Gadolinium-153, which is used to detect osteoporosis, has a half-life of 242 days. Which value is closest to the percentage of the Gd-153 left in a patient's system after 2 years (730 days).

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25.0 %

33.0 %

12.5 %

6.25 %

answer is C.

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Detailed Solution

$t_{1} or t_{50 %}$

for 1st order

$\begin{array}{l} t_{75 %} = 2 \times t_{2} or \frac{t_{j}}{t_{2}} = 2 \\ t_{87.5 %} = 3 \times t_{\frac{1}{2}} \\ t_{93.75 %} = 4 \times t_{2} \\ t_{l} = 242 days \end{array}$

for 2 year

$\frac{\ln 2}{242} \times 730 = \ln [\frac{100}{x}] \Rightarrow x = 12.5 %$

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