Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 30 - Quantum Physics - Problems and Conceptual Exercises - Page 1075: 56

Answer

Please see the work below.

Work Step by Step

(a) $\Delta \lambda \propto \frac{1}{m}$. This shows that mass and wavelength are inversely proportional to each other. We know that the mass of the helium atom is more than the electron, therefore, the change in wavelength of the x-ray is greater in case of the electron. (b) We know that $\Delta \lambda=\frac{2h}{mc}$ For the electron, we plug in the known values to obtain: $\Delta \lambda=\frac{2(6.63\times 10^{-34}J.s)}{(9.1\times 10^{-31}Kg)(3\times 10^8m/s)}$ $\Delta \lambda=4.85\times 10^{-12}m=4.85pm$ Now for the helium atom $\Delta \lambda=\frac{2(6.63\times 10^{-34}J.s)}{(6.6\times 10^{-27}Kg)(3\times 10^8m/s)}$ $\Delta \lambda=0.67fm$ Thus, the change in wavelength for the electron is $4.85pm$ and for helium is $0.67fm$
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