Decay Problem

[Jason76]

A bone of bison is found to contain 40% of the original amount of \(\displaystyle C^{14}\) Taking into consideration that the half life of \(\displaystyle C^{14}\) is \(\displaystyle 5,700\) years, calculate how long ago did the animal die.

\(\displaystyle k = \dfrac{\ln(2)}{5,700}\)

\(\displaystyle y = y_{0}e^{-kt}\)


\(\displaystyle (.35)(1.0) = (1.0)e^{-(1.216 X 10^{-04})t}\) ( y is .35 percent of original) (\(\displaystyle y_{0}\) is the original is 100 percent)

\(\displaystyle y = (.35)e^{-(1.216 X 10^{-04})t}\)

 

[daon2]

A bone of bison is found to contain 40% of the original amount of \(\displaystyle C^{14}\) Taking into consideration that the half life of \(\displaystyle C^{14}\) is \(\displaystyle 5,700\) years, calculate how long ago did the animal die.

\(\displaystyle k = \dfrac{\ln(2)}{5,700}\)

\(\displaystyle y = y_{0}e^{-kt}\)


\(\displaystyle (.35)(1.0) = (1.0)e^{-(1.216 X 10^{-04})t}\)

\(\displaystyle y = (.35)e^{-(1.216 X 10^{-04})t}\)

You are tasked with finding how long it takes for 60% of the carbon to break down. That is, find \(\displaystyle T\) such that \(\displaystyle y(T) = 0.4y(0)\).

\(\displaystyle y(T) = 0.4y(0)\iff e^{-kT} = 0.4\iff T = -\ln(0.4)/k = \ln(2.5)/k\)

 

[Jason76]

(Typo with regards to (.35) value in first post on thread.)

A bone of bison is found to contain 40% of the original amount of \(\displaystyle C^{14}\) Taking into consideration that the half life of \(\displaystyle C^{14}\) is \(\displaystyle 5,700\) years, calculate how long ago did the animal die.

\(\displaystyle k = \dfrac{\ln(2)}{5,700}\)

\(\displaystyle y = y_{0}e^{-kt}\)

\(\displaystyle y = \)(percentage of)\(\displaystyle (y_{0}) \) that is the present amount

\(\displaystyle y_{0} = \)(percentage of)\(\displaystyle (y_{0}) \) that is the intial amount (usually 100 percent of 1)

\(\displaystyle (.40)(1) = (100.0)(1)e^{-(1.216 X 10^{-04})t}\)

\(\displaystyle (.40) = e^{-(1.216 X 10^{-04})t}\)

\(\displaystyle \ln[(.40)] = \ln[e^{-(1.216 X 10^{-04})t}]\)

\(\displaystyle -0.916290731= -1.216 X 10^{-04}t\)

\(\displaystyle \dfrac{-0.916290731}{-1.216 X 10^{-04}} = t\)

 

[HallsofIvy]

(Typo with regards to (.35) value in first post on thread.)

A bone of bison is found to contain 40% of the original amount of \(\displaystyle C^{14}\) Taking into consideration that the half life of \(\displaystyle C^{14}\) is \(\displaystyle 5,700\) years, calculate how long ago did the animal die.

\(\displaystyle k = \dfrac{\ln(2)}{5,700}\)

\(\displaystyle y = y_{0}e^{-kt}\)

\(\displaystyle y = \)(percentage of)\(\displaystyle (y_{0}) \) that is the present amount

\(\displaystyle y_{0} = \)(percentage of)\(\displaystyle (y_{0}) \) that is the intial amount (usually 100 percent of 1)

\(\displaystyle (.40)(1) = (100.0)(1)e^{-(1.216 X 10^{-04})t}\)

\(\displaystyle (.40) = e^{-(1.216 X 10^{-04})t}\)

\(\displaystyle \ln[(.40)] = \ln[e^{-(1.216 X 10^{-04})t}]\)

\(\displaystyle -0.916290731= -1.216 X 10^{-04}t\)

\(\displaystyle \dfrac{-0.916290731}{-1.216 X 10^{-04}} = t\)

Why did you stop there? What is your final answer?

By the way, all exponentials are equivalent. Since this talks about "half life", I would have been inclined to use "1/2" instead of "e":
Saying that the half life of C¹⁴ is 5700 years means that every 5700 years, the amount is multiplied by 1/2. In "t" years there are t/5700 periods of 5700 years so the amount will be multiplied by 1/2 t/5700 times: if the initial amount was A then in t years it will be \(\displaystyle A(1/2)^{t/5700}\).

In order to have only 40% left, we must have \(\displaystyle A(1/2)^{t/5700}= 0.40A\). Divide by A to get \(\displaystyle (1/2)^{t/5700}= 0.4\). Solve that for t.