# can you help with this exponential decay question?

Mathematics Asked by Perfectoid on December 19, 2020

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 $$t$$.

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) -

$$displaystyle A_t = A_0e^{-kt} ,$$ ...(i)

or $$, displaystyle ln (frac{A_t}{A_0}) = -kt$$

As $$A_0 = 100, A_t = 80, t = 20, -> , displaystyle ln (frac{80}{100}) = -20k$$

$$displaystyle k = -frac{ln 0.8}{20}$$ ...(ii)

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

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

Answered by Math Lover on December 19, 2020

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

Answered by Aryadeva on December 19, 2020

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