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, 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

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, 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Quizplus · Accepted Answer

The correct answer is A because the distance of closest approach is given by:$$r = \frac{2kq^2}{mv^2}$$where k is Coulomb's constant, q is the charge of the alpha particle, m is the mass of the alpha particle, and v is the speed of the alpha particle. Plugging in the given values, we get:$$r = \frac{2(8.99 	imes 10^9 N \cdot m^2/C^2)(2e)^2}{(4 	imes 1.66 	imes 10^{-27} kg)(8.20 	imes 10^6 eV)(1.60 	imes 10^{-19} J/eV)} = 2.81 	imes 10^{-14} m$$

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

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