Question #7bdb7

1 Answer
Oct 18, 2017

Answer:

Here's what I got.

Explanation:

The idea here is that a radioactive isotope's nuclear half-life tells you the time needed for half of an initial sample to undergo radioactive decay.

In your case, you know that it takes #3# minutes for half of any amount of polonium-218 that you have to undergo radioactive decay.

If you take #A_0# to be the initial mass of polonium-238 and #A_t# to be the amount that remains undecayed after a given period of time #t#, you can say that you have

#A_t = A_0 * (1/2)^n#

Here

  • #n# is the number of half-lives that pass in the given period of time #t#

In your case, you know that it takes #30# minutes to transport the sample of polonium-238, which implies that you have

#n = (30 color(red)(cancel(color(black)("minutes"))))/(3color(red)(cancel(color(black)("minutes")))) = 10#

So if #10# half-lives pass in #30# minutes and you know that you must end up with #"0.10 g"# of polonium-238, you can say that you have

#"0.10 g" = A_0 * (1/2)^10#

Rearrange to solve for #A_0#

#A_0 = "0.10 g" * 2^10 = "102.4 g"#

Now, I'll leave the answer rounded to two sig figs, but you could round it to one significant figure based on the value you have for the half-life of the isotope.

#color(darkgreen)(ul(color(black)("amount needed" = 1.0 * 10^2color(white)(.)"g")))#