What is the half-life of Iodine-131?

[Coco12]

Homework Statement

0ops.. The title should read solving logs..
Mod note: Fixed.

the original dosage contains 280 MBq of Iodine-131. If none is lost from the body, then after 6 hr there are 274 MBq of iodine-131. What is the half life of iodine I-131?

Homework Equations

Logcx=y

The Attempt at a Solution

I tried to put
274=280(1/2)^t/6 in and solve for t. But now I know it can't be right because it is not halving every 6 hrs.. However I don't know how you would figure it out. Can someone give me a hint?

 

[Chestermiller]

The concentration varies with time according to the equation $I=I_0e^{-kt}$, where k is a constant that you need to determine from the data: -kt = ln(I/I₀). Once k is known, find the time at which I is equal to half of I₀: -kt_1/2=ln(1/2)

 

[Ray Vickson]

Homework Statement

0ops.. The title should read solving logs..
Mod note: Fixed.

the original dosage contains 280 MBq of Iodine-131. If none is lost from the body, then after 6 hr there are 274 MBq of iodine-131. What is the half life of iodine I-131?

Homework Equations

Logcx=y

The Attempt at a Solution

I tried to put
274=280(1/2)^t/6 in and solve for t. But now I know it can't be right because it is not halving every 6 hrs.. However I don't know how you would figure it out. Can someone give me a hint?

Put $D = D_0 (1/2)^{t/c},$ where $D =$ dosage after t hours, $D_0 =$ initial dosage (at t = 0) and $c$ is a constant. You are given $D_0 = 280$, and at t = 6 you have $D = 274$. You have enough information to find $c$.

 

[Dick]

Homework Statement

0ops.. The title should read solving logs..
Mod note: Fixed.

the original dosage contains 280 MBq of Iodine-131. If none is lost from the body, then after 6 hr there are 274 MBq of iodine-131. What is the half life of iodine I-131?

Homework Equations

Logcx=y

The Attempt at a Solution

I tried to put
274=280(1/2)^t/6 in and solve for t. But now I know it can't be right because it is not halving every 6 hrs.. However I don't know how you would figure it out. Can someone give me a hint?

The radioactivity will follow the decay law 280*(1/2)^(t/k). k is the half life. You want to solve for k, not put k=6.

 

[Coco12]

Put $D = D_0 (1/2)^{t/c},$ where $D =$ dosage after t hours, $D_0 =$ initial dosage (at t = 0) and $c$ is a constant. You are given $D_0 = 280$, and at t = 6 you have $D = 274$. You have enough information to find $c$.

Thank you!