Chapter 29, Problem 26P

(a)

To determine

The number of $\text{C}$ sample.

Answer to Problem 26P

The number of $\text{C}$ sample is $3.54\times {10}^9$ nuclei.

Explanation of Solution

Given info: Time elapsed is 1730 days. Half-life of $\text{C}$ is $1.10\times {10}^4\text{days}$. Half-life of $\text{C}$ is 734 days. The combined activity of both the samples is 11.0 Bq.

Formula to calculate the combined activity is,

$R=\left(\frac{N_0}{t_{1/2}^{137}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]+\left(\frac{N_0}{t_{1/2}^{134}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]$      (I)

• $t_{1/2}^{137}$ is the half-life of $\text{C}$.
• $t_{1/2}^{134}$ is the half-life of $\text{C}$.
• $N_0$ is the initial amount of sample.
• t is the time elapsed.

From Equation (I), the initial amount of sample is,

$N_0=\frac{R}{\left(\frac{1}{t_{1/2}^{137}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]+\left(\frac{1}{t_{1/2}^{134}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]}$      (II)

Formula to calculate the number of $\text{C}$ sample is,

$N=N_0\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]$      (III)

Substitute Equation (II) in (III).

$N=\frac{R\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]}{\left(\frac{1}{t_{1/2}^{137}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]+\left(\frac{1}{t_{1/2}^{134}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]}$

Substitute 11.0 Bq for R, $1.10\times {10}^4\text{days}$ for $t_{1/2}^{137}$, 734 days for $t_{1/2}^{134}$ and 1730 days for t in the above equation to get N.

$\begin{array}{c}N=\frac{\left(11.0\text{Bq}\right)\exp \left[-\left(\frac{1730\text{days}}{1.10\times {10}^4\text{days}}\right)\ln 2\right]}{\left(\frac{1}{1.10\times {10}^4\text{days}}\right)\exp \left[-\left(\frac{1730\text{days}}{1.10\times {10}^4\text{days}}\right)\ln 2\right]+\left(\frac{1}{734\text{days}}\right)\exp \left[-\left(\frac{1730\text{days}}{734\text{days}}\right)\ln 2\right]}\\ =3.54\times {10}^9\end{array}$

Conclusion:

The number of $\text{C}$ sample is $3.54\times {10}^9$ nuclei.

(b)

To determine

The number of $\text{C}$ sample.

Answer to Problem 26P

The number of $\text{C}$ sample is $7.72\times {10}^8$ nuclei.

Explanation of Solution

Given info: Time elapsed is 1730 days. Half-life of $\text{C}$ is $1.10\times {10}^4\text{days}$. Half-life of $\text{C}$ is 734 days. The combined activity of both the samples is 11.0 Bq.

From Equation (II) of (a),

$N_0=\frac{R}{\left(\frac{1}{t_{1/2}^{137}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]+\left(\frac{1}{t_{1/2}^{134}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]}$

Formula to calculate the number of $\text{C}$ sample is,

$N=N_0\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]$      (IV)

Substitute Equation (II) in (IV).

$N=\frac{R\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]}{\left(\frac{1}{t_{1/2}^{137}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{137}}\right)\ln 2\right]+\left(\frac{1}{t_{1/2}^{134}}\right)\exp \left[-\left(\frac{t}{t_{1/2}^{134}}\right)\ln 2\right]}$

Substitute 11.0 Bq for R, $1.10\times {10}^4\text{days}$ for $t_{1/2}^{137}$, 734 days for $t_{1/2}^{134}$ and 1730 days for t in the above equation to get N.

$\begin{array}{c}N=\frac{\left(11.0\text{Bq}\right)\exp \left[-\left(\frac{1730\text{days}}{734\text{days}}\right)\ln 2\right]}{\left(\frac{1}{1.10\times {10}^4\text{days}}\right)\exp \left[-\left(\frac{1730\text{days}}{1.10\times {10}^4\text{days}}\right)\ln 2\right]+\left(\frac{1}{734\text{days}}\right)\exp \left[-\left(\frac{1730\text{days}}{734\text{days}}\right)\ln 2\right]}\\ =7.72\times {10}^8\end{array}$

Conclusion:

The number of $\text{C}$ sample is $7.72\times {10}^8$ nuclei.