Radioactive Decay Problem

An unknown radioactive element decays into non-radioactive substances. In https://mathit.unco.edu/webwork2_fil...673f443751.png days the radioactivity of a sample decreases by https://mathit.unco.edu/webwork2_fil...7f8ea2bd81.png percent. What is the half life of the element in days?

AND

[IMG]file:///C:/Users/WILS58%7E1.003/AppData/Local/Temp/moz-screenshot.png[/IMG][IMG]file:///C:/Users/WILS58%7E1.003/AppData/Local/Temp/moz-screenshot-1.png[/IMG]How long will it take for a sample of 100 mg to decay down to 60 mg? How many days?

Re: Radioactive Decay Problem

First of all we can do some estimation. 51% is near half so we're expecting an answer of around 900 days. Other than that we'd need the decay constant (except if the assumption I make later about the initial condition is valid)

**Short Way **

Half life is related to the decay constant by the following equation: $\displaystyle t_{1/2} = \dfrac{\ln(2)}{\lambda}$ where $\displaystyle \lambda$ is the decay constant.

**Long Way**

Since you've posted this in differential equations we have the separable DE: $\displaystyle \dfrac{dN}{dt} = -\lambda t$

As said, this is separable: $\displaystyle \dfrac{dN}{N} = -\lambda \ dt$

$\displaystyle \int \dfrac{dN}{N} = -\lambda \int \ dt \Leftrightarrow \ \ln |N| = -\lambda t + C_1$ (where $\displaystyle C_1$ is the constant of integration)

$\displaystyle |N| = e^{-\lambda t + C_1} = e^{C_1}e^{-\lambda t} = Ce^{-\lambda t}$ where $\displaystyle C = e^{C_1}$ (also a constant)

I am also going to assume the initial condition of when t =0 then $\displaystyle N=N_0$ (I'm not sure how valid this assumption is but without being told the decay constant it's the only way to find it)

Sub in the initial condition and you find that $\displaystyle N = N_0e^{-\lambda t}$ and rearranged it is $\displaystyle \lambda = -\dfrac{1}{t} \ln \left(\dfrac{N}{N_0}\right)$

In your case: $\displaystyle \lambda = -\dfrac{1}{900} \ln \left(\dfrac{0.49N_0}{N_0}\right) \approx 7.926 \cdot 10^{-4} \text{ day}^{-1} $

Finally: $\displaystyle t_{1/2} = \dfrac{\ln(2)}{7.926 \cdot 10^{-4}\text{ day}^{-1}} = 874.5 \text{ days}$

We can check this against what we'd expect and it's close enough to be correct. You'll need to sort out the attachment on your second one but try the same thing again.