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In a certain experiment, \isotope[32][15]{P} (half life \SI{14.26}{d}) is injected into a medium containing a culture of bacteria. After $\SI{45}{min}$ the cells are washed and a detector that is 75\% efficient (i.e. counts 75\% percent of emitted $\upbeta$ rays) records 990 counts per minute from all the cells. Calculate the activity of the original Phosphorsource under the assumption, that the cells took up 5.4\% of the Phosphor.
\begin{empheq}[box=\Gegeben]{align} T &= \SI{14.26}{d} = \SI{1.232e6}{s}\\ t &= \SI{45}{min} = \SI{2700}{s}\\ \eta' &= 75\% = 0.75\\ A_d &= \SI[permode=reciprocal]{990}{\per\minute} = \SI{16.50}{Bq}\\ \eta &= 5.4\% = 0.054 \end{empheq} \begin{empheq}[box=\Gesucht]{align} \text{Initial activity, } [A_0]=\si{Bq} \end{empheq} After \SI{45}{min}, the Phosphor's acitivty within the cells  since the detector registers only 75%  is: \begin{align} A_c &= \frac{A_d}{\eta'}\\ &= \frac{\SI{16.50}{Bq}}{0.75}\\ &= \SI{22}{Bq} \end{align} And because the cells took only 5.4\% of the Phosphor up, the overall Phosphor's activity (including the Phosphor that wasn't taken up by the cells) must be: \begin{align} A_t &= \frac{A_c}{\eta} = \frac{A_d}{\eta' \cdot \eta}\\ &= \frac{\SI{22}{Bq}}{0.054}\\ &= \SI{407.4}{Bq} \end{align} This activity was \SI{45}{min} earlier: \begin{align} A_0 &= A_t \cdot \mathrm{e}^{+\frac{\ln 2}{T}t} = A_t \cdot 2^{+\frac{t}{T}} = \frac{A_d}{\eta' \cdot \eta} \cdot \mathrm{e}^{+\frac{\ln 2}{T}t}\\ &= \SI{408}{Bq} \end{align} \begin{empheq}[box=\Lsgbox]{align} A_0 &= \frac{A_d}{\eta' \cdot \eta} \cdot \mathrm{e}^{+\frac{\ln 2}{T}t}\\ &= \SI{408}{Bq} \end{empheq}
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