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

Quizplus · Accepted Answer

The closest approach can be calculated using the formula for the electrostatic potential energy at the closest distance, which equals the initial kinetic energy of the alpha particle. The formula is $U = \frac{kq_1q_2}{r}$, where $k = 1/4\pi\epsilon_0$, $q_1$ and $q_2$ are the charges of the alpha particle and the nucleus, respectively, and $r$ is the distance of closest approach. Given that the charge of the alpha particle is $2e$ and the charge of the nucleus is $80e$, and converting the initial kinetic energy from MeV to Joules, we can solve for $r$.

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

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