Question

Situation:A 33 gram sample of a substance that'sused for drug research has a k-value of0.1124.-ktN = NoeNo initial mass (at time t = 0)=N= mass at time tk = a positive constant that depends onthe substance itself and on the unitsused to measure timet = time, in daysFind the substance's half-life, in days..Round your answer to the nearest tenth

Answer 1

Answer:

6.2 days

Explanation:

The half-life occurs when the mass N is equal to half of the initial mass, so to find the half-life, we need to replace N with No and solve for t

$\begin{gathered} N=N_oe^{-kt} \\ N=N_oe^{-0.1124t} \\ \\ \text{ Replacing N = No/2, we get} \\ \frac{N_o}{2}=N_oe^{-0.1124t} \\ \\ \text{ Solving for t} \\ \frac{N_o}{2N_o}=e^{-0.1124t} \\ \\ \frac{1}{2}=e^{-0.1124t} \\ \\ ln(\frac{1}{2})=ln(e^{-0.1124t}) \\ \\ -0.6931=-0.1124t \\ \\ \frac{-0.6931}{-0.1124}=\frac{-0.1124t}{-0.1124} \\ \\ 6.2\text{ days = t} \end{gathered}$

Therefore, the answer is

t = 6.2 days