College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 28 - Quantum Mechanics and Atomic Physics - Learning Path Questions and Exercises - Exercises - Page 962: 5


$1.5×10^{4}$ V

Work Step by Step

Recall the relation λ= $\frac{1.227}{\sqrt V}nm$ where λ is the wavelength and V is the magnitude of accelerating potential. Thus we get, V=$(\frac{1.227}{ λ(nm)})^{2}$= $(\frac{1.227}{0.010})^{2}$= $1.5×10^{4}$ volts
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