# Half-life of Am-$241$, $3$ micrograms decays over $9$ years, how much if left?

$3$ micrograms of Americium-$241$, which has a half life of $432$ years. After $9$ years how much will remain?

I'm not sure of the formula to use or how to calculate it. I'm assuming it's exponential decay since it's a radioactive substance.

• Yes, it is. Compute the constant from half life definition and apply the formula. Commented May 10, 2016 at 9:08

Denote mass of substance by $M(t)$ at any given time $t$.

So $M(t) =$ (initial mass) $\times 2 ^{\frac{-t}{\text{(half life)}}}$.

From your question, $M(9) = 3 \times 2^{\frac{-9}{432}} \approx 2.957$.

Edit for further explanation: In case the above formula for $M(t)$ isn't clear, observe than when $t=0$, $M(t)$ equals the initial mass - obviously (since $2^{0}=1$). When $t=$(half life), $M(t)$ equals the initial mass times $\frac{1}{2}$, i.e. it has divided by one half - exactly what "half-life" means. When $t = 2 \times$(half life), $M(t)$ equals the initial mass times $\frac{1}{4}$ (since $2^{-2} = \frac{1}{4}$), i.e. it has halved twice. Make sense? It is exponentially decaying because of the exponent $2^{-t/432}$, as $t$ gets bigger this quantity gets smaller exponentially fast.

• Thank you! That made perfect sense. Done some practises with the formula when t = 0, t= 432, worked well cheers Commented May 10, 2016 at 10:39

HINT:

After $432x$ years, you are left with $\frac{1}{2^x}$ of the original amount.

So calculate the value of $x$ in the equation $432x=9$, and plug it into $[\frac{1}{2^x}\times\text{original amount}]$.

Another way to derive the formula is as follows. Suppose that the initial amount of the element is $N_0$ micrograms.

• In $1\times 432$ years there will be left $N_0\cdot\frac{1}{2}=\frac{N_0}{2}$ micrograms.

• In $2 \times 432$ years there will be left $\frac{N_0}{2}\cdot \frac{1}{2} = N_0\cdot \left(\frac{1}{2}\right)^2$ micrograms.

..................

• In $n\times 432$ years there will be left $N_0\cdot \left(\frac{1}{2}\right)^n$ micrograms.

Setting $x = n\times 432\implies n = \frac{x}{432}$. Substituting to the last formula we have:

• In $x$ years there will be left $N_0\cdot \left(\frac{1}{2}\right)^{\frac{x}{432}}$ micrograms.

Substitute $x$ and $N_0$ and you will reach the answer.

You start wit $$m(0) = m_0$$ where $m_0 = 3$ micrograms. After the time $t = T$ has passed. with $T=432$y you get $$m(T) = (1/2) m_0$$ After $t = 2T$ $$m(2T) = (1/2) m(T) = (1/2)^2 m_0$$ After $t = 3T$ $$m(3T) = (1/2) m(2T) = (1/2)^2 m(T) = (1/2)^3 m_0$$ and so on. This leads to $$m(kT) = m_0 (1/2)^k$$ and with $t = k T \iff k = t / T$ finally to $$m(t) = m_0 (1/2)^{t / T}$$

You are looking for $m(9)$.