The activity of a sample of Ba-137m has decreased by 75% from its initial activity after 5.10 minutes. What is the half-life of Ba-137m? What fraction of activity remains after 6.25 minutes?
(A=Aoe (-kt); k = .693/t1/2)
ln(No/N) = kt
No = 100
N = 100-75 = 25
k = ?
t = 5.10 minutes.
Solve for k, then
k = 0.693/(t1/2>
To determine the half-life of Ba-137m, we can use the formula for exponential decay: A = A₀e^(-kt), where A is the current activity, A₀ is the initial activity, k is the decay constant, and t is the time elapsed.
In this case, we are given that the activity has decreased by 75% after 5.10 minutes, which means the remaining activity (A) is 25% of the initial activity (A₀). Thus, we can write the equation as:
0.25A₀ = A₀e^(-k * 5.10)
To find the value of k, we can rearrange the equation:
e^(-k * 5.10) = 0.25
Taking the natural logarithm (ln) of both sides:
-k * 5.10 = ln(0.25)
Dividing both sides by 5.10:
k = -ln(0.25) / 5.10
Using the given relationship k = 0.693 / t1/2 (where t1/2 is the half-life), we can solve for the half-life:
0.693 / t1/2 = -ln(0.25) / 5.10
Solving for t1/2:
t1/2 = - (0.693 * 5.10) / ln(0.25)
Calculating this expression:
t1/2 ≈ 3.37 minutes
Therefore, the half-life of Ba-137m is approximately 3.37 minutes.
To determine the fraction of activity remaining after 6.25 minutes, we can use the formula A = A₀e^(-kt) again. Plugging in the values:
A = A₀e^(-k * 6.25)
Since no information is given regarding the initial activity (A₀), we cannot calculate the exact fraction remaining.
To find the half-life of Ba-137m, we can use the formula A = A0e^(-kt), where A is the final activity, A0 is the initial activity, k is the decay constant, and t is the time elapsed.
Given that the activity of Ba-137m has decreased by 75% after 5.10 minutes, we can write:
A = 0.25A0 (since the activity has decreased by 75% which means only 25% is remaining)
Since the initial activity A0 cancels out in the formula, we can rearrange it to solve for k:
0.25 = e^(-5.10k)
Taking the natural logarithm (ln) on both sides, we get:
ln(0.25) = -5.10k
Now we can solve for k:
k = ln(0.25) / -5.10 = -0.693 / t1/2
Rearranging the last equation, we can solve for the half-life t1/2:
t1/2 = -0.693 / k
Substituting the value of k we found earlier, we get:
t1/2 = -0.693 / (ln(0.25) / -5.10)
t1/2 ≈ 2.89 minutes
Therefore, the half-life of Ba-137m is approximately 2.89 minutes.
Now, to find the fraction of activity remaining after 6.25 minutes, we can use the formula A = A0e^(-kt) again.
Given t = 6.25 minutes and the initial activity A0, we want to find A.
Using the formula A = A0e^(-kt) and substituting the known values, we have:
A = A0e^(-kt)
A = A0e^(-k * 6.25)
We can now substitute the value of k we found earlier:
A = A0e^(-1.353 * 6.25)
A ≈ A0 * 0.064
Therefore, the fraction of activity remaining after 6.25 minutes is approximately 0.064, or 6.4%.