Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 31 - Think and Solve: 25


The wavelength is $6.6 \times 10^{-30} m$, which is about $3 \times 10^{-19}$ times smaller than the electron wavelength in the previous problem.

Work Step by Step

The de Broglie wavelength is Planck’s constant/momentum, as stated on page 590. First find the ball's momentum. $$p = mv = (0.1 kg)(0.001 \frac{m}{s}) = 1 \times 10^{-4} \frac{kg m}{s}$$ Now find the de Broglie wavelength. $$\lambda = \frac{h}{p} = (6.63 \times 10^{-34} J s )/(1 \times 10^{-4} \frac{kg m}{s} )$$ $$= 6.6 \times 10^{-30} m$$ This is ridiculously small, over a billion billion times smaller than the wavelength of the electron in the previous example.
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