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%.