Conceptual Physics (12th Edition)

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

Chapter 31 - Think and Solve: 24

Answer

$2.4 \times 10^{-11} m$. This is less than an atomic diameter.

Work Step by Step

The de Broglie wavelength is Planck’s constant/momentum, as stated on page 590. First find the electron’s momentum. $$p = mv = (9.1 \times 10^{-31} kg)(3.0 \times 10^{7} \frac{m}{s}) $$ $$= 2.7 \times 10^{-23} \frac{kg m}{s}$$ Now find the de Broglie wavelength. $$\lambda = \frac{h}{p} = (6.63 \times 10^{-34} J s )/(2.7 \times 10^{-23} \frac{kg m}{s} )$$ $$= 2.4 \times 10^{-11} m$$
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