Chapter 29, Problem 41P

(a)

To determine

The decay constant for ${}^{14}\text{C}$.

Answer to Problem 41P

The decay constant for ${}^{14}\text{C}$ is $3.83\times {10}^{-12}{\text{ s}}^{-1}$.

Explanation of Solution

The activity of ${}^{14}\text{C}$ in a living sample is $0.25\text{Bq}/\text{g}$ of carbon. The half-life of  ${}^{14}\text{C}$ is $5730\text{ yr}$

Write the formula for half-life

$T_{1/2}=\tau \ln 2$                                                                                           (I)

Here, $T_{1/2}$ is the half-life and $\tau$ is the time constant.

Write the formula for the decay constant

$\lambda =\frac{1}{\tau }$                                                                                                  (II)

Here, $\lambda$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $\lambda$

$\lambda =\frac{\ln 2}{T_{1/2}}$

Substitute $5730\text{ yr}$ for $T_{1/2}$ in the above relation to find the value of $\lambda$. [Note: $1\text{yr}=3.156\times {10}^7\text{ s}$

$\begin{array}{l}\lambda =\frac{\ln 2}{5730\text{ yr}\times 3.156\times {10}^7\text{s}/\text{yr}}\\ \lambda =3.83\times {10}^{-12}{\text{ s}}^{-1}\end{array}$

Thus, the decay constant for ${}^{14}\text{C}$ is $3.83\times {10}^{-12}{\text{ s}}^{-1}$.

(b)

To determine

Calculate the number of ${}^{14}\text{C}$ atoms in $1.00\text{ g}$ of carbon.

Answer to Problem 41P

The number of ${}^{14}\text{C}$ atoms in $1.00\text{ g}$ of carbon is $6.5\times {10}^{10}\text{ atoms}$.

Explanation of Solution

One mole of carbon atom has $12.011\text{ g}$ and the relative abundance of ${}^{14}\text{C}$ is $1.3\times {10}^{-12}$. The Avogadro number $N_A=6.022\times {10}^{23}\text{atoms}/\text{mol}$.

The number of ${}^{14}\text{C}$ atoms in $1.00\text{ g}$ of carbon is the total number of nuclei in $1.00\text{ g}$ of carbon times the relative abundance of ${}^{14}\text{C}$.

$N=\frac{\text{mass}}{\text{mass per mol}}\times N_A\times \text{relative abundance}$

Substitute $6.022\times {10}^{23}\text{atoms}/\text{mol}$ for $N_A$, $1.3\times {10}^{-12}$ for the relative abundance, $1.00\text{ g}$ for the mass and $12.011\text{g}/\text{mol}$ for mass per mol.

$\begin{array}{l}N=\frac{1.00\text{ g}}{12.011\text{g}/\text{mol}}\times 6.022\times {10}^{23}\text{atoms}/\text{mol}\times 1.3\times {10}^{-12}\\ N=6.5\times {10}^{10}\text{ atoms}\end{array}$

Thus, the number of ${}^{14}\text{C}$ atoms in $1.00\text{ g}$ of carbon is $6.5\times {10}^{10}\text{ atoms}$.

(c)

To determine

The activity of ${}^{14}\text{C}$ per gram of carbon in a living sample.

Answer to Problem 41P

The activity of ${}^{14}\text{C}$ per gram of carbon in a living sample is $0.25\text{Bq}/\text{g}$.

Explanation of Solution

Write the formula for the activity

$R=\lambda N$                                                                                            (III)

Here,$\lambda$ is the decay constant and $N$ is the number of nuclei after time $t$.

Conclusion:

Substitute $3.83\times {10}^{-12}{\text{ s}}^{-1}$ for $\lambda$ and $6.5\times {10}^{10}\text{ atoms}$ for $N$ in equation (III) to find the value of $R$

$\begin{array}{l}R=3.83\times {10}^{-12}{\text{ s}}^{-1}\times 6.5\times {10}^{10}\text{ atoms}\\ R=0.25\text{ Bq}\end{array}$

Divide by $1.00\text{ g}$ on both sides to find the value of activity per gram

$\begin{array}{l}\frac{R}{1.00\text{ g}}=\frac{0.25\text{ Bq}}{1.00\text{ g}}\\ \frac{R}{1.00\text{ g}}=0.25\text{Bq}/\text{g}\end{array}$

Thus, the activity of ${}^{14}\text{C}$ per gram of carbon in a living sample is $0.25\text{Bq}/\text{g}$.