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
Stefan V.
Dec 23, 2016

Here's what I got.

Explanation:

As you know, the nuclear half-life of a radioactive nuclide, t_"1/2"t1/2 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 138138 days, so right from the start you know that with every 138138 days that pass, the mass of the sample gets halved.

You can say that you have

  • 1/2 * A_0 = A_0/2 ->12⋅A0=A02→ after one half-life
  • 1/2 * A_0/2 = A_0/4 ->12⋅A02=A04→ after two half-lives
  • 1/2 * A_0/4 = A_0/8 ->12⋅A04=A08→ 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.

Related questions
See all questions in Nuclear Half-Life Calculations
Impact of this question
15246 views around the world
You can reuse this answer
Creative Commons License