# Homework Help: Half life and decay differential EQ problem

1. Sep 14, 2014

### Sneakatone

1. The problem statement, all variables and given/known data
The Shroud of Turin, which shows the negative image of the body of a man who appears to have been crucified, is believed by many to be the burial shroud of Jesus of Nazareth. See the figure below. In 1988 the Vatican granted permission to have the shroud carbon-dated. Three independent scientific laboratories analyzed the cloth and concluded that the shroud was approximately 660 years old,† an age consistent with its historical appearance. Using this age, determine what percentage of the original amount of C-14 remained in the cloth as of 1988. (The half-life of C-14 is approximately 5730 years. Round your answer to the nearest percent.)

2. Relevant equations
P(t)=[P_o]e^(kt) where P(t) is the percentage and t is time in years.
[P_o] is the initial percentage of C-14.

3. The attempt at a solution
50=[P_o]e^(k5730)

I decided to take a the second half like which is P(11460)=25 (i dont know if that is do able)
25=[P_o]e^(k11460)

with the two equations I solved for [P_o] and set them equal to each other to solve for k. which is k= -(log(2))/5730.

when I plug it back in I get [P_o]=100 which I dont feel is right.

Also I dont know what to do with the 660 years given.

and got 94% but it is still incorrect.

2. Sep 14, 2014

### Ray Vickson

I don't understand what you are doing. The general decay equation is
$A(t) = A_0 e^{-kt}$, where $A_0$ is the initial amount of C14 present, $A(t)$ is the amount present at time $t$ (years) and $k > 0$ is the decay constant (in 1/yr). You need $1/2 = e^{-k 5730}$ (half of the initial amount), so you can get $k$.

BTW: of course $P_0 = 100$ is right, because you are considering $P(t)$ to be the percentage of C14 present as compared with its amount at $t = 0$; in other words, at $t=0$ the amount of C14 present is 100% of the amount of C14 present.

3. Sep 14, 2014

### Sneakatone

I see now, I had slightly the wrong equation. I used the one for population instead. I solved for k=ln(2)/5730

then did A(660)=e^(-ln(2)/5730*660) = .92 => 92%.

thanks alot!