Half-life very difficult!

  • Thread starter dagg3r
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  • #1
dagg3r
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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
200/(2^3)
=25 gms is that right?

3. convert 4000 counts per min into bq's
how do i do that?
 

Answers and Replies

  • #2
quantumdude
Staff Emeritus
Science Advisor
Gold Member
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Originally posted by dagg3r
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?

Yes.

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

Yes.

3. convert 4000 counts per min into bq's
how do i do that?

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

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