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!
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!