Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises: 6.47b


The wavelength of the bullet is $\lambda=2.65\times10^{-34}m$.

Work Step by Step

*The de Broglie relationship: $$\lambda=\frac{h}{mv}$$ $\lambda$: wavelength of object $h$: Planck's constant ($h=6.626\times10^{-34} J.s$) $m$: mass of object $v$: velocity of object 1) Find the known variables - Mass of the bullet: $m=10.0g=10^{-2}kg$ - Velocity of the bullet: $v=250m/s$ 2) Calculate the wavelength of the bullet $$\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}}{10^{-2}\times250}\approx2.65\times10^{-34}m$$
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.