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 = ??

0.0231?

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.

ok thanks

To find out how much Cesium-137 will be left in 2063, we can use the concept of half-life.

The half-life of Cesium-137 is 30 years, which means that every 30 years, the amount of Cesium-137 is reduced by half.

Since the lab stored a 30-mCi sample in 1973, we need to find out how many half-lives have passed from 1973 to 2063.

To do that, we calculate the number of half-lives that have passed by dividing the total time elapsed (2063 - 1973 = 90 years) by the half-life of Cesium-137 (30 years):

Number of half-lives = (2063 - 1973) / 30 = 90 / 30 = 3 half-lives

Therefore, three half-lives have passed from 1973 to 2063.

To find out how much will be left in 2063, we need to calculate the remaining amount after three half-lives:

Remaining amount = Initial amount / (2^(Number of half-lives))

Remaining amount = 30 mCi / (2^3) = 30 mCi / 8 = 3.75 mCi

So, in 2063, there will be approximately 3.75 milliCuries (mCi) of Cesium-137 left.