Chapter 39: Problem 1
The de Broglie wavelength, \(\lambda\), of a 5-MeV alpha particle is \(6.4 \mathrm{fm}\), as shown in this chapter, and the closest distance, \(r_{\text {min }}\), to the gold nucleus this alpha particle can get is \(45.5 \mathrm{fm}\) (calculated in Example 39.1). Based on the fact that \(\lambda \ll r_{\text {min }}\), one can conclude that, for this Rutherford scattering experiment, it is adequate to treat the alpha particle as a a) particle. b) wave.
Short Answer
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Key Concepts
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