The half-life of cesium-137 is 30 years. Suppose that we start with 50 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.)
Let n be the number of half-lives. After n half-lives, the amount of cesium-137 remaining is given by the formula:
Amount_remaining = Initial_amount * (1/2)^n
In this case, the initial amount is 50 grams and we want to find n when the amount remaining is 10 grams. So, we can set up the equation:
10 = 50 * (1/2)^n
Now we want to solve for n. We can divide both sides of the equation by 50:
10/50 = (1/2)^n
Simplifying the left-hand side:
0.2 = (1/2)^n
Now we can take the logarithm of both sides, which allows us to bring down the exponent:
log(0.2) = log((1/2)^n)
Using the property of logarithms that log(a^b) = b*log(a):
log(0.2) = n * log(1/2)
Now we can solve for n by dividing both sides by log(1/2):
n = log(0.2) / log(1/2)
Using a calculator to find the logarithms and divide, we get:
n ≈ 2.32
So it takes approximately 2.32 half-lives for there to be 10 grams of cesium-137 remaining in the storage pool.