Finding Mass of Isotope After Radioactive Decay by Half-Life

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the remaining mass of a radioactive isotope after a specified period, given its half-life. Participants explore methods for solving the problem and share their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about whether to treat the half-life as a normal number for calculations.
  • Another participant suggests a formula for calculating the remaining mass after a certain number of half-lives, indicating it involves the expression 2.25 times 1/2^(n/3.8).
  • A participant shares their calculation method and arrives at a mass of approximately 0.303g after applying the formula.
  • There is a confirmation from another participant that the calculated answer appears correct and meets the significant figures requirement.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculation and the final answer, but there is no explicit consensus on the initial confusion regarding the treatment of half-lives.

Contextual Notes

Some participants reference the need for clarity on the application of the half-life concept, and there is mention of the textbook's limitations in providing guidance on the problem.

Who May Find This Useful

This discussion may be useful for students studying nuclear chemistry or those encountering similar problems related to radioactive decay and half-life calculations.

rakeru
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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!
 
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welcome to pf!

hi rakeru! welcome to pf! :smile:
rakeru said:
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 1/2n/3.8 …

so you need to find 211/3.8 :wink:​
 
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 × 1/211/3.8

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!
 
hi rakeru! :smile:
rakeru said:
I did: 2.25g × 1/211/3.8

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! :smile:
 
Yup! I got it right :) Thank you.
 

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