The activity of a radioactive material is 2.56 × 10–3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 × 10–5 Ci
Question
The activity of a radioactive material is 2.56 × 10–3 Ci.
If the half-life of the material is 5 days, after how many days will the activity become 2 × 10–5 Ci?
Solution
To solve this problem, we can use the formula for radioactive decay:
A = A₀ * (1/2)^(t / T)
Where: A is the final activity A₀ is the initial activity t is the time elapsed T is the half-life of the material
We are given that the initial activity (A₀) is 2.56 × 10–3 Ci and the final activity (A) is 2 × 10–5 Ci. We need to find the time elapsed (t).
Let's substitute the given values into the formula:
2 × 10–5 = 2.56 × 10–3 * (1/2)^(t / 5)
Now, let's solve for t. First, divide both sides of the equation by 2.56 × 10–3:
(2 × 10–5) / (2.56 × 10–3) = (1/2)^(t / 5)
Next, take the logarithm of both sides of the equation to solve for t:
log[(2 × 10–5) / (2.56 × 10–3)] = log[(1/2)^(t / 5)]
Using logarithm properties, we can bring down the exponent:
log[(2 × 10–5) / (2.56 × 10–3)] = (t / 5) * log(1/2)
Now, divide both sides of the equation by log(1/2):
[t / 5] = log[(2 × 10–5) / (2.56 × 10–3)] / log(1/2)
Finally, multiply both sides of the equation by 5 to solve for t:
t = 5 * [log[(2 × 10–5) / (2.56 × 10–3)] / log(1/2)]
Using a calculator, we can evaluate the right side of the equation to find the value of t.
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