$^{131} I$ is an isotope of Iodine that $β$ decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with $^{131} I$ is injected into the blood of a person. The activity of the amount of $^{131} I$ injected was $2.4 \times 1 0^{5}$ Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After $11.5$ hours, $2.5 m l$ of blood is drawn from the person's body, and gives an activity of $115$ Bq. The total volume of blood in the person's body, in liters is approximately (you may use $e^{x} \approx 1 + x$ for $∣ x ∣ < < 1$ and $l n 2 \approx 0.7$ ).

Question

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Answer 1

$t_{\frac{1}{2}} = 8 d a y s = 8 \times 24 h r$
$R_{0} = 2.4 \times 1 0^{5} B q$
$U s i n g \frac{0 . 6 9 1}{t _{\frac{1}{2}}} \times t = l n \frac{R _{0}}{R}$
$\frac{0 . 6 9 2}{8 \times 2 4} \times 11.5 = l n \frac{2 . 4 \times 1 0 ^{5}}{R}$
$\frac{2 . 4 \times 1 0 ^{5}}{R} = e^{0.041} = 1 + 0.041$
$R = \frac{2 . 4 \times 1 0 ^{5}}{1 . 0 4 1} = 2.3 \times 1 0^{5}$
$115 B q i s i n v o l u m e = 2.5 m l$
$2.3 \times 1 0^{5} B q i s i n v o l u m e = \frac{2 . 5}{1 1 5} \times 2.3 \times 1 0^{5}$
$= 0.05 \times 1 0^{5}$
$= 5 \times 1 0^{3} m l$
$= 5 l i t r e s$

Answer 2

t_{\frac{1}{2}} = 8 d a y s = 8 \times 24 h r

R_{0} = 2.4 \times 1 0^{5} B q

U s i n g \frac{0 . 6 9 1}{t _{\frac{1}{2}}} \times t = l n \frac{R _{0}}{R}

\frac{0 . 6 9 2}{8 \times 2 4} \times 11.5 = l n \frac{2 . 4 \times 1 0 ^{5}}{R}

\frac{2 . 4 \times 1 0 ^{5}}{R} = e^{0.041} = 1 + 0.041

R = \frac{2 . 4 \times 1 0 ^{5}}{1 . 0 4 1} = 2.3 \times 1 0^{5}

115 B q i s i n v o l u m e = 2.5 m l

2.3 \times 1 0^{5} B q i s i n v o l u m e = \frac{2 . 5}{1 1 5} \times 2.3 \times 1 0^{5}

= 0.05 \times 1 0^{5}

= 5 \times 1 0^{3} m l

= 5 l i t r e s