In an alpha-particle scattering experiment, a 5.5 MeV alpha-particle approaches a gold nucleus (Z =
79). What is the approximate distance of closest approach? (Take \( \frac{1}{4\pi\epsilon_0} = 9 \times
10^9 \, \text{N·m}^2/\text{C}^2 \), \( e = 1.6 \times 10^{-19} \, \text{C} \), 1 MeV = \( 1.6 \times
10^{-13} \, \text{J} \))
Energy conservation: \( K = \frac{2Ze^2}{4\pi\epsilon_0 d} \).
\( d = \frac{2Ze^2}{4\pi\epsilon_0 K} \).
\( K = 5.5 \, \text{MeV} = 5.5 \times 1.6 \times 10^{-13} = 8.8 \times 10^{-13} \, \text{J} \).
\( d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times 9 \times 10^9}{8.8 \times 10^{-13}} \).
\( d = \frac{158 \times 2.56 \times 10^{-38} \times 9 \times 10^9}{8.8 \times 10^{-13}} \approx 4.13
\times 10^{-14} \, \text{m} \approx 41 \, \text{fm} \).
Rounded to 40 fm for simplicity.