The half-life of oxygen-15 is 124 s. If a sample of oxygen-15 has an activity of 4800 Bq, how many minutes will elapse before it has an activity of 600 Bq?
6.2
k = 0.693/t1/2
Substitute k into the below.
ln(No/N) = kt
No = 4800
N = 600
k from above
Solve for t(which will be in seconds) and convert to minutes.
To determine the time it takes for a sample of oxygen-15 to reach a certain activity, we can use the concept of radioactive decay with the half-life. The half-life represents the time it takes for half of the radioactive substance to decay.
First, let's calculate the decay constant (λ) using the half-life:
λ = ln(2) / half-life
Given that the half-life of oxygen-15 is 124 seconds, we can calculate λ:
λ = ln(2) / 124
Next, we can use the decay equation to find the time required for the oxygen-15 sample to reach a specific activity:
N(t) = N0 × e^(-λt)
Where:
N(t) = Final activity
N0 = Initial activity
λ = Decay constant
t = Time
In this case, the initial activity (N0) is 4800 Bq, and the final activity (N(t)) is 600 Bq. We need to solve for t.
Let's set up the equation:
600 = 4800 × e^(-λt)
Now, we can solve for t by rearranging the equation:
t = (ln(600/4800)) / -λ
Let's substitute the value of λ we calculated earlier:
t = (ln(600/4800)) / -[ln(2) / 124]
Now, we can calculate the value of t:
t = (ln(600/4800)) / -[ln(2) / 124]
Using ln(600/4800) ≈ -0.847 and ln(2) ≈ 0.693:
t ≈ (-0.847) ÷ -[0.693 / 124]
t ≈ 10.72 minutes
Therefore, it will take approximately 10.72 minutes for the oxygen-15 sample to reach an activity of 600 Bq.