# Can you help with this exponential decay question?

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 $$t$$.

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

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: $$N=\dfrac {N_0}2$$ $$\dfrac {N_0}2=N_0e^{-\lambda T}$$ $$\implies \dfrac 12=e^{-\lambda T}$$ $$T=\dfrac {\ln 2}{\lambda}$$

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).

• Thank you so much! Commented Sep 18, 2020 at 14:48