Answer:
47.6 days
Step-by-step explanation:
Given formula:
[tex]N(t)=N_0e^{-kt}[/tex]
where:
- N₀ = initial mass (at time t = 0)
- N = mass (at time t)
- k = a positive constant
- t = time (in days)
First, find the decay constant (k) for chromium-51.
Given its half-life is about 28 days:
[tex]\implies \dfrac{1}{2}=1e^{-28k}[/tex]
[tex]\implies \dfrac{1}{2}=e^{-28k}[/tex]
[tex]\implies \ln \dfrac{1}{2}= \ln e^{-28k}[/tex]
[tex]\implies \ln1 - \ln 2= -28k \ln e[/tex]
[tex]\implies -\ln 2= -28k[/tex]
[tex]\implies k=\dfrac{-\ln 2}{-28}[/tex]
[tex]\implies k=\dfrac{1}{28}\ln 2[/tex]
To determine how long it will take for a 650 mg of a sample of chromium-51 to decay to 200 mg, substitute the following values into the formula and solve for t:
- N₀ = 650 mg
- N = 200 mg
- [tex]k= \dfrac{1}{28}\ln 2[/tex]
[tex]\implies 200=650e^{(-\frac{1}{28}t\ln 2)}[/tex]
[tex]\implies \dfrac{200}{650}=e^{(-\frac{1}{28}t\ln 2)}[/tex]
[tex]\implies \dfrac{4}{13}=e^{(-\frac{1}{28}t\ln 2)}[/tex]
[tex]\implies \ln \dfrac{4}{13}=\ln e^{(-\frac{1}{28}t\ln 2)}[/tex]
[tex]\implies \ln \dfrac{4}{13}=-\dfrac{1}{28}t\ln 2\ln e[/tex]
[tex]\implies \ln \dfrac{4}{13}=-\dfrac{1}{28}t\ln 2[/tex]
[tex]\implies -28 \ln \dfrac{4}{13}=t\ln 2[/tex]
[tex]\implies t=\dfrac{-28 \ln \dfrac{4}{13}}{\ln 2}[/tex]
[tex]\implies t=47.61231211[/tex]
Therefore, it will take 47.6 days (nearest tenth) for a 650 mg of a sample of chromium-51 to decay to 200 mg.
