Physics: Principles with Applications (7th Edition)

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

Chapter 31 - Nuclear Energy; Effects and Uses of Radiation - Problems - Page 912: 50

Answer

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Work Step by Step

a) We can see in appendix B, that the Iodine $\rm ^{131}_{53}I$ decays by $\beta^-$ to Xenon $\rm ^{131}_{54}Xe$. Thus, $$\rm ^{131}_{53}I\rightarrow\;\;\rm ^{131}_{54}Xe+\beta^-+\nu$$ _____________________________________________ b) We need to find the time it takes our iodine sample to have only 5% of $\rm ^{131}_{53}I$; in other words, when $N=0.05N_0$. Thus, $$N=N_0e^{-\lambda t}$$ $$0.05\color{red}{\bf\not}N_0=\color{red}{\bf\not}N_0e^{-\lambda t}$$ $$0.05= e^{-\lambda t}$$ Hence, $$-\lambda t=\ln 0.05$$ $$ t=\dfrac{\ln 0.05}{-\lambda}$$ where $\lambda=\dfrac{\ln 2}{T_{\rm {\frac{1}{2}}}}$ $$ t=\dfrac{-T_{\rm {\frac{1}{2}}}\ln 0.05}{ \ln 2 }$$ Plugging the known; $$ t=\dfrac{-(8)\ln 0.05}{ \ln 2 }=\color{red}{\bf 34.68}\;\rm day$$ _____________________________________________ c) First, we need to find the number of nuclei. So $$\dfrac{\Delta N}{\Delta t}=\lambda N$$ $$N=\dfrac{\Delta N}{\lambda\Delta t} =\dfrac{T_{\rm {\frac{1}{2}}}\Delta N}{ \ln 2\Delta t}$$ $$N =\dfrac{T_{\rm {\frac{1}{2}}}}{ \ln 2}\dfrac{\Delta N}{\Delta t}$$ Plugging the known; $$N =\dfrac{8\times 24\times 60^2}{ \ln 2}\times 1\times 10^{-3}\times 3.7\times 10^{10}$$ $$N = 3.7\times10^{13}\;\rm nuclei$$ Thus, the mass of this number of nuclei is given by $$m=\rm 3.7\times10^{13}\times 130.906126\times 1.67\times 10^{-27} $$ $$m=\color{red}{\bf 8.07\times10^{-12}}\;\rm kg$$
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