The amount of radioactive substance after 38 days

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Homework Help Overview

The discussion revolves around a problem related to radioactive decay, specifically calculating the remaining amount of a radioactive substance after a specified duration, given its half-life. The original poster presents an equation for exponential decay and attempts to determine the correct answer without fully solving the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods for calculating the remaining amount of substance, including using half-life and exponential decay equations. Questions arise regarding notation, specifically the use of a raised dot for multiplication versus a decimal point.

Discussion Status

There is an ongoing dialogue about the correct interpretation of mathematical notation and the implications for solving the problem. Some participants provide insights into the decay constant and its derivation, while others express confusion about the notation used in the original post.

Contextual Notes

The problem involves specific values such as a half-life of 3.8 days and an initial amount of 10.24 grams. There is a focus on ensuring clarity in mathematical expressions and notation throughout the discussion.

Pushoam
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Homework Statement


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Homework Equations

The Attempt at a Solution



Is there anyway to choose the correct option without solving it?

## \lambda = \frac { 0.693 } { t_{\frac 1 2}} ##
[/B]
## N = N_0 e^{- 0.693 {\frac { t } { t_{\frac 1 2 }}}} ##

## N = 10.24 e^{- \frac { 0.693 * 38 } { 3.8} } = 10^{-2} ~ gm##So, the correct option is (b).

Is this correct?
 

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Why the raised dot between those digits? If to mean decimal point, that is not the way to show it.
Half-life, 3.8 days;
Want how much after 38 days;
Begin with 10.24 grams

If model is for exponential decay, then A=10.24*e^(-k*t).
Find k using half-life information.
(1/2)=1*e^(-k*3.8)
(1/2)=e^(-3.8k)
ln(1/2)=ln(e^(-3.8k))
ln(1/2)=-3.8k*1
k=-ln(1/2)/3.8
k=ln(2)/3.8
k=0.1824

Revise for the more specific decay equation:
A=10.24*e^(-0.1824t)

How much remaining after 38 days?
Let t=38 and evaluate A.
 
38 days is 10 half-lives. 2^10 = 1024. 10.24/1024 = 0.01.
 
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symbolipoint said:
Why the raised dot between those digits? If to mean decimal point, that is not the way to show it.
It is the symbol of multiplication. Please, see the keyboard for this. You, too, have used it in the following expression,
symbolipoint said:
A=10.24*e^(-k*t).

symbolipoint said:
k=ln(2)/3.8
I have written ##\lambda## instead of k.
The above is nothing but, k = 0.693 /t½.
 
Pushoam said:
It is the symbol of multiplication. Please, see the keyboard for this. You, too, have used it in the following expression,
I have written ##\lambda## instead of k.
The above is nothing but, k = 0.693 /t½.
I looked at the linked document you posted in the beginning post. There was a raised dot but that is to be used to signify multiplication. We want the character to be at the botton of the digits levels to read it as a decimal point. If I use a raised dot, I use it as a multiplication operation. That is the normal, common usage. The middle quoted passage shows the asterisk also often used for multiplication, and common and usually understood when using plain text to write an algebraic expression; it might appear as a raised dot, or it might appear exactly the asterisk. The 10.24 shows the decimal point, and the "three point eight" as the decimal number should be written as 3.8 (notice the dot at the bottom-level).
 
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