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A 7.7 MeV alpha particle heads straight at a gold nucleus. What is its closest approach?
AAbout 3.0 × 10⁻¹⁶ m
BAbout 6 × 10⁻¹⁵ m
CAbout 3.0 × 10⁻¹⁴ m
DAbout 3.0 × 10⁻¹² m
Answer & Solution
Correct answer: C. About 3.0 × 10⁻¹⁴ m
Setting kinetic energy 1.2 × 10⁻¹² J equal to 2Ze²/(4πε₀d) with Z = 79 gives d near 3.0 × 10⁻¹⁴ m, or 30 fm.
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