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Half-life very difficult!

  1. May 15, 2003 #1
    a radioactive with a half-life of 5 days, has an initial activity of 4000 counts/min

    determine the activity after 10 days

    well because 1 half life is 5 days, so 10 days must be 2
    so 4000/(2^2)
    =1000 counts/min is that right?

    2. If the initial quantitiy of radioactive material is 200gms determine the amount left after 15 days have elapsed
    no of half lives = 3
    =25 gms is that right?

    3. convert 4000 counts per min into bq's
    how do i do that?
  2. jcsd
  3. May 15, 2003 #2

    Tom Mattson

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    You do it by looking up the definition of a bq in your book.
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