Math 131

[Steven G]

I will work out a similar one for you.
The half-life of cesium-137 is 20 years. Suppose that we start with 100 grams of cesium-137 in a storage pool.
How many half-lives will it take for there to be 10 grams of cesium-137 in the storage pool?

The formula is A = A₀(.5)^t/20

Solve 10 = 100*(.5)^t/20 for t
.1 = (.5)^t/20
ln(.1) = (t/20)ln(.5)
20ln(.1)/ln(.5) = t

 

[Steven G]

sonra6, out of curiosity what makes you think that you can decide on how this forum runs? I am anxiously awaiting your answer.

1st you come on this forum and post a problem without sharing any work that is a requirement to receive help here. I guess that you were too bothered to read the posting guideline. Then you have the nerve to decide who gets to help you. What makes you feel so entitled.

I know that you seem to have no idea how to do these problems so i decided to work one out for you. The more I think about that decision the more I realized that I should not have. I will leave the post up in case some more deserving student will benefit my it.

 

[JeffM]

The half-life of cesium-137 is 30 years. Suppose that we start with 90 grams of cesium-137 in a storage pool.

How many half-lives will it take for there to be 10 grams of cesium-137 in the storage pool? (Round your answer to two decimal places.)
9.46 half-lives << this answer is right

How many years is that? (Round your answer to one decimal place.)
? yr

Who says that answer is correct?

If you already know the answer, why ask?

 

[Steven G]

Ok I’m just asking a question for help, just like everyone else is....

No, others ask a question and show their work. They want help with their problem. You just want an answer.

 

[HallsofIvy]

Half of 90 is 45. Half of 45 is 22.5. Half of 22.5 is 11.25. Half of 11.25 is 5.625 which is less than 10 so it will take slightly more than 3 "half lives". To get "two decimal" accuracy you need to use logarithms. A "half life" divides by two so 90, after x half lives is reduced to \(\displaystyle 90(1/2)^x= 10\)
\(\displaystyle (1/2)^x= 10/90= 1/9\)
\(\displaystyle log((1/2)^x)= x log(1/2)= log(1/9)\)
\(\displaystyle x= \frac{log(1/9)}{log(1/2)}= \frac{-log(9)}{-log(2)}= \frac{log(9)}{log(2)}\)