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