Question #fca61

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Question #fca61

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Answer 1 · 2014-12-29T19:27:43

The answer is $62 d a y s$ $62 days$ .

We know that an exponential decay can be expressed mathematically by

$A (t) = A_{0} \cdot {(\frac{1}{2})}^{\frac{t}{t_{1/2}}}$ $A(t) = A_0 * (1/2)^(t/t_("1/2"))$ , where

$A (t)$ $A(t)$ - the amount left after t years;
$A_{0}$ $A_0$ - the initial quantity of the substance that will undergo decay;
$t_{1/2}$ $t_("1/2")$ - the half-life of the decaying quantity.

$Ra-223$ $"Ra-223"$ has a molar mass of approximately $223 \frac{g}{m o l}$ $223g/(mol)$ , which means that the sample's initial mass and the final mass will be

$A_{0} = 0.240$ $A_0 = 0.240$ $m o l e s \cdot 223 \frac{g}{m o l} = 53.5 g$ $mol es * 223 g/(mol) = 53.5g$

$A (t) = 7.50 \cdot 10^{- 3}$ $A(t) = 7.50 * 10^(-3)$ $m o l e s \cdot 223 \frac{g}{m o l} = 1.67 g$ $mol es * 223 g/(mol) = 1.67g$

So,

$1.67 = 53.5 \cdot {(\frac{1}{2})}^{\frac{t}{12.4}} \to 0.0312 = {(\frac{1}{2})}^{\frac{t}{12.4}}$ $1.67 = 53.5 * (1/2)^(t/12.4) -> 0.0312 = (1/2)^(t/12.4)$

$\frac{t}{12.4} = {log}_{0.5} (0.0312) = 5.002$ $t/(12.4) = log_(0.5)(0.0312) = 5.002$ , which means that

$t = 5.002 \cdot 12.4 = 62$ $t = 5.002 * 12.4 = 62$ days.

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Question #fca61

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