Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 30 - Nuclear Physics and Radioactivity - Problems - Page 882: 40

Answer

0.1.

Work Step by Step

Use equations 30–4 and 30–6 to find the fraction left after 2.5 years, or 30 months. $$N=N_0e^{-\lambda t}$$ $$\frac{N}{N_0}= e^{-\lambda t}=e^{-(ln\;2) t/T_{1/2}}$$ $$= e^{-(ln\;2) (30\;months)/(9\;months)}=0.0992\approx 0.1$$ $$\frac{N}{N_0}= 0.1$$ Only 1 significant figure is kept because the problem stated "about 9 months".
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