If I 131 has a half-life of 8 days, what fraction will be left after a year?
k = 0.693/t1/2. Solve for k and substitute into the below equation.
ln(No/N) = kt.
No = 100 (or any number you choose)
N = Solve for this.
k = from above
t = 365 days.
Solve for N, then 100/N = fraction remaining. Another way is to simply solve for No/N and take the inverse of that number. You should get the same answer either way. Check my work.
1/131
To determine the fraction of I-131 that will be left after a year, we need to consider its half-life. The half-life of an isotope is the time it takes for half of the initial amount of the substance to decay.
In this case, the half-life of I-131 is 8 days. To find out how many half-lives occur in a year, we can divide the number of days in a year by the half-life:
365 days per year / 8 days per half-life
This gives us approximately 45.625 half-lives in a year. We can round this down to 45 for simplicity.
Now, we need to calculate the fraction left after 45 half-lives have occurred. Each half-life reduces the amount of I-131 by half. So, after 1 half-life, we have 1/2 (or 0.5) remaining. After 2 half-lives, we have 1/2 * 1/2 = 1/4 (or 0.25) remaining. And so on.
Using this pattern, we can calculate the fraction left after 45 half-lives:
(1/2)^45
Mathematically, this is equivalent to dividing 1 by 2 raised to the power of 45:
1 / (2^45) ≈ 3.55271 x 10^(-14)
Therefore, approximately 3.55271 x 10^(-14) fraction of I-131 will be left after a year.