What percent of the original sample of strontium-90 (half-life = 28.9 years) will remain after the following time periods?after 18 years?
after 47.5 years?
after 1008 years?
k = 0.693/t1/2
ln(No/N) = kt
Substitute into equation 1 and solve for k, the substitute into equation 2 and solve for N. No will be 18, 47.5, 1008 years.
.....it's actually asking for the % left after so many years
So N = amount left.
% = (N/No)*100 = ?
To determine the percent of the original sample of strontium-90 that will remain after a given time period, we can use the concept of half-life.
The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, the half-life of strontium-90 is given as 28.9 years.
To calculate the percent remaining after a specific time period, we need to consider how many half-lives have passed.
1. After 18 years:
To find the number of half-lives that have occurred in 18 years, divide the time period (18 years) by the half-life (28.9 years).
Number of half-lives = 18 years ÷ 28.9 years = 0.622 half-lives
To calculate the percent remaining, we need to raise 0.5 (representing the half that decays each half-life) to the power of the number of half-lives and multiply by 100.
Percent remaining = (0.5)^(number of half-lives) × 100
Now, substitute the calculated number of half-lives (0.622) into the equation:
Percent remaining after 18 years = (0.5)^(0.622) × 100 = 77.99%
Therefore, approximately 77.99% of the original sample of strontium-90 will remain after 18 years.
2. After 47.5 years:
Using the same process as above, divide 47.5 years by the half-life of 28.9 years to find the number of half-lives:
Number of half-lives = 47.5 years ÷ 28.9 years = 1.642 half-lives
Apply the formula:
Percent remaining after 47.5 years = (0.5)^(1.642) × 100 = 67.41%
Approximately 67.41% of the original sample of strontium-90 will remain after 47.5 years.
3. After 1008 years:
Divide 1008 years by the half-life of 28.9 years:
Number of half-lives = 1008 years ÷ 28.9 years = 34.8 half-lives
Using the formula:
Percent remaining after 1008 years = (0.5)^(34.8) × 100 = 0.012%
After 1008 years, approximately 0.012% of the original sample of strontium-90 will remain.