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 - General Problems - Page 883: 81

Answer

$3.46897\times 10^{18}\;\rm s$

Work Step by Step

We know that the half-life for a decay process is given by $$R=\lambda N=\dfrac{\ln 2}{T_{\frac{1}{2}}}N_{\rm ^{147}_{62}Sm}$$ Solving for $T_{\frac{1}{2}}$; $$T_{\frac{1}{2}}=\dfrac{\ln 2}{R}N_{\rm ^{147}_{62}Sm}\tag 1$$ Now we need to find $N_{\rm ^{147}_{62}Sm}$ and we know that its amount in a natural sample is about $15\%$. So that $$N_{\rm ^{147}_{62}Sm}=0.15 N_{natural }$$ Plugging the known; $$N_{\rm ^{147}_{62}Sm}=0.15\left(\dfrac{1\times 6.02\times 10^{23}}{150.36}\right)$$ where 150.36 is the samarium atomic mass in g/mole and $6.02\times 10^{23}$ is Avogadro's number which is the number of atoms in one mole. $$N_{\rm ^{147}_{62}Sm}=\bf 6.0056\times 10^{20}\rm nuclei$$ Plugging into (1) and plug the known as well; $$T_{\frac{1}{2}}=\dfrac{\ln 2}{120} \times 6.0056\times 10^{20} $$ $$T_{\frac{1}{2}}= \color{red}{\bf 3.46897\times 10^{18}}\;\rm s=1.09925\times 10^{11}\;\rm year$$
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