Answer:
[tex]A_0\left(\dfrac{1}{2}\right)^{\dfrac{5}{8.1}}[/tex]
[tex]A_0\cdot e^{-\dfrac{5\ln(2)}{8.1}}[/tex]
Step-by-step explanation:
Method 1: Half-life formula
To determine how much of a radioactive iodine will be present after 5 days, given it has a half-life of about 8.1 days, we can use the half-life formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Half Life Formula}}\\\\A(t)=A_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{\frac12}}}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A(t)$ is the quantity remaining.}\\\phantom{ww}\bullet\;\textsf{$A_0$ is the initial quantity.}\\ \phantom{ww}\bullet\;\textsf{$t$ is the time elapsed.}\\\phantom{ww}\bullet\;\textsf{$t_{\frac12}$ is the half-life of the substance.}\end{array}}[/tex]
In this case:
- [tex]t_{\frac{1}{2}}=8.1\; \sf days[/tex]
- [tex]t = 5\;\sf days[/tex]
Substitute the given values into the formula:
[tex]A(5)=A_0\left(\dfrac{1}{2}\right)^{\dfrac{5}{8.1}}[/tex]
Therefore, the amount present after 5 days is:
[tex]A_0\left(\dfrac{1}{2}\right)^{\dfrac{5}{8.1}}[/tex]
[tex]\hrulefill[/tex]
Method 1: Exponential decay formula
To determine how much of a radioactive iodine will be present after 5 days, given it has a half-life of about 8.1 days, we can use the exponential decay formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Exponential Decay Formula}}\\\\A(t)=A_0\cdot e^{-kt}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A(t)$ is the quantity remaining.}\\\phantom{ww}\bullet\;\textsf{$A_0$ is the initial quantity.}\\ \phantom{ww}\bullet\;\textsf{$t$ is the time elapsed.}\\\phantom{ww}\bullet\;\textsf{$k$ is a constant.}\end{array}}[/tex]
Given that the half-life is 8.1 days, then A(8.1) = A₀/2:
[tex]A(8.1)=\dfrac{A_0}{2}\\\\\\A_0\cdot e^{-8.1k}=\dfrac{A_0}{2}[/tex]
Solve for k:
[tex]e^{-8.1k}=\dfrac{1}{2}\\\\\\\ln\left(e^{-8.1k}\right)=\ln\left(\dfrac{1}{2}\right)\\\\\\-8.1k\ln(e)=\ln(1)-\ln(2)\\\\\\-8.1k=-\ln(2)\\\\\\k=\dfrac{\ln(2)}{8.1}[/tex]
Therefore:
[tex]A(t)=A_0\cdot e^{-\dfrac{t\ln(2)}{8.1}}[/tex]
To determine how much radioactive iodine will be present after 5 days, substitute t = 5 into the equation:
[tex]A(5)=A_0\cdot e^{-\dfrac{5\ln(2)}{8.1}}[/tex]
Therefore, the amount present after 5 days is:
[tex]A_0\cdot e^{-\dfrac{5\ln(2)}{8.1}}[/tex]
