume the gold nucleus remains at rest during the collision (b) What minimum initial speed must the alpha particle have to approach as close as 3.00 x 10-18m to the gold nucleus?

Answer 1

(a) The initial kinetic energy of the alpha particle must equal the electrostatic potential energy at the distance of closest approach.

$K_{i}= {U_{f}= \frac{K_{e}qQ}{r_{min}}}$ , therefore,

$r_{min}= \frac{8.99\times10^{9}\times2\times79\times(1.60\times10^{-19})^{2}}{0.500\times1.60\times10^{-13}}= 4.55\times10^{-13}m$

(b) we know that, $K_{i}=\frac{1}{2}m_{\alpha}V^{2}_{i}=\frac{K_{e}qQ}{r_{min}}$

$V_{i}= \sqrt\frac{2K_{e}qQ}{m_{\alpha}r_{min}}$

$= \sqrt\frac{2(\times8.99\times10^{9})\times2\times79(1.660\times10^{-19})^{2}}{4.00\times1.66\times10^{-27}\times3.0\times10^{-18}} = 1.91\times10^{9}m/s$