posted by Angie on .
Cesium-137 has a half-life of 30 years. Suppose a lab stored a 30-mCi sample n 1973. How much will be left in 2063?
Use the half life to calculate the k in
k = 0.693/(t1/2).
Then use k in the following equation.
ln(Co/C) = kt
Co = 30 mCi
Solve for C. The answer will be in mCi.
k = from above
t = years from dates listed.
Post your work if you get stuck.
what does k stand for?
k is the proportionality constant for the equation:
ln(Co/C) = kt and it is evaluated from k = 0.693/t(1/2). In this case, its
k=0.693/30 years = ??
Yes, k = 0.0231 years^-1. Now put that into the ln formula I provided and solve for C following the instructions I gave you.
why is it to the -1?
Are you putting me on?
1/x = x^-1.
so 0.693/30 = 0.0231 years^-1. It's just a unit and the -1 means years is in the denominator. It's the same thing as saying wavenumber = 1/5 cm = 0.2 cm^-1 or in words it is 0.2 reciprocal centimeters.