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: 82

Answer

a) $1.63\%$ b) $0.6631\%$

Work Step by Step

a) Let's assume that the sample amount is $m$ since the author did not give us a definite mass of the sample. So, $$m_{{235}}=\dfrac{0.72}{100}m=\color{blue}{0.0072}\;m$$ and $$m_{ {238} }=\left[1-\dfrac{0.72}{100}\right]m=\color{blue}{0.9928}\; m$$ Now we need to find the ratio between two amounts 1.0 billion years ago. We know that $$N=N_0e^{-\lambda t}$$ where $N$ is for today and $N_0$ is for 1.0 billion years ago where both of them are for $\rm ^{235}_{92}U$. Solving for $N_0$; $$\dfrac{N_{235}}{e^{-\lambda t}}=N_{0,235}$$ $$ N_{0,235}=e^{\lambda t} N_{235} $$ Recalling that $\lambda=\dfrac{\ln 2}{T_{\frac{1}{2}}}$; $$ N_{0,235}=e^{\frac{(\ln 2)t}{T_{\frac{1}{2}}}} N_{235} $$ Plugging the known; $$ N_{0,235}=e^{\frac{ (\ln 2) (1\times 10^9)}{ (7.04\times 10^8)} }\times0.0072m=\color{blue}{0.01927}m $$ By the same approach; $$ N_{0,238}=e^{\frac{ (\ln 2) (1\times 10^9)}{ (4.468\times 10^9)} }\times 0.9928m=\color{blue}{1.1594}m $$ Now we can easily find the percentage of $\rm ^{235}_{92}U$ by $${\rm ^{235}_{92}U\%}=\dfrac{N_{0,235}}{N_{0,238}+N_{0,235}}\times 100\%$$ $${\rm ^{235}_{92}U\%}=\dfrac{0.01927m}{1.1594m+0.01927m}\times 100\%=\color{red}{\bf1.63\%}$$ _______________________________________________ b) Applying the same approach as we did above but for $10^8$ years in the future. We need to use the amount of today as $N_0$. $$N_{235}=N_{0,235}e^{-\lambda t}$$ $$N_{235}=N_{0,235}e^{\left[-\frac{\ln 2}{T_{\frac{1}{2}}}t\right]}$$ Plugging the known; $$ N_{235}=e^{-\frac{ (\ln 2) (1\times 10^8)}{ (7.04\times 10^8)} }\times 0.0072m=\color{blue}{6.5249\times10^{-3}}m $$ By the same approach; $$ N_{238}=e^{-\frac{ (\ln 2) (1\times 10^8)}{ (4.468\times 10^9)} }\times0.9928 m=\color{blue}{977.52\times10^{-3}}m $$ Hence; $${\rm ^{235}_{92}U\%}=\dfrac{N_{0,235}}{N_{0,238}+N_{0,235}}\times 100\%$$ $${\rm ^{235}_{92}U\%}=\dfrac{6.5249\times10^{-3}m}{977.52\times10^{-3}m+6.5249\times10^{-3}m}\times 100\%$$ $${\rm ^{235}_{92}U\%}=\color{red}{\bf 0.6631\%}$$
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