Half life Problem which I am right and the teacher is wrong
I am dealing with Half Lives in chemistry. In case you don't know, a half life is the time it takes for half of a sample of some element to decay into other types of elements.
I want someone to check this over, review my explanations, and tell me if you either agree or disagree with me. If you disagree with me I would appreciate it if you would tell me why.
Here is the question:
The choices are:
Geologists use the decay of potassium-40 in volcanic rocks to determine their age. Potassium-40 has a half life of 1.26E^9 years, so it can be used to date very old rocks. If a sample of rock 3.15E^8 years old contains 2.73E^-7g of potassium-40 today, how much potassium-40 was originally present in the rock?
a) 2.30E^-7g b) 3.25E^-7g
c) 1.17E^-8g d) 4.37E^-6
Okay, now I said 'b' for my answer, and my teacher insists it is 'a'. I don't know exactly how she got 'a' for an answer. However I will explain how I got b for an answer and how I can prove it.
First of all you can find the number of half lives by taking the total duration that you are looking at and dividing it by the half life duration.
This means the number of half lives in this problem is:
3.15E^8 years / 1.26E^9 years = .25 half lives
In order to find the amount remaining you do:
Original amount * .5 ^#halflives
This is the same as saying take the original amount, and multiply it by one half by the number of half lives that it has undergone.
But with this problem we are doing it backwards. Instead of trying to find the amount left with the original amount given we are instead given the amount left and we have to find the original amount.
This means we have to do something like:
AmountLeft * 2 ^#halflives
2.73E^-7g(2) ^ .25
When I do this, I get B for an answer, which is 3.25E-7g
Now, even if you didn't follow along up until I can prove my answer. Remember Original amount * .5 ^#halflives equals the amount remaining.
When you do:
3.25E-7(.5) ^ .25 you get 2.73E-7g as the amount left
Thank you to any and all who help in any way!