Exponential Decay Question

jw1180

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Apr 19, 2021
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Hello!

I am working through a problem in a study guide that asks:

"The half-life of a certain isotope is 5.5 years. If there were 20 grams of one such isotope left after 22 years, what was the original weight?"

To set this up, there are 4 halvings (22 years/5.5 years) = t. The rate, r, is 0.5 (for half-life). The final mass, y, is 20g. Using y = a(1 - r)t:

20 = a(1 - 0.5)4 ; 20 = a(0.5)4 ; 20 = a(0.0625)

a = 320 grams original weight

Straightforward enough. To understand this process better, I tried using the equation A(t) = Ae-kt

where

A(t) = 0.5A
t = 5.5

solve for k:

0.5A = Ae-k(5.5)
0.5 = e-k(5.5)
ln(0.5) = -k(5.5)
-0.693 = -k(5.5)
k = 0.126

Returning to the earlier equation and using t = 22 with this calculated per-year rate, I tried:

20 = a(1 - 0.126)22 ; 20 = a(0.874)22 ; 20 = a(0.0517) ; a = 386.85. Similarly,

20 = 320(1 - 0.126)t ; 0.0625 = (1- 0.126)t ; 0.0625 = (0.874)t ; log0.874(0.0625) = 20.58 years.

Why don't these two methods match up? What assumption or math error am I making that I shouldn't?

Thanks!
 

Dr.Peterson

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Returning to the earlier equation and using t = 22 with this calculated per-year rate, I tried:

20 = a(1 - 0.126)22 ; 20 = a(0.874)22 ; 20 = a(0.0517) ; a = 386.85. Similarly,

20 = 320(1 - 0.126)t ; 0.0625 = (1- 0.126)t ; 0.0625 = (0.874)t ; log0.874(0.0625) = 20.58 years.

Why don't these two methods match up? What assumption or math error am I making that I shouldn't?
Why do you use 20 = a(1 - 0.126)22 ? That is a different equation than A(t) = Ae-kt , which is your "earlier equation", and its k has a different meaning than the k in A(t) = a(1 - k)t.

Try again using A(t) = Ae-kt.
 

Subhotosh Khan

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24,691
Hello!

I am working through a problem in a study guide that asks:

"The half-life of a certain isotope is 5.5 years. If there were 20 grams of one such isotope left after 22 years, what was the original weight?"

To set this up, there are 4 halvings (22 years/5.5 years) = t. The rate, r, is 0.5 (for half-life). The final mass, y, is 20g. Using y = a(1 - r)t:

20 = a(1 - 0.5)4 ; 20 = a(0.5)4 ; 20 = a(0.0625)

a = 320 grams original weight

Straightforward enough. To understand this process better, I tried using the equation A(t) = Ae-kt

where

A(t) = 0.5A
t = 5.5

solve for k:

0.5A = Ae-k(5.5)
0.5 = e-k(5.5)
ln(0.5) = -k(5.5)
-0.693 = -k(5.5)
k = 0.126

Returning to the earlier equation and using t = 22 with this calculated per-year rate, I tried:

20 = a(1 - 0.126)22 ; 20 = a(0.874)22 ; 20 = a(0.0517) ; a = 386.85. Similarly,

20 = 320(1 - 0.126)t ; 0.0625 = (1- 0.126)t ; 0.0625 = (0.874)t ; log0.874(0.0625) = 20.58 years.

Why don't these two methods match up? What assumption or math error am I making that I shouldn't?

Thanks!
Hint:

22 = 5.5 * 4 → (1/2)4 = 1/16 → 20 * 16 = 320
 

jw1180

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Apr 19, 2021
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Why do you use 20 = a(1 - 0.126)22 ? That is a different equation than A(t) = Ae-kt , which is your "earlier equation", and its k has a different meaning than the k in A(t) = a(1 - k)t.

Try again using A(t) = Ae-kt.
Thank you for your response, Dr.Peterson.

My rationale was that if I could solve the equation using the number of half-lives (4) and the half-life rate (-0.5), I could also solve the equation using the number of years (22) and the per-year rate (-0.126). That these results differ suggest to me that I am wrong about this assumption and I'm curious as to why. My sense of it is that is has to do with e using an infinite number of compoundings, whereas the half-life equation uses only four, but I'm not certain that's accurate.
 

jw1180

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Messages
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Hint:

22 = 5.5 * 4 → (1/2)4 = 1/16 → 20 * 16 = 320
Thank you, Subhotosh Khan.

I looks as though you are referring to my initial strategy of using the number of half-lives (4) and the half-life rate (-0.5) to calculate the starting amount. I would like to follow up on this and use a different method to arrive at the same result.

As a side note, the portion of your response following the 'arrow' did not seem to appear in your initial response, but does appear in the quoted string in this reply.
 

Subhotosh Khan

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Staff member
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Messages
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Thank you, Subhotosh Khan.

I looks as though you are referring to my initial strategy of using the number of half-lives (4) and the half-life rate (-0.5) to calculate the starting amount. I would like to follow up on this and use a different method to arrive at the same result.

As a side note, the portion of your response following the 'arrow' did not seem to appear in your initial response, but does appear in the quoted string in this reply.
Where did you get the equation:

y = a(1 - r)t ................................................... (1)

I guess you got that from "compound" interest equation. That compounding happens in discrete steps. In chemical decay - the process is continuous. So equation (1) does not work - you have to use "e" form.
 
Last edited:

Dr.Peterson

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Thank you for your response, Dr.Peterson.

My rationale was that if I could solve the equation using the number of half-lives (4) and the half-life rate (-0.5), I could also solve the equation using the number of years (22) and the per-year rate (-0.126). That these results differ suggest to me that I am wrong about this assumption and I'm curious as to why. My sense of it is that is has to do with e using an infinite number of compoundings, whereas the half-life equation uses only four, but I'm not certain that's accurate.
You can solve it using any of the three ways; but you can't confuse them with one another as you did!

If you use A(t) = Ae-kt, as you showed you get k = 0.126; then you have to use the same equation to get A:

20 = Ae-0.126*22
A = 20/e-0.126*22 = 20/e-2.772 = 319.8

If you use A(t) = a(1 - r)t, you have to do this:

0.5 = 1(1 - r)5.5
0.51/5.5 = 1 - r​
r =1 - 0.51/5.5 = 0.1184

to get the (effective annual) rate, and then

20 = a(1 - 0.1184)22
a = 20/(1 - 0.1184)22 = 319.93

Both results are a little off because I used rounded values; both would give 320 exactly if I used the entire value from the calculator.

Observe that k and r are different numbers.
 

jw1180

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Thank you Dr.Peterson and Subhotosh Kahn! This clears it up for me considerably.
 
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