In a head-on collision,an alpha particle (Z = 2)of energy 8.20 MeV bounces straight back from a nucleus of charge 80.0 e.How close were the centers of the objects at closest approach?

Question

Quizplus · Accepted Answer

The correct answer is A because the alpha particle has a charge of +2e and the nucleus has a charge of +80e. The distance of closest approach is given by the equation:$$d = \frac{2ke^2}{E}$$where k is Coulomb's constant, e is the elementary charge, and E is the energy of the alpha particle. Plugging in the values, we get:$$d = \frac{2(9 	imes 10^9 	ext{ Nm}^2/	ext{C}^2)(1.6 	imes 10^{-19} 	ext{ C})^2}{8.20 	imes 10^6 	ext{ eV}}$$$$d = 2.81 	imes 10^{-14} 	ext{ m}$$

In a Head-On Collision,an Alpha Particle (Z = 2)of Energy