Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 38 - Quantization - Exercises and Problems - Page 1114: 16

Answer

(a) $v = 2.8\times 10^8~m/s$ (b) $v = 7.3\times 10^5~m/s$ (c) $v = 730~m/s$ (d) $v = 0.73~m/s$

Work Step by Step

(a) We can find the speed: $\lambda = \frac{h}{\gamma ~m~v}$ $\gamma~v = \frac{h}{m~\lambda}$ $\frac{1}{\sqrt{1-v^2/c^2}}~v = \frac{h}{m~\lambda}$ $\frac{v^2}{1-v^2/c^2} = (\frac{h}{m~\lambda})^2$ $v^2~c^2 = (\frac{h}{m~\lambda})^2(c^2-v^2)$ $v^2~[c^2+(\frac{h}{m~\lambda})^2] = (\frac{h~c}{m~\lambda})^2$ $v^2 = \frac{(\frac{h~c}{m~\lambda})^2}{c^2+(\frac{h}{m~\lambda})^2}$ $v^2 = \frac{(h~c)^2}{(m~\lambda~c)^2+h^2}$ $v = \frac{h~c}{\sqrt{(m~\lambda~c)^2+h^2}}$ $v = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{\sqrt{[(9.109\times 10^{-31}~kg)~(1.0\times 10^{-12}~m)~(3.0\times 10^8~m/s)]^2+(6.626\times 10^{-34}~J~s)^2}}$ $v = 2.8\times 10^8~m/s$ (b) We can find the speed: $\lambda = \frac{h}{m~v}$ $v = \frac{h}{m~\lambda}$ $v = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)~(1.0\times 10^{-9}~m)}$ $v = 7.3\times 10^5~m/s$ (c) We can find the speed: $\lambda = \frac{h}{m~v}$ $v = \frac{h}{m~\lambda}$ $v = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)~(1.0\times 10^{-6}~m)}$ $v = 730~m/s$ (d) We can find the speed: $\lambda = \frac{h}{m~v}$ $v = \frac{h}{m~\lambda}$ $v = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)~(1.0\times 10^{-3}~m)}$ $v = 0.73~m/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.