Calculating Radioactive Decay: Finding the Time for 80% Decay

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Homework Help Overview

The problem involves calculating the time required for 80% decay of a radioactive substance, given its half-life of 194 days. The context is centered around radioactive decay and the mathematical relationships governing it.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of logarithmic equations and exponential decay formulas to determine the time for 80% decay. There are attempts to clarify the setup of the equations and the validity of the calculations presented.

Discussion Status

The discussion includes various approaches to the problem, with some participants suggesting double-checking the original equations used. There is an indication of productive dialogue, with different methods being explored without reaching a consensus on the final answer.

Contextual Notes

Some participants question the initial equation setup and the interpretation of the decay process, indicating a need for clarity on the mathematical relationships involved.

SETHOSCOTT
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Homework Statement


The half-life of a radioactive substance is 194 days. How many days will it take for 80% of the substance to decay?


Homework Equations





The Attempt at a Solution


I had to make 1/2n=1/5. I did this by finding log25=n. After this, I needed to only multiply by the number of days it took the half-life to occur, which was 194, and got 451 (approx.) as a reasonable answer.
 
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There is absolutely enough information there. I could tell you the answer right now (but I won't, that's what you get to do ;-)

I'd advise you to go back and double-check that equation, though.
 
In one "half life", T, (1/2)C= CekT. The Cs cancel, and you solve that for k. After you know that you can solve 0.8C= Cekt for t.

But you don't need to use "e". If T is the half life then [itex]C(t)= C(0)(1/2)^{t/T}[/itex] because every time "T" days, you multiply by1/2. Solve [itex].8C= C(1/2)^{t/194}[/itex].
 
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SETHOSCOTT said:

The Attempt at a Solution


I had to make 1/2n=1/5. I did this by finding log25=n. After this, I needed to only multiply by the number of days it took the half-life to occur, which was 194, and got 451 (approx.) as a reasonable answer.
Looks good...
 
=) omg! Thanks!
 

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