iodine 131 has a half life of 8 days. If 5 mg are stored for a week, how much is left? How many days does it take before only 1 mg remains?
An explanation for each step would be nice please. I want to learn how to do these problems for my upcoming exam.
every time 8 days pass, you multiply by 1/2. So, after t days, 8 days have passed t/8 times, and the amount left is
5*(1/2)^(t/8)
After 7 days, then, you have
5*0.5^(7/8) = 2.726mg
1mg is 1/5 of the starting amount. 1/5 is a bit less than 1/4 = (1/2)^2, so it should take a little more than 2 half-lives, or 16 days.
5(0.5)^(t/8) = 1
0.5^(t/8) = 0.2
t/8 = log0.2/log0.5
t = 8 log0.2/log0.5 = 18.575 days
To determine how much Iodine-131 is left after a week, we need to understand the concept of half-life. The half-life of a substance is the time it takes for half of the substance to decay or transform into another element.
In this case, the half-life of Iodine-131 is 8 days. This means that every 8 days, the initial amount of Iodine-131 will decrease by half.
Now let's solve the first part of the problem:
If we start with 5 mg of Iodine-131 and want to know how much is left after a week (7 days), we need to determine how many half-lives have passed within that time.
First, divide the number of days that have passed (7 days) by the half-life of Iodine-131 (8 days):
7 days / 8 days = 0.875
This means that approximately 0.875 half-lives have passed within a week.
To find out how much Iodine-131 is left, we multiply the initial amount (5 mg) by 0.5 raised to the power of the number of half-lives (0.875):
5 mg * (0.5)^0.875 ≈ 3.206 mg
So, after a week, approximately 3.206 mg of Iodine-131 remains.
Now, let's move on to the second part of the problem:
We want to determine how many days it takes until only 1 mg of Iodine-131 remains.
Using the same approach, we set up the equation:
5 mg * (0.5)^n = 1 mg
Here, 'n' represents the number of half-lives it will take to reach 1 mg. To solve for 'n', we need to isolate it by taking the logarithm of both sides of the equation:
log(5 mg * (0.5)^n) = log(1 mg)
Applying the logarithmic property, we can move 'n' out of the exponent:
log(5 mg) + log((0.5)^n) = log(1 mg)
Now, rearranging the equation:
log((0.5)^n) = log(1 mg) - log(5 mg)
Using the property that log(x^y) = y * log(x), we simplify the equation:
n * log(0.5) = log(1 mg) - log(5 mg)
Divide both sides by log(0.5):
n = (log(1 mg) - log(5 mg)) / log(0.5)
Calculating the right side of the equation using logarithmic properties:
n ≈ (0 - log(5 mg)) / log(0.5)
Taking the values:
n ≈ (-log(5)) / log(0.5)
Using a calculator, we can determine:
n ≈ 4.643
This means that it takes approximately 4.643 half-lives for only 1 mg of Iodine-131 to remain. Since each half-life is 8 days:
4.643 * 8 days ≈ 37.144 days
Therefore, it takes approximately 37.144 days for only 1 mg of Iodine-131 to remain.