Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 37 - Relativity - Problems - Page 1152: 95

Answer

$m = 4.00~u$ The particle is an alpha particle, which is the nucleus of a helium atom.

Work Step by Step

We can find $\gamma$: $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ $\gamma = \frac{1}{\sqrt{1-0.710^2}}$ $\gamma = 1.42$ We can find the mass $m$: $r = \frac{\gamma ~m~v}{\vert q \vert B}$ $m = \frac{r~\vert q \vert B}{\gamma~v}$ $m = \frac{(6.28~m)~(2)(1.6\times 10^{-19}~C) (1.00~T)}{(1.42)(0.710)(3.0\times 10^8~m/s)}$ $m = 6.644\times 10^{-27}~kg$ $m = (6.644\times 10^{-27}~kg)(\frac{1~u}{1.66\times 10^{-27}~kg})$ $m = 4.00~u$ We can assume that this particle has two protons and two neutrons. Therefore, the particle is an alpha particle, which is the nucleus of a helium atom.
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