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Suppose that 100 kg of a radioactive substance decays to 80 kg in 20 years.

a) Find the half-life of the substance (round to the nearest year).

b) Write down a function $y(t)$ ($t$ in years) modeling the amount (in kg) of the radioactive substance at time $t$.

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  • $\begingroup$ I have solved your problem. Please show your attempt before I share the solution. $\endgroup$
    – Math Lover
    Commented Sep 18, 2020 at 13:56
  • $\begingroup$ I did and got this 80/100=ln2/T Where T is T=lnsqrt32 $\endgroup$
    – Perfectoid
    Commented Sep 18, 2020 at 14:31
  • $\begingroup$ I posted the answer. You can confirm your working. $\endgroup$
    – Math Lover
    Commented Sep 18, 2020 at 14:39

2 Answers 2

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Hint: You have the radioactive decay law: $$N(t)=N_0e^{-\lambda t}$$ You have $N=80$ the time $t$ and $N_0=100$ deduce $\lambda$. Then half life $T$ is: $$N=\dfrac {N_0}2$$ $$\dfrac {N_0}2=N_0e^{-\lambda T}$$ $$\implies \dfrac 12=e^{-\lambda T}$$ $$ T=\dfrac {\ln 2}{\lambda}$$

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If $A_t$ is the amount left at time $t$, $A_0$ is the initial amount and $k$ is the rate constant for the first order reaction, we know for the radioactive decay (which is first order reaction) - \begin{align} A_t &= A_0e^{-kt} \tag i \\ \text{or } \ln \left(\frac{A_t}{A_0}\right) &= -kt. \end{align}

As $A_0 = 100$, $A_t = 80$, $t = 20$, \begin{align} \implies \ln \left(\frac{80}{100}\right) &= -20k. \\ k &= -\frac{\ln 0.8}{20} \tag {ii} \end{align}

At half life, $\displaystyle \ln \left(\frac{50}{100}\right) = -kt_{1/2}$ or $$ t_{1/2} = - \frac {\ln (0.5)}{k} = 20 \times \frac {\ln (0.5)} {\ln(0.8)} \approx 62.1 \text{ years}.$$

Now, find $k$ using equation (ii) and plug $k$ and $A_0 = 100$ into equation (i) and that is your answer to (b).

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Perfectoid
    Commented Sep 18, 2020 at 14:48

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