00:01
In this problem we have been given that there is a radioactive material and that has 10 to power 15 atoms.
00:08
And for this radioactive material, the activity is observed to be 6 times 10 raised to 11 pectoral.
00:16
So let's represent the activity using a and the number of atoms using m.
00:21
So here for any radioactive element, the activity and the number of atoms that can be related using the expression, a is equal to lambda types n.
00:33
So lambda here is the decay constant, and this is in accordance to the equation whereby activity is observed to be directly proportional to the number of nuclei.
00:44
And here we need to determine the half -life.
00:46
So we already know that the half -life of the sample represented by t -half, that's log 2 to base e, divided by the constant of decay.
00:57
So that will be equal to log 2 divided by instead of decay constant we can use a by m because a was equal to lambda times m.
01:07
So that will be n times log of 2 to base e divided by a...