Radioactive substances decay by first-order kinetics. How many years would be required for a sample containing strontium-90 to decrease to 58.32% of its initial activity? The half-life of strontium-90 is 2.88e1 years.
k = 0.693/t1/2
Substitute k into the below.
ln(No/N) = kt
No = 100
N = 58.32
K from above.
Solve for t in years.
To find out the number of years required for a sample containing strontium-90 to decrease to 58.32% of its initial activity, we can use the concept of half-life and first-order kinetics.
First, let's understand the concept of half-life. The half-life is the amount of time it takes for half of the radioactive substance to decay. In this case, the half-life of strontium-90 is given as 2.88e1 years, which means that after every 2.88e1 years, the activity of strontium-90 reduces to half of its initial value.
Now, we need to find out how many half-lives it would take for the activity to decrease to 58.32% of its initial value. To do that, we can use the formula:
N = N0 * (1/2)^(t / t1/2)
Where:
N - Final activity as a fraction of initial activity (58.32% or 0.5832 in decimal form)
N0 - Initial activity
t - Number of years
t1/2 - Half-life
We need to solve this equation for t.
0.5832 = 1 * (1/2)^(t / 2.88e1)
Taking the logarithm (base 2) of both sides to remove the exponent:
log2(0.5832) = log2((1/2)^(t / 2.88e1))
Simplifying further:
log2(0.5832) = (t / 2.88e1) * log2(1/2)
Using the properties of logarithms, we can simplify the right-hand side:
log2(0.5832) = -(t / 2.88e1) * log2(2)
Since log2(2) = 1, the equation becomes:
log2(0.5832) = -t / 2.88e1
Now, we can solve this equation for t:
t = -2.88e1 * log2(0.5832)
Evaluating this expression using a calculator, we find that t is approximately 3.84e1 years.
Therefore, it would take approximately 3.84e1 years for a sample containing strontium-90 to decrease to 58.32% of its initial activity.