Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 31 - Nuclear Physics and Radioactivity - Problems - Page 900: 33

Answer

$f=\frac{1}{16}$

Work Step by Step

Suppose for sample $ A$ initial number of nuclei presents is $N_{0,A}$ radioactive decay constant is $\lambda_{A}$ half life is $ T_{1/2,A}$ Similarly for sample $ B$ initial number of nuclei presents is $N_{0, B}$ radioactive decay constant is $\lambda_{ B}$ half life is $ T_{1/2, B}$ Given that $ T_{1/2, B}=\frac{1}{2} T_{1/2, A}$........equation(1) From equation 31.6 $T_{1/2}=\frac{ln2}{\lambda}$ $ T_{1/2,A}=\frac{ln2}{\lambda_{A}}$ $ T_{1/2,B}=\frac{ln2}{\lambda_{B}}$ putting these values to equation(1) $ \frac{ln2}{\lambda_{B}}=\frac{1}{2} \frac{ln2}{\lambda_{A}}$. $\lambda_{B}=2\lambda_{A}$..............................equation(2) Suppose in time $t$ Sample $A$ decreases to one fourth of its initial number. so after time $t$, $N_{t,A}=\frac{N_{0,A}}{4} $ from decay equation $N=N_{0}e^{-\lambda t}$ after time $t$ $N_{t,A}=\frac{N_{0,A}}{4}= N_{0,A}e^{-\lambda_{A} t}$ $\frac{1}{4}=e^{-\lambda_{A} t}$ $e^{\lambda_{A} t}=4$ taking natural log both side $ln(e^{\lambda_{A} t})=ln4$ $\lambda_{A} t=2ln2$ $t=\frac{2ln2}{\lambda_{A}}$.................................equation(3) in the same time Sample $B$ decrease to a fraction $f$ of the number present initially so after time $t$ , $N_{t,B}=fN_{0,B} $ from decay equation $N=N_{0}e^{-\lambda t}$ after time $t=\frac{2ln2}{\lambda_{A}}$. $N_{t,B}=fN_{0,B}=N_{0,B}e^{-\lambda_{B}\frac{2ln2}{\lambda_{A}}}$ $f=e^{-\lambda_{B}\frac{2ln2}{\lambda_{A}}}$ now putting $\lambda_{B}=2\lambda_{A}$ from equation (2) we will get $f=e^{-2\lambda_{A}\frac{2ln2}{\lambda_{A}}}$ $f=e^{-4ln2}$ $f=e^{ln(2)^{-4}}$ $f=(2)^{-4}$ $f=\frac{1}{16}$
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