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

Answer:

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.