Jump to content

Half-Lives (Chemistry)

Recommended Posts

Qs: If it takes 24 days for 75% of an isotope to undergo a first order decay reaction what is the half-life for the isotope?

I understand half-lives, however, I'm just not sure what the question is asking for me to solve. I thought it meant that 75% of an element (that happened to be an isotope of an element) took 24 days to decay by 75%, meaning 25% was still left. So I thought I would figure out the original amount of time it took at that rate, which would be 32 days. Then I halved that (because it is the half life of it) and got 16 days, however, that is not an answer choice.

I don't know if there is a formula to solve this, but this has got me quite confused for the last 30 minutes or so. Please help, thanks!

Share on other sites

The decay function is not linear as you assume here, but exponential. Kind of like exponential growth.

Share on other sites

• 2 weeks later...

To say that something has a "half life" of, say, T means that half of it will be gone (so half is still left) after time T.  Since half is still left, if we start with amount A, after time T, we will have (1/2)A.  After another time T, we will have half of that, (1/2)(1/2)A= (1/2)^2A.  After still another time T, we will have half of that, (1/2)((1/2)^2A)= (1/2)^3A.   Continuing that, after "n" periods of T, will we have left (1/2)^n A left.

But, since this is a continuous process, n does not have to be an integer.  Given any time, t, we can think of that as t/T periods of t whether that is an integer or not.  After time t, we will have left

(1/2)^(t/T)A.

You are told "it takes 24 days for 75% of an isotope to undergo a first order decay reaction".

So (1/2)^(24/T)A= (3/4)A.  Obviously "A" cancels leaving (1/2)^(24/T)= 3/4.  Since the unknown, T, is an exponent, use logarithms- (24/T) log(1/2)= log(3/4). (T/24)log(3/4)= log(1/2), T/24= (log(1/2))/(log(3/4)), and, finally, T= (24log(1/2))/(log(3/4)) days.

Share on other sites

The numbers here lend themselves to a more straightforward analysis than using logarithms though.  The target amount can be reached by halving the amount a certain number of times (I'm guessing the more complex logarithm analysis isn't expected given the simplicity of the numbers).

Edited by Juno

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
• Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.