Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 41 - Atomic Physics - Exercises and Problems - Page 1207: 22


$t = 27.4~ns$

Work Step by Step

We can find the number of excited atoms that remain: $N = (1.00\times 10^6)-(8.0\times 10^5) = 2.0\times 10^5$ We can find the time $t$: $N = N_0~e^{-t/\tau}$ $2.0\times 10^5 = (1.00\times 10^6)~e^{-t/\tau}$ $0.20 = e^{-t/\tau}$ $ln(0.20) = -\frac{t}{\tau}$ $t = -ln(0.20)~(\tau)$ $t = -ln(0.20)~(17~ns)$ $t = 27.4~ns$
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