Question

(a) Calculate the distance of closest approach for a head-on collision between an alpha particle having an 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

The initial kinetic energy of the alpha particle should be equal to the electrostatic potential energy at the distance of closest approach.

(a) $(a) K_{i}=U_{f}=\frac{K_{e}qQ}{r_{min}}$$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) having $K_{i}=\frac{1}{2}m_{\alpha}V^{2}_{i} = \frac{K_{e}qQ}{r_{min}}$

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

= $\sqrt\frac{2(8.99\times10^{9})\times2\times79\times(1.60\times10^{-19})^{2}}{4.00\times1.66\times10^{-27}\times3.00\times10^{-18}} = \sqrt{3.65 \times10^{18}} = 1.91\times10^{9}m/s$