A sample has a activity of 0.0065 Bq per gram of carbon. (a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq. (b) Evidence suggests that the value of 0.23 Bq might have been as much as 49% larger. Repeat part (a), taking into account this 49% increase.

To calculate the age of the sample, we need to use the concept of radioactive decay and the formula for radioactive decay:

Activity(t) = Activity(0) * e^(-λt)

Where:
- Activity(t) is the activity at time t
- Activity(0) is the initial activity at time t0 = 0
- λ is the decay constant
- t is the time passed

(a) Assuming a constant activity per gram of carbon in a living organism of 0.23 Bq, we can set up the equation as follows:

0.0065 Bq/g = 0.23 Bq/g * e^(-λt)

To solve for t, we need to isolate the exponential term on one side:

0.0065 / 0.23 = e^(-λt)

Now, we can take the natural logarithm (ln) of both sides to solve for t:

ln(0.0065 / 0.23) = -λt

To find the decay constant λ, we need to know the half-life of the radioactive isotope. Once we have λ, we can solve for t.

(b) Taking into account the 49% increase in the value of 0.23 Bq, the new activity is:

0.23 Bq * (1 + 0.49) = 0.23 Bq * 1.49 = 0.3427 Bq

Following the same procedure as in part (a), we can set up the equation:

0.0065 Bq/g = 0.3427 Bq/g * e^(-λt)

Now, we can solve for t by isolating the exponential term:

0.0065 / 0.3427 = e^(-λt)

Taking the natural logarithm (ln) of both sides gives:

ln(0.0065 / 0.3427) = -λt

Again, we need the decay constant λ, which depends on the half-life of the radioactive isotope, in order to solve for t.

To find the age of the sample, we can use the radioactive decay law which states that the activity of a radioactive substance decreases exponentially over time. The equation is given by:

A = A0 * e^(-λt)

Where:
A = current activity
A0 = initial activity
λ = decay constant
t = time

Let's solve part (a) first:

(a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq.

Given:
A = 0.0065 Bq/g (current activity)
A0 = 0.23 Bq (initial activity)

To find the decay constant (λ), we can use the relationship between the initial activity and decay constant:

A0 = A * e^(λt)
0.23 = 0.0065 * e^(λt)
e^(λt) = 0.23 / 0.0065
e^(λt) = 35.385

Now, let's solve for λt:

λt = ln(35.385)
λt ≈ 3.56

To find the age (t), we need the decay constant (λ):

λ = ln(2) / half-life

We can use the half-life of carbon-14, which is approximately 5730 years:

λ ≈ ln(2) / 5730
λ ≈ 1.21 x 10^(-4) year^(-1)

Now, let's solve for the age (t):

3.56 = 1.21 x 10^(-4) * t
t ≈ 29379 years

Therefore, the age of the sample is approximately 29379 years.

Now, let's move on to part (b):

(b) Evidence suggests that the value of 0.23 Bq might have been as much as 49% larger. Repeat part (a), taking into account this 49% increase.

To take into account the 49% increase, we need to calculate the new initial activity (A0_new):

A0_new = A0 * (1 + 0.49)
A0_new = 0.23 * (1 + 0.49)
A0_new = 0.23 * 1.49
A0_new ≈ 0.3427 Bq

Now, let's repeat the calculations using the new initial activity (A0_new):

A0 = 0.0065 Bq/g (current activity)
A0_new = 0.3427 Bq (new initial activity)

e^(λt) = A0_new / A
e^(λt) = 0.3427 / 0.0065
e^(λt) = 52.723

λt = ln(52.723)
λt ≈ 3.96

λ ≈ ln(2) / 5730
λ ≈ 1.21 x 10^(-4) year^(-1)

3.96 = 1.21 x 10^(-4) * t
t ≈ 32694 years

Therefore, taking into account the 49% increase in initial activity, the age of the sample would be approximately 32694 years.