Question

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If a radiopharmaceutical has an effective half life of 75 seconds and a biological half life of 6 minutes what is the physical half life in minutes?

Answer 1

$Suppose\,\, t_e,\,\,t_p\,\,and\,\,t_b\,\,represent\,\,effective \,\,half\,\,life,\,\,physical\,\,half\,\,life\,\,and\,\,\\biological\,\,half\,\,life\,\,respectively\,\,then,$

$Mathematically\,\,\,\,\,\, \frac{1}{}t_e\,\,=\frac{1}{}t_p\,\,+\,\,\frac{1}{}t_b\,\,$

${\color{Blue} Problem\,\,(1)}$

$Given,\,\,\\ t_e=75\,\,seconds=\frac{75}{60}\,\,minutes\,\, \\t_b=6\,\,minutes\\\\ \therefore \,\,\frac{1}{\frac{75}{60}}=\frac{1}{t_p}+\frac{1}{6}\\\\ \frac{1}{t_p}=\frac{60}{75}-\frac{1}{6}=\frac{120-25}{150}\\\\ t_p=\frac{150}{95} \,\,minutes=1.57895\,\, minutes\\\\ t_p=\frac{150}{95}\times60 \,\,seconds=94.73684\,\,seconds$

${\color{Blue} Problem\,\,(2)}$

$Given,\,\, \\ t_e=190\,\,seconds=\frac{190}{60}\,\,minutes=\frac{19}{6}\,\,minutes\,\, \\t_p=42\,\,minutes\\\\ \therefore \,\,\frac{1}{\frac{19}{6}}=\frac{1}{42}+\frac{1}{t_b}\\\\ \frac{1}{t_b}=\frac{6}{19}-\frac{1}{42}=\frac{6\times42-19}{19\times42}\\\\ t_b=\frac{798}{233} \,\,minutes=3.42489\,\, minutes\\\\ t_b=\frac{798}{233}\times60 \,\,seconds=205.49356\,\,seconds$

${\color{Blue} Problem(3)}$

$Given,\,\, \\ t_e=110\,\,minutes \\t_p=150\,\,minutes\\\\ \therefore \,\,\frac{1}{110}=\frac{1}{150}+\frac{1}{t_b}\\\\ \frac{1}{t_b}=\frac{1}{110}-\frac{1}{150}=\frac{150-110}{110\times150}\\\\ t_b=\frac{16500}{40} \,\,minutes=412.5\,\, minutes\\\\$