00:01
Okay, in this question, we are told that a sample of an isotope is measured to have an activity a nought of 5 ,500 becquerels.
00:09
So a nought is equal to 5 ,000, right then, probably 5 ,500.
00:15
It's awfully written.
00:16
Let's try that again.
00:19
5 ,500 becorels.
00:22
And then after three hours, the count is 480 becquerels.
00:27
So what is the half -life of the isotope? in order to calculate the half -life t -half here, we're going to need to calculate the decay constant.
00:36
And to calculate the decay constant, we can use this equation here, where a, which is going to be the activity or the count rate after a certain period of time, is equal to the initial count rate, which we have here, multiplied by e to the minus lambda t, and t will be three hours.
00:50
So we can say that a over a -0 equals e to the minus lambda t, and we can then take the natural log of both sides.
01:00
L 'n a over a nought, put into brackets, is equal to minus lambda t, and therefore lambda is equal to minus len a over a nought divided by t.
01:14
So we can then substitute in all of our values...