Question

According to the text, the de Broglie wavelength, \(\lambda,\) of a \(5-\mathrm{MeV}\) alpha particle is \(6.4 \mathrm{fm},\) and the closest distance, \(r_{\min },\) to the gold nucleus this alpha particle can get is \(45.5 \mathrm{fm}\) (calculated in Example 39.1 ). Based on the fact that \(\lambda \ll r_{\min },\) 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

Answer: (a) particle

Step-by-step solution

Understand the Rutherford Scattering Experiment

In the Rutherford Scattering Experiment, alpha particles are fired at a gold foil. The way these particles are deflected gives us information about the structure of gold atoms. For our analysis, we need to determine if de Broglie's wave-particle duality concept is applicable in this situation.

Definition of De-Broglie Wavelength

According to the wave-particle duality concept, every particle has a wave-like property known as its de Broglie wavelength or \(\lambda\). It is given by the formula: \(\lambda = \dfrac{h}{p}\), where \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.

Compare De-Broglie Wavelength (\(\lambda\)) and Closest Distance (\(r_{min}\))

Let's compare the de Broglie wavelength (\(\lambda = 6.4~\mathrm{fm}\)) to the closest distance (\(r_{min} = 45.5~\mathrm{fm}\)). Given the fact that \(6.4~\mathrm{fm} \ll 45.5~\mathrm{fm}\), we can determine whether to treat the alpha particle as a particle or wave.

Determine if Alpha Particle is Treated as a Particle or Wave

Since the de Broglie wavelength \(\lambda\) is much smaller than the closest distance \(r_{min}\), it means that the wave-like character of the alpha particle is weak in this situation. Therefore, the alpha particle behaves more like a particle rather than a wave.

Conclusion

Based on the comparison of the de Broglie wavelength (\(\lambda\)) and the closest distance (\(r_{min}\)), and the fact that the wave-like character of the alpha particle is weak in this situation, we conclude that it is appropriate to treat the alpha particle as a particle in the Rutherford scattering experiment. Thus, the correct answer is (a) particle.