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Tags bequerel, biology, cells, curie, dosimetry, incorporation, nuclear physics, physics, radiation, radioactivity, radiometry Difficulty Points 2 Language Type Quantity Last modified 9 months ago Initial Version by Urs Zellweger (urs)	AG - Radioactivity taken up by cells (urs) Biologische Wirkung (Radiometrie) (urs)	0

Exercise

In a certain experiment, $\SI{0.016}{\micro Ci}$ ($\SI{1}{Ci}=\SI{3.7e10}{Bq}$) of \isotope[32][15]{P} (half life \SI{14.26}{d}) is injected into a medium containing a culture of bacteria. After $\SI{1.0}{h}$ the cells are washed and a detector that is 70\% efficient (counts 70\% percent of emitted $\upbeta$ rays) records 720 counts per minute from all the cells. What percentage of the original \isotope[32][15]{P} was taken up by the cells?

Short Solution

2.9\% was incorporated

Solution

\begin{empheq}[box=\Gegeben]{align} A_0 &= \SI{0.016}{\micro Ci} = \SI{592}{Bq}\\ T &= \SI{14.26}{d} = \SI{1.232e6}{s}\\ t &= \SI{1.0}{h} = \SI{86400}{s}\\ \eta' &= 70\% = 0.70\\ A_d' &= \SI[per-mode=fraction]{720}{\per\minute} = \SI{12}{Bq} \end{empheq} \begin{empheq}[box=\Gesucht]{align} \text{Precentage, } [\eta]=\si{/} \end{empheq} The injected Phosphor has an initial activity of $A_0 = \SI{0.016}{Ci} = \SI{592}{Bq}$. This activity decays with the half-life of \isotope[32][]{P} ($\SI{14.26}{d}$). After one hour, no significant change in the activity is expected, since this one hour is much less than the half-life. A short calculation can confirm that: \begin{align} A_t &= A_0 \cdot \mathrm{e}^{-\frac{\ln 2}{T} t} = A_0 \cdot 2^{\frac{t}{T}}\\ &= \SI{591}{Bq} \end{align} Observation of the cell culture shows 720 counts per minute, i.e. $\SI{12}{Bq}$, where only 70\% is registered. Hence there is a real activity of \begin{align} A_d &= \frac{A_d'}{\eta'}\\ &= \frac{\SI{12}{Bq}}{0.70}\\ &= \SI{17}{Bq} \end{align} in the cell culture. This is only \begin{align} \eta &= \frac{A_d}{A_t} = \frac{\frac{A_d'}{\eta'}}{A_0 \cdot \mathrm{e}^{-\frac{\ln 2}{T} t}} = \frac{A_d'}{\eta' \cdot A_0 \cdot \mathrm{e}^{-\frac{\ln 2}{T} t}} \approx \frac{A_d'}{\eta' \cdot A_0}\\ &= \frac{\SI{17}{Bq}}{\SI{591}{Bq}}\\ &= 0.029 = 2.9\% \end{align} of the theoretical expected activity, which means that only 2.9\% of the Phosphorus was taken up by the cells. \begin{empheq}[box=\Lsgbox]{align} \eta &= \frac{A_d'}{\eta' \cdot A_0 \cdot \mathrm{e}^{-\frac{\ln 2}{T} t}} \\ &= 0.029 = 2.9\% \end{empheq}

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22:19, 22. May 2018	lsg boxed	Urs Zellweger (urs)	Current Version
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