Half-life with formula help?

Question

My teacher only ever gave us half life problems that worked out evenly and never bothered to teach us the formula or how it works. So now I some problems that I have to work out and I need to use the formula... please help! Here are the problems:

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July 16, 1945. What percentage of the strontium-90 (t1/2 = 28.8 years) originally produced by that explosion still remained as of July 16, 2000?

A patient is administered 20 mg of iodine-131. How much of this isotope will remain in the body after 20 days if the half-life for iodine-131 is 8 days?

Thanks!

Answer 1

The formula is ln (Ao/A) = kt, where k = 0.693/ (t1/2)

A is the concentration at time t, Ao is the initial concentration

As for your questions:

1. Time change is 55 years, and k = 0.693/28.8 or 0.0241
ln (Ao/A) = 0.0241 * 55 = 1.326.
Ao/A = e^1.326 = 3.77.
A/Ao = 1/3.77 = 0.266 or 26.6% of the original amount remains.

2. Time change is 20 days, and k = 0.693/8 = 0.0866.
ln(Ao/A) = 0.0866 * 20 = 1.732.
20/A = e^1.732 = 5.65.
A = 20/5.65 = 3.54 mg.

Answer 2

Just use the equation

Yt=Yo(1/2)^(t / t-half)

where

Yt = the value after time t
Yo =the original value
t = the time
t-half = the half life

So for the first part

t=2000-1945=55
t-half=28.9
Yo=100%
Yt=?

Yt=100%(1/2)^(55/28.9)

For the second part
Yo=20x10^-3 g
t= 20 days
t-half= 8 days
Yt=?

Yt=(20x10^-3g)(1/2)^(20/8)

After you plug in these values, you should get the answer.