00:01
Given that iodine 125 has a half -life of 60 .1 days, we could go ahead and set up that we have half of that amount.
00:08
And that's equal to the original amount again, which we could divide out.
00:12
And then we want to solve for this k value.
00:15
But we could take the natural log of both sides.
00:17
We get rid of the exponential on the right side.
00:20
So ln of 1� is equal to ln of e would cancel.
00:24
And then that would leave us with, well, i'd just like to leave it as a regular k value, because it's going to turn into a negative here anyway.
00:32
And then divide by the 60 .1, which is a positive, and one of one half will be a negative.
00:39
So this will give us a negative k value, which is approximately equal to zero.
00:45
Oops, sorry, let's put a negative now.
00:48
Negative 0 .0 .1, 153, 3, 3, 2, 3, 2, 3...