In a head-on collision, an alpha particle (Z = 2)of energy $8.20 \mathrm { MeV }$ bounces straight back from a nucleus of charge $80.0 \mathrm { e. }$ How close were the centers of the objects at closest approach? (1 eV = 1.60 × 10^-19 J, e = 1.60 × 10^-19 C, 1/4??₀ = 8.99 × 10⁹ N ? m²/C²)

Question

Quizplus · Accepted Answer

The correct answer is A because the distance of closest approach is given by:$$r = \frac{1}{4\pi \varepsilon_0} \frac{2Ze^2}{K}$$where:* r is the distance of closest approach* Z is the atomic number of the nucleus* e is the elementary charge* K is the kinetic energy of the alpha particlePlugging in the given values, we get:$$r = \frac{1}{4\pi (8.99 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2)} \frac{2(2)(1.60 	imes 10^{-19} 	ext{ C})^2}{8.20 	imes 10^6 	ext{ eV}}$$$$r = 2.81 	imes 10^{-14} 	ext{ m}$$

In a Head-On Collision, an Alpha Particle (Z = 2)of 8.20MeV8.20 \mathrm { MeV }8.20MeV

In a Head-On Collision, an Alpha Particle (Z = 2)of $8.20 \mathrm { MeV }$