I have $44 mg$ of Actinium now, but by this time tomorrow it decays to $32 mg$ . I'll model the amount of Actinium left with $Q (t) = Q_{0} e^{k t}$ , where $t$ is time in days and $Q (t)$ is Actinium in milligrams. What is the value of $k$ ?
32
0.0912
44
$- 0.3185$

Answer

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Answer

So, the decay constant $k$ is approximately $- 0.3185$ per day. $k \approx - 0.3185$

Steps

Step 1 ：Given that $Q_{0} = 44$ mg, $Q (t) = 32$ mg, and $t = 1$ day, we can substitute these values into the equation to solve for $k$ .

Step 2 ：Substitute the given values into the equation: $32 = 44 * e^{k * 1}$ .

Step 3 ：Divide both sides by 44 to isolate $e^{k}$ : $\frac{32}{44} = e^{k}$ .

Step 4 ：Take the natural logarithm of both sides to solve for $k$ : $\ln (\frac{32}{44}) = k$ .

Step 5 ：Calculate the value of $k$ : $k = \ln (0.72727)$ .

Step 6 ： $k \approx - 0.3185$ .

Step 7 ：So, the decay constant $k$ is approximately $- 0.3185$ per day. $k \approx - 0.3185$