A) \[2.25\times {{10}^{9}}\]
B) \[4.5\times {{10}^{9}}l\,\,n3\]
C) \[4.5\times {{10}^{9}}\frac{ln\left( \frac{3}{2} \right)}{ln2}\]
D) \[2.25\times {{10}^{9}}ln\left( \frac{3}{2} \right)\]
Correct Answer: C
Solution :
[c] Suppose an initial radionuclide I decays to a final product F with a half- life \[{{T}_{1/2}}.\] At any time, \[{{N}_{1}}={{N}_{0}}{{e}^{-\lambda t}}\] Number of product nuclei \[={{N}_{F}}={{N}_{0}}-{{N}_{I}}\] \[\frac{{{N}_{F}}}{{{N}_{I}}}=\frac{{{N}_{0}}-{{N}_{I}}}{{{N}_{I}}}=\left( \frac{{{N}_{0}}}{{{N}_{I}}}-I \right)\] \[\frac{{{N}_{0}}}{{{N}_{I}}}=\left( 1+\frac{{{N}_{F}}}{{{N}_{I}}} \right)=1+0.5=1.5\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{T}_{1/2}}\,\,In\,\,(1.5)}{In\,\,2}=4.5\times {{10}^{9}}\frac{\ell n\left( \frac{3}{2} \right)}{\ell n2}year\]
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