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? $( 1 \mathrm { eV } = 1.60$ $\times 10 ^ { - 19 } \mathrm {~J} , e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , 1 / 4 \pi \varepsilon _ { 0 } = 8.99 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )
A) $3.39 \times 10 ^ { - 12 } \mathrm {~m}$
B) $6.56 \times 10 ^ { - 15 } \mathrm {~m}$
C) $2.81 \times 10 ^ { - 14 } \mathrm {~m}$
D) $2.17 \times 10 ^ { - 14 } \mathrm {~m}$

Question

Quizplus · Accepted Answer

The closest approach distance can be calculated using the formula for electrostatic potential energy: $ k \frac{Z_1 Z_2 e^2}{r} = E $, where $ k = 1 / 4 \pi \varepsilon_0 $, $ Z_1 = 2 $, $ Z_2 = 80 $, and $ E = 8.20 \, \text{MeV} $. Solving for $ r $ gives $ r = \frac{k \cdot 2 \cdot 80 \cdot e^2}{8.20 \times 10^6 \times 1.60 \times 10^{-19}} $, which results in approximately $ 2.81 \times 10^{-14} \, \text{m} $.

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

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