Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 18 - Additional Exercises - Page 750d: 65

Answer

1975

Work Step by Step

$\frac{A}{A_{0}}=\frac{17}{100}=0.17$ where $A_{0}$ is the original amount and $A$ is the amount after time $t$. Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{12.3\,y}=0.05634\,y^{-1}$ Recall that $\ln(\frac{A}{A_{0}})=-kt$ $\implies \ln(0.17)=-1.77=-(0.05634\,y^{-1})(t)$ $\implies t=\frac{-1.77}{-0.05634\,y^{-1}}=31.4\,y$ The watch can be used until (1944+31)=1975
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.