Posted by Manu on Saturday, January 26, 2013 at 6:29pm.
A radioactive isotope, used in the agricultural industry has a half life of 8 days. On application, the radiation is emitted is measured at 12500 counts per sec (cps). If the area that it is applied to cannot be used until the radiation strength is less than 50 cps how long will it take before the site can be used again? (Hint assume initial strength is 100%)

Science  Radiation  Devron, Saturday, January 26, 2013 at 6:46pm
(12500)*1/2^n=500
solve for n
n=ln(12500)/ln(500)
n* the number of days (8) =should give you the time it takes to before the site can be used again. 
Science  Radiation  Manu, Saturday, January 26, 2013 at 6:51pm
Thank you for your answer. Why is it 500? Sorry I am just trying to understand

Science  Radiation  Devron, Saturday, January 26, 2013 at 6:55pm
Opps, should say
n=ln([12500/500)]/ln(2)
n* the number of days (8) =should give you the time it takes to before the site can be used again. 
Science  Radiation  Devron, Saturday, January 26, 2013 at 6:56pm
And you are right it should say 50 not 500.
Typos 
Science  Radiation  Devron, Saturday, January 26, 2013 at 7:02pm
So, about 63.7 days or 64 days.

Science  Radiation  Manu, Saturday, January 26, 2013 at 7:06pm
Thanks Devron

Science  Radiation  Devron, Saturday, January 26, 2013 at 7:13pm
It's not 500, you are correct it is 50. I made a typo when I plugged in the numbers. I started to show you how to rearrange the equation, and decided not to halfway through it. I didn't catch it until you mentioned it, so replace 500 with 50 and ignore the final rearrangement for my initial post and look at my second post. Just replace 500 with 50 and multiply your anser for n, which should be about 8, and multiply it by 8, the number of days for 1/2 of the sample to decay. The final rearrangement to give you n is in my second post, just replace 500 with 50, since you want to know the number of halflives to get to 50. That is why the initial equation is set up as (1/2^n)*12500=50. Like I said, its a typo.