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: 55

Answer

$\approx 7920\;\rm year$

Work Step by Step

Since we know that $$\dfrac{\rm ^{14}C}{\rm ^{12}C}=1.3\times 10^{-12}$$ which means that $\rm ^{14}C$ is so small. Now we need to find the number of the $\rm ^{12}C$ carbon atoms in this 73 g. Thus, $$N_{\rm ^{12}C}=\dfrac{73}{12}\times N_A=\dfrac{73}{12}\times 6.02\times 10^{23}$$ $$N_{\rm ^{12}C}=\bf 3.66\times10^{24}\;\rm atom$$ And hence, the number of the $\rm ^{14}C$ carbon atoms is given by $$ \rm ^{14}C =1.3\times 10^{-12}\;\;{\rm ^{12}C}$$ Plugging from above; $$ \rm ^{14}C =1.3\times 10^{-12}(3.66\times10^{24})$$ $$N_{\rm ^{14}C}=\bf 4.76\times10^{12}\;\rm nuclei$$ We know that the activity is given by $$R=\left|\dfrac{\Delta N}{\Delta t}\right|_{\rm today}=\left|\dfrac{\Delta N}{\Delta t}\right|_0e^{-\lambda t}=7\;\rm decays /second$$ And hence, $$ 7=\left|\lambda N_{\rm ^{14}C}\right|_0e^{-\lambda t}$$ $$ \dfrac{7}{\left|\lambda N_{\rm ^{14}C}\right|_0}=e^{-\lambda t}$$ Thus, $$\ln\left[ \dfrac{7}{\left|\lambda N_{\rm ^{14}C}\right|_0}\right] =-\lambda t$$ Solving for $t$; $$t=\dfrac{-1}{\lambda}\;\ln\left[ \dfrac{7}{\left(\lambda N_{\rm ^{14}C}\right)_0}\right]$$ where $\dfrac{-1}{\lambda}=\tau=\dfrac{T_{\frac{1}{2}}}{\ln 2}$; $$t=\dfrac{-T_{\frac{1}{2}}}{\ln 2}\;\ln\left[ \dfrac{7}{\left(\dfrac{\ln 2} { T_{\frac{1}{2}}}N_{\rm ^{14}C}\right)_0}\right]$$ Plugging the known; $$t=\dfrac{-5730\times 365.25\times 86400}{\ln 2}\;\ln\left[ \dfrac{7}{\dfrac{\ln 2} {5730\times 365.25\times 86400} \times4.76\times10^{12}} \right]$$ $$t=\color{red}{\bf 2.499\times 10^{11}}\;\rm s\approx \color{red}{\bf 7920}\;\rm year$$
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