Exponential Decay
The rate of decay of radioactive material is proportional to its actual mass. When solving the resulting equation, we get the mathematical model as follows:
![m(t)=m_oe^{-\lambda\mathrm{}t}](https://img.qammunity.org/2023/formulas/mathematics/college/kuioskdf00vkebwif3mvb7cfvli5228zf8.png)
Where mo is the initial mass, λ is a constant, and t is the time.
The half-life time is the time it takes for the initial mass to be halved, i.e., the remaining mass is mo/2. Substituting into the formula for t=14 days:
![(m_o)/(2)=m_oe^(-14\lambda)](https://img.qammunity.org/2023/formulas/mathematics/college/91rjd6lifxd1le4w2gwp8o4vx8230mbk59.png)
Simplifying by mo and solving for λ:
![\begin{gathered} (1)/(2)=_{}e^(-14\lambda) \\ \ln (1)/(2)=-14\lambda \\ \lambda=(\ln 2)/(14) \\ Calculate\colon \\ \lambda=0.04951 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ftvqzap3pk1sh8wbdyf1zzitfd3sogm928.png)
Now our model is complete:
![m(t)=m_oe^(-0.04951t)](https://img.qammunity.org/2023/formulas/mathematics/college/nq5nrhopopyf6edhozyke9m8s0eplj4io7.png)
Now we are given the initial mass of a sample mo = 500 grams. It's required to calculate the remaining mass after t = 5 weeks = 5*7 = 35 days.
Substitute values:
![\begin{gathered} m(35)=500e^(-0.04951\cdot35) \\ \text{Operating:} \\ m(35)=500\cdot0.17678 \\ \boxed{m(35)=88.39} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/odfs0pq8e920ssmhf4ovnc2stb7b0jwcc2.png)
Approximately 88 grams of the sample will remail after 5 weeks