Finding Mass of Isotope After Radioactive Decay by Half-Life

[rakeru]

Homework Statement

A radioactive sample contains 2.25g of an isotope with a half-life of 3.8 days.
How much of the isotope in grams will remain after 11.0 days?

Homework Equations

The Attempt at a Solution

Hi! I've just started college this semester. I'm taking Introductory Chemistry. Right now we are doing nuclear chemistry. This problem is confusing, though, because I'm not sure if I can just divide the half-life like a normal number.. My professor doesn't teach. Literally.

I tried doing this problem this way:

It starts off with 2.25g.
After 3.8 days (one half-life) it's 1.125g.
After 7.6 days (two half lives) it's 0.5625g.
But in another half-life, 11.4 days would have passed. More than 11. In 11.4 days it would be 0.28125g.

Is this correct? How would I calculate this? My book doesn't tell me.. Is there a specific way? It only tells me that it's "beyond the scope of this book". Please help!

 

[tiny-tim]

welcome to pf!

hi rakeru! welcome to pf!

It starts off with 2.25g.
After 3.8 days (one half-life) it's 1.125g.
After 7.6 days (two half lives) it's 0.5625g.
But in another half-life, 11.4 days would have passed. More than 11. In 11.4 days it would be 0.28125g.

in other words, after 3.8n days it's 2.25 times 1/2ⁿ

so after n days it's 2.25 times 1/2^n/3.8 …

so you need to find 2^11/3.8 ​

 

[rakeru]

Hi!

So I was looking up similar questions previously and it turns out that one person said the same thing as you.

I tried it first with others just in case and it worked. :D

I did: 2.25g × 1/2^11/3.8

And I got 0.302537977g.

The question had also told me to give the answer in three significant figures so: 0.303g, right?

Uhm, I hope it works! I'm going to submit my answer now.. :) Thank you so much!

 

[tiny-tim]

hi rakeru!

I did: 2.25g × 1/2^11/3.8

And I got 0.302537977g.

The question had also told me to give the answer in three significant figures so: 0.303g, right?

yup, looks fine!

 

[rakeru]

Yup! I got it right :) Thank you.