Question

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?

Answer 1

Here's what I got.

Explanation:

As you know, the nuclear half-life of a radioactive nuclide, `t_"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

• `1/2 * A_0 = A_0/2 ->` after one half-life
• `1/2 * A_0/2 = A_0/4 ->` after two half-lives
• `1/2 * A_0/4 = A_0/8 ->` after three half-lives
• `vdots`

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

`color(blue)(ul(color(black)(A_t = A_0 * 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

`"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_"276 days" = "26 g" * 1/2^color(red)(2)`

`color(darkgreen)(ul(color(black)(A_"276 days" = "6.5 g")))`

In other words, only `"6.5 g"` of polonium-210 will remain undecayed after `276` days.