00:01
Hi there.
00:02
So for this problem, we are given that the half -life of sisyon -137 is equal to 30 years.
00:10
And suppose that we have initially an amount, well, the initial amount that we're going to call by zero, that is equal to 170 milligrams.
00:21
So for part a of this problem, we are asked about to find the mass that remains after t years.
00:26
Okay, so first of all, we know what we need to consider is that this is an exponential decay, so that will be the initial amount that is 170, this times the exponential of the concept of proportionality times the time.
00:43
Now, to obtain the value of the concept of proportionality, we can use the half -life that we are given for this.
00:50
So remember that the half -life is the time that it takes for the initial amount to, half, so that will be half of 170, so that will be 85.
01:03
So we will have 85 is equal to 170 times the exponential of the consumption proportionality times the half -line, that is 30.
01:12
Then if we divide this in here, we will have 1 divided by 2, then we can elevate both sides of this to 1 divided by 30.
01:20
So we will have the exponential of k.
01:23
So with this, you can answer then that the amount at any given time is then equal to the initial amount that is 170, and this times 1 divided by 2 times the time divided by 30.
01:38
So that's a solution for the first part of this problem.
01:42
Now, for the second part of this, we are asked about how much of the sample remains after 80 years.
01:47
So what we need to do is to evaluate the previous expression at 80, so that will be 170 times 1 divided by 2, elevated to m80 divided by 30.
01:59
So let's see what value we obtained from this.
02:11
So the value that we obtained from this is equal to approximately.
02:16
Now we need to round to the nearest to two decimal places.
02:22
So that will be 26 .77 milligrams.
02:28
So that's a solution for our b of this problem...