Question #e7a7d

1 Answer
Sep 15, 2017

Answer:

Here's what I got.

Explanation:

For a given radioactive isotope, the nuclear half-life, #t_"1/2"#, tells you the time needed for half of an initial sample of this isotope to undergo radioactive decay.

This means that with every half-life that passes, the mass of the sample will be halved.

http://www.schoolphysics.co.uk/age14-16/Nuclear%20physics/text/Half_life/index.html

In your case, you know that the isotope has a half-life of #"120 s"#. If you start with an #"80-g"# sample, you will end up with

  • #"80 g" * 1/2 = "80 g"/2^color(red)(1) -> # after #color(red)(1)# half-life passes
  • #"80 g"/2^color(red)(1) * 1/2 = "80 g"/2^color(red)(2) -># after #color(red)(2)# half-lives pass
  • #"80 g"/2^color(red)(2) * 1/2 = "80 g"/2^color(red)(3) -># after #color(red)(3)# half-lives pass
    #vdots#

and so on. You can thus say that your sample will contain

  • #"After 120 s"#

#(120 color(red)(cancel(color(black)("s"))))/(120 color(red)(cancel(color(black)("s")))) = color(red)(1) implies "80 g"/2^color(red)(1) = "40 g"#

  • #"After 240 s"#

#(240 color(red)(cancel(color(black)("s"))))/(120 color(red)(cancel(color(black)("s")))) = color(red)(2) implies "80 g"/2^color(red)(2) = "20 g"#

  • #"After 480 s"#

# (480 color(red)(cancel(color(black)("s"))))/(120 color(red)(cancel(color(black)("s")))) = color(red)(3) implies "80 g"/2^color(red)(3) = "10 g"#