 Let's solve a question on half life. Here we have a sample of an isotope of Francium. Initially it has 128 moles of nucleates at 0 seconds. Only 8 moles of Francium nucleates remain at 20 seconds. At what time does the sample have only 2 moles of Francium nucleates left? And we have to choose one answer out of these 4 options. Alright as always pause the video and try this question on your own first. Okay hopefully you have given us a short. Now in this question we need to figure out at what time does the sample have only 2 moles of Francium nucleates left? So we have the data till 20 seconds and till 20 seconds we know that 8 moles remain. But we need to figure out, we need to go ahead and find out that time when only 2 moles remain. If we try and think about how many half lives are between 8 moles and 2 moles. So from 8 moles, from 8 moles when it goes to 4 moles, when only 4 moles remain, then one half life is spent. Then from 4 to 2 moles, then one more half life is spent. So it turns out that we need to figure out 2 half lives, we need to figure out 2 into t half. This is the value that we need to figure out. Once we do figure out this, then we can add 20, we can add 20 to it to find the total time that it takes when the sample has only 2 moles left. So our aim is to figure out the half life. But from this much data we might not be able to figure out what half life is. So let's go back to the information that the question is providing us. We are starting with 128 moles of nucleates and then after 20 seconds we have 8 moles remaining. From this data alone we can try and find what half life is. Because we also know the time that it took for Francium to go from 128 to 8 moles. So one relation that we know is that the number of nucleates that remain, this is equal to the initial number of nuclei divided by 2 to the power n. And here n is the number of half lives that are spent. We can also very quickly try and derive this as well. So let's say initially if we have n not number of nuclei which are radioactive. After one half life n not by 2 remain. So I am shading and showing how many nuclei remain. So initially this is all shaded. So n not by 2 remain after one half life. After one more half life we have n not by 4 remaining. So we can write 4 as 2 square. This is n not by 2 square. After one more half life we have n not by 8 remaining. And we can write that as n not divided by 2 to the power 3 which is 8. So you see what is happening. After one half life this is one half life spent. This is one more half life. This is one more. So after one half life we have n not by 2 remaining. After two half lives we have n not by 2 to the power 2 remaining. After 3 we have n not by 2 to the power 3 remaining. So after n half lives we will have n not by 2 to the power n nuclei remaining. So now let's go back to this information. We have 8 remaining, 8 nuclei remaining. We start off with 128 and we need to figure out after how many half lives is this happening. So if we take 2 to the power n to the left hand side. This will be 2 to the power n equals to 128 divided by 8. This would be 64 divided by 4. This would be 16 and we can write 16 as 2 to the power 4. This is 2 to the power 4. So after 4 half lives, after 4 half lives we have 8 moles remaining. After 4 half lives we have 8 moles remaining. And we know the time that it took that is 20 seconds. So this means that 4 half lives, 4 half lives this is equal to 20 seconds. So 1 half life this is equal to 5 seconds. This is 5 seconds and we are interested in figuring out 2 half lives. So this is 10 and from the beginning we already spent 20 seconds. Then after 20 seconds, 10 seconds are spent more and then 2 moles of Francium remain. So this is, so this will be 20. This will be 20, 20 plus plus 1 half life. That is 5 plus 1 more half life. Again 5 and this comes out to be equal to 30 seconds. So the right option is option A. Alright, you can try more questions from this exercise in the lesson and if you're watching on YouTube, do check out the exercise link which is added in the description.