# 4. Suppose a laboratory has a 26 g sample of polonium 210. The half-life of Polonium 210 about 138 days. How many half-lives of polonium 210 occur in 276 days? How much polonium 210 is in the sample 276 days later?

##### 1 Answer
Dec 23, 2016

Here's what I got.

#### Explanation:

As you know, the nuclear half-life of a radioactive nuclide, ${t}_{\text{1/2}}$ represents the time needed for half of an initial sample to decay.

This essentially means that a nuclide's half-life tells you how much time must pass in order for your sample to be reduced to half of its initial value.

In this particular case, you know that polonium-210 has a half-life of $138$ days, so right from the start you know that with every $138$ days that pass, the mass of the sample gets halved.

You can say that you have

• $\frac{1}{2} \cdot {A}_{0} = {A}_{0} / 2 \to$ after one half-life
• $\frac{1}{2} \cdot {A}_{0} / 2 = {A}_{0} / 4 \to$ after two half-lives
• $\frac{1}{2} \cdot {A}_{0} / 4 = {A}_{0} / 8 \to$ after three half-lives
• $\vdots$

and so on. The half-life equation can be written as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{A}_{t} = {A}_{0} \cdot \frac{1}{2} ^ n}}}$

Here

• ${A}_{t}$ is the amount that remains undecayed after in $t$ time interval
• ${A}_{0}$ is the initial mass of the sample
• $n$ is the number of half-lives that pass in the $t$ time interval

Now, notice that

$\text{276 days" = color(red)(2) xx "138 days}$

which means that $2$ half-lives pass in the given time period. Consequently, you can say that your sample will decay to

${A}_{\text{276 days" = "26 g}} \cdot \frac{1}{2} ^ \textcolor{red}{2}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{A}_{\text{276 days" = "6.5 g}}}}}$

In other words, only $\text{6.5 g}$ of polonium-210 will remain undecayed after $276$ days.