Radioactive Dating: Calculating Half-Life & Remaining Mass

Click For Summary

Discussion Overview

The discussion revolves around the calculations involved in radioactive dating, specifically focusing on determining the remaining mass of a substance after a certain period, using half-life and decay equations. Participants are attempting to solve a homework problem related to the decay of lead-210 (Pb210) and its conversion to mercury-206 (Hg206).

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their calculations for the remaining mass of Pb210 after 17.5 years, arriving at 4.36 grams, but questions the correctness of their answer.
  • Another participant points out that the question asks for the amount of Hg206, not Pb210, suggesting a misinterpretation of the problem.
  • A participant proposes that the remaining mass of Hg206 should be calculated as 7.5 grams minus the amount of Pb210 remaining, leading to a value of 3.14 grams, but expresses uncertainty as they expected 3.09 grams.
  • Questions arise about the mass distribution between Pb210 and Hg206 after a full half-life, with one participant suggesting that after 22.5 years, the masses would be equal.
  • Another participant notes the density difference between Pb and Hg, expressing confusion about how this affects the calculations.
  • There is a question regarding the decay mode of Pb210, with a participant asserting it undergoes alpha decay.
  • Further discussion includes the implications of alpha decay on mass changes, with one participant suggesting that the mass of the nucleus changes by 4 due to this decay mode.
  • A later reply critiques the problem as potentially misleading, suggesting that the structure of the question may lead to confusion and misinterpretation.
  • Another participant expresses a preference for non-multiple choice tests, indicating that the reasoning process should be valued over simply selecting an answer.

Areas of Agreement / Disagreement

Participants express various interpretations of the problem and its calculations, with no consensus reached on the correct approach or final answer. Disagreements exist regarding the interpretation of the question and the implications of decay on mass.

Contextual Notes

Participants highlight potential traps in the problem's wording and the impact of mass changes due to decay, indicating that assumptions about mass distribution and decay processes may not be fully resolved.

toothpaste666
Messages
517
Reaction score
20

Homework Statement



chemproblem.png

Homework Equations


N= (N_0) e^(-kT)
k = ln(2)/(T_1/2)

The Attempt at a Solution


N0 = 7.5g
T_1/2 = 22.3 years
T = 17.5 years

k = ln(2)/22.3 = .031

N = 7.5 e^(-.031 * 17.5)
N = 4.36 g

So I chose A, but the correct answer was B. can't figure out where I went wrong
 
Physics news on Phys.org
You calculated the amount of Pb210 after 17.5a, you were asked for the amount of Hg206
 
then wouldn't it be 7.5 - 4.36 = 3.14? because you start with 7.5 and end up with 4.36 and the rest was converted to Hg206. I am still doing something wrong though because I am supposed to end up with 3.09
 
Do you think that after 22.5years, 7.5g of Pb will become 3.75g of Pb and 3.75g of Hg?
Does Hg have the same density as Pb?
 
Pb is slightly more dense. I'm not sure how to include this in the calculation though. if at the beginning all 7.5 grams is Pb and after the time has elapsed 4.36 of the 7.5 is Pb then I have a hard time understanding how Hg won't be the remainder of the 7.5 grams
 
What is the decay mode?
 
i am pretty sure it undergoes alpha decay
 
I would look for atomic numbers of isotopes mentioned.

But in general, if it is alpha decay, you can assume mass of the nucleus changes by 4.
 
  • Like
Likes   Reactions: toothpaste666
Bit of a trick question IMHO. Firstly you misread it, but I'd call that a set trap.:devil: Then you made your calculation and, once you realize the first trap the result is clearly nearer one figure, D, than another one. But the second trap :devil: is there is a 2% difference due to the mass change (which was given in the problem).

It would be fair to ask questions as a quiz when it is possible to get answers by principles or ballpark reasoning. It would be fair to ask the question with no answers given, since you can eliminate some answers by ballpark, but you cannot choose between B and D without calculation. But they have set you up by two answers meant to confirm you in mistakes!

These people are evil. :devil: :devil:
 
  • Like
Likes   Reactions: toothpaste666
  • #10
yeah i like better when tests are not multiple choice and have partial credit. That way they can grade you on your method of solving instead of just running it through a scantron =\

so the reason its B) is because of the mass that becomes the alpha particle?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
Replies
4
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K