Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 32 - Elementary Particles - Problems - Page 944: 7

Answer

The alpha particle is better,

Work Step by Step

25 MeV is small compared to the rest energies of the alpha particle and the proton, so these are not relativistic particles. Calculate the wavelength with equation 32–1. $$\lambda_p=\frac{h}{p}=\frac{h}{\sqrt{2 m_p KE}}$$ $$\lambda_p=\frac{6.626\times10^{-34}J \cdot s}{\sqrt{2 (1.67\times10^{-27}kg) (25MeV)(1.6\times10^{-13}J/MeV)}}$$ $$\lambda_p=5.73\times10^{-15}m$$ $$\lambda_{\alpha}=\frac{h}{p}=\frac{h}{\sqrt{2 m_{\alpha} KE}}$$ $$\lambda_{\alpha}=\frac{6.626\times10^{-34}J \cdot s}{\sqrt{2(4) (1.67\times10^{-27}kg) (25MeV)(1.6\times10^{-13}J/MeV)}}$$ $$\lambda_{\alpha}=2.87\times10^{-15}m$$ Use equation 30–1 to see that the diameter of a nucleon is $\approx 2.4\times10^{-15}m$. The alpha particle’s wavelength is closer to the size of a nucleon, and 25-MeV alpha particles are better than 25-MeV protons at resolving details of the nucleus.
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