Ok - so the 1:10 ratio will be obtained somewhere between 3 and 4 half-lives, right? 2^3 = 8, 2^4 = 16... I would say something like 3.3 half-lives...?
To the second problem - if I know the final size has to be 10^-10 meters (the size of an atom), the equation should be as follows:
10^(-10) =...
Oh, thanks! So, if I understand it correctly the answer to the first one will be after 5 half-lives, am I right?
Because 1/2^5 = 1/32 -> 1/32 remains radioactive while 31/32 have decayed, i.e. they are not radioactive -> the ratio is 1:31. Correct?
To the second exercise - I understand the...
Hey guys,
I've been given a worksheet with a couple of questions concerning radioactivity, half-life etc. It is high school level and for me, it seems like the first year doing physics as our current teacher is the first one who really knows something about the subject. :) So please be patient...