Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 38 - Quantization - Exercises and Problems - Page 1152: 1

Answer

$6.25\times 10^{13}~electrons/s$

Work Step by Step

On the graph, we can see that the current is $10~\mu A$ We can find the number of photoelectrons which are ejected per second: $\frac{10\times 10^{-6}~A}{1.6\times 10^{-19}~C} = 6.25\times 10^{13}~electrons/s$
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