Isotope half-life of 8 days: when will 5% of original mass remain?

PabloPeterson99

New member
Joined
Jan 15, 2018
Messages
3
Hi! I need some help in how to calculate when an isotope (Iodine 131I) has left 5% of its original mass, if the half-life period is 8 days.

By writing down on paper, I noticed that after 32 days, it will have 6.25% left of its original mass and after 40 days only 3.125%.

How do I count that though using a (simple) formula? When will it have 5% left of the original mass (100%) if the half-life period is 8 days?

Thank you!
 
Why do you think there is a "simple" formula?

Really, though, all you need is a continuous form of what you already have done. You will need logarithms. Have you been introduced to logarithms?
 
I'm high school first year student. Just learning logarithms but I have a little bit of difficulties in understanding them sometimes. I've tried logarithm formulas that I've used for other exercises but they weren't correct. One was for bacteria doubling once in an hour and I needed to find out after how many hours there'd be million. I got that right, but I can't seem to understand how to calculate this one. I don't understand what formula I'm supposed to use or how to come up with a formula.
 
It helps to think of each 8 day period as a separate, discrete piece, but, as you have noticed, this will take you only so far.

32 Days - Four Periods
\(\displaystyle \left(\dfrac{1}{2}\right)^{4} = 0.0625\)

40 Days - Five Periods
\(\displaystyle \left(\dfrac{1}{2}\right)^{5} = 0.03125\)

If you consider the more fundamental unit - one day - you can get closer. Exactly where do that 4 and 5 come from?


32 Days
\(\displaystyle \left(\dfrac{1}{2}\right)^{32/8} = 0.0625\)

33 Days
\(\displaystyle \left(\dfrac{1}{2}\right)^{33/8} = 0.05731\)

34 Days
\(\displaystyle \left(\dfrac{1}{2}\right)^{34/8} = 0.05256\)

35 Days
\(\displaystyle \left(\dfrac{1}{2}\right)^{35/8} = 0.04819\)

The logartihms will allow you to consider ALL possible values, not just integers.

\(\displaystyle \left(\dfrac{1}{2}\right)^{n/8} = 0.05\)

Introduce the logarithm

\(\displaystyle \log\left(\dfrac{1}{2}\right)^{n/8} = log(0.05)\) -- Use your best algebra (logarithm rules) to solve for 'n'.

Let's see what you get.
 
It helps to think of each 8 day period as a separate, discrete piece, but, as you have noticed, this will take you only so far.

The logartihms will allow you to consider ALL possible values, not just integers.

\(\displaystyle \left(\dfrac{1}{2}\right)^{n/8} = 0.05\)

Introduce the logarithm

\(\displaystyle \log\left(\dfrac{1}{2}\right)^{n/8} = log(0.05)\) -- Use your best algebra (logarithm rules) to solve for 'n'.

Let's see what you get.
My problem is that I don't know how to do it from here. I just don't understand how/what I'm supposed to do with the "n/8". I try but I can't make it work.
 
My problem is that I don't know how to do it from here. I just don't understand how/what I'm supposed to do with the "n/8". I try but I can't make it work.
Using laws of logarithm

\(\displaystyle \displaystyle{log\left(\dfrac{1}{2} \right)^\frac{n}{8} \ = \dfrac{n}{8} * log\left(\dfrac{1}{2} \right)}\)

Then

\(\displaystyle \displaystyle{\dfrac{n}{8} * log\left(\dfrac{1}{2} \right) \ = \ log(0.05)}\)

Continue....
 
I'm high school first year student. Just learning logarithms but I have a little bit of difficulties in understanding them sometimes. I've tried logarithm formulas that I've used for other exercises but they weren't correct. One was for bacteria doubling once in an hour and I needed to find out after how many hours there'd be million. I got that right, but I can't seem to understand how to calculate this one. I don't understand what formula I'm supposed to use or how to come up with a formula.
They were supposed to have covered this material before assigning homework on it. To learn the basic equation as well as how to set up and solve these sorts of exercises, try here. ;)
 
Top