Chapter 44, Problem 42P

(a)

To determine

The number of carbon atoms in the given sample.

Answer to Problem 42P

The number of carbon atoms in the given sample is $1.05\times {10}^{21}$.

Explanation of Solution

Write the equation to find the number of carbon atoms.

    $N=\frac{m}{M_C}{N}_{ava}$

Here, $N$ is the number of carbon atoms, $m$ is the given mass in gram, $M_C$ is the atomic mass of carbon, and $N_{ava}$ is the Avogadro number.

Conclusion:

Substitute $0.021 0\text{g}$ for $m$, $\text{12}\text{.0 }\text{g}/\text{mol}$ for $M_C$, $6.02\times {10}^{23}\text{atoms}/\text{mol}$ for $N_{ana}$ in the above equation to find $N$.

    $\begin{array}{c}N=\frac{0.021 0\text{g}}{\text{12}\text{.0 }\text{g}/\text{mol}}\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)\\ =1.05\times {10}^{21}\end{array}$

Therefore, the number of carbon atoms in the given sample is $1.05\times {10}^{21}$.

(b)

To determine

The number of carbon atoms in the given sample.

Answer to Problem 42P

The number of carbon-14 atoms in the given sample is $1.37\times {10}^9$.

Explanation of Solution

Write the equation to find the number of carbon-14 atoms.

    ${\left(N_0\right)}_{\text{C-14}}=\frac{N}{N_{stable}}$

Here, ${\left(N_0\right)}_{\text{C-14}}$ is the number of carbon-14 atoms and $N_{stable}$ is the number of stable atoms which one carbon-14 is found.

Conclusion:

Substitute $1.05\times {10}^{21}$ for $N$ and $7.70\times {10}^{11}$ for $N_{stable}$ in the above equation to find ${\left(N_0\right)}_{\text{C-14}}$.

    $\begin{array}{c}{\left(N_0\right)}_{\text{C-14}}=\frac{1.05\times {10}^{21}}{7.70\times {10}^{11}}\\ =1.37\times {10}^9\end{array}$

Therefore, the number of carbon-14 atoms in the given sample is $1.37\times {10}^9$.

(c)

To determine

The decay constant for carbon-14 in inverse seconds.

Answer to Problem 42P

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

Explanation of Solution

Write the equation to find the half-life time of carbon-14.

    ${\lambda }_{\text{C-14}}=\frac{\ln 2}{t_{1/2}}$

Here, ${\lambda }_{\text{C-14}}$ is the decay constant and $t_{1/2}$ is the half life time.

Conclusion:

Substitute $5730\text{yr}$ in the above equation to find ${\lambda }_{\text{C-14}}$.

    $\begin{array}{c}{\lambda }_{\text{C-14}}=\frac{\ln 2}{5730\text{yr}\left(\frac{3.16\times {10}^7\text{s}}{1\text{yr}}\right)}\\ =3.83\times {10}^{-12}{\text{ s}}^{-1}\end{array}$

Therefore, the decay constant for carbon-14 in inverse seconds is $3.83\times {10}^{-12}{\text{ s}}^{-1}$.

(d)

To determine

The number of initial number of decays in a week immediately after the death of species.

Answer to Problem 42P

The number of initial number of decays in a week immediately after the death of species is $3.17\times {10}^3\text{decays}/\text{week}$.

Explanation of Solution

Write the equation to find the decay rate.

    $R={\lambda }_{\text{C-14}}N$

Here, $R$ is the number of decay and $N$ is the number of C-14 atoms at time $t$.

Write the equation to find $N$.

  $N={\left(N_0\right)}_{\text{C-14}}{e}^{-\lambda t}$

Here, $t$ is the time.

Rewrite the equation for $R$ by substituting the above relation for $N$.

  $R={\lambda }_{\text{C-14}}\left({\left(N_0\right)}_{\text{C-14}}{e}^{-\lambda t}\right)$

To find the $R$ at initial time, rewrite the above relation by substituting $0\text{s}$ for $t$.

  $\begin{array}{c}R={\lambda }_{\text{C-14}}\left({\left(N_0\right)}_{\text{C-14}}{e}^{-\lambda \left(0\text{s}\right)}\right)\\ ={\lambda }_{\text{C-14}}{\left(N_0\right)}_{\text{C-14}}\end{array}$

Conclusion:

Substitute $3.83\times {10}^{-12}{\text{s}}^{-1}$ for ${\lambda }_{\text{C-14}}$ and $1.37\times {10}^9$ for ${\left(N_0\right)}_{\text{C-14}}$ in the above equation to find $R$

    $\begin{array}{c}R=\left(3.83\times {10}^{-12}{\text{s}}^{-1}\right)\left(1.37\times {10}^9\left(\frac{7\left(86 400\text{ s}\right)}{1\text{week}}\right)\right)\\ =3.17\times {10}^3\text{decays}/\text{week}\end{array}$

Therefore, the number of initial number of decays in a week immediately after the death of species is $3.17\times {10}^3\text{decays}/\text{week}$.

(e)

To determine

The new number of decays in a week in the current sample.

Answer to Problem 42P

The new number of decays in a week in the current sample is $951\text{decays}/\text{week}$.

Explanation of Solution

Write the equation to find the new number of decays in a week in the current sample.

    $R=\frac{n_{counts}}{\eta }$

Here, $n_{counts}$ is the number of counts accumulated in a week and $\eta$ is the counting efficiency.

Conclusion:

Substitute $837$ for $n_{counts}$ and $88\%$ for $\eta$ in the above equation to find $R$.

    $\begin{array}{c}R=\frac{837\text{decays}/\text{week}}{\left(\frac{88\%}{100\%}\right)}\\ =\frac{837\text{decays}/\text{week}}{0.88}\\ =951\text{decays}/\text{week}\end{array}$

Therefore, the new number of decays in a week in the current sample is $951\text{decays}/\text{week}$.

(f)

To determine

Lifetime of specimen in years using the results from part (c) and (d).

Answer to Problem 42P

Lifetime of specimen is $9.95\times {10}^3\text{yr}$.

Explanation of Solution

Write the equation to find the fraction of decay.

    $\frac{R}{R_0}=e^{-\lambda t}$

Apply logarithm on both sides.

  $\begin{array}{l}\ln \frac{R}{R_0}=\ln e^{-\lambda t}\\ \ln \frac{R}{R_0}=-\lambda t\end{array}$

Rewrite the above equation in terms of $t$.

  $t=\frac{-1}{\lambda }\ln \left(\frac{R}{R_0}\right)$

Conclusion:

Substitute $1.21\times {10}^{-4}{\text{ yr}}^{-1}$ for $\lambda$, $951\text{decays}/\text{week}$ for $R$, and $3.17\times {10}^3\text{decays}/\text{week}$ for $R_0$ in the above equation to find $t$.

  $\begin{array}{c}t=\frac{-1}{\left(1.21\times {10}^{-4}{\text{ yr}}^{-1}\right)}\ln \left(\frac{951\text{decays}/\text{week}}{3.17\times {10}^3\text{decays}/\text{week}}\right)\\ =-\left(8264.46\text{yr}\right)\left(\ln 0.3\right)\\ =-\left(8264.46\text{yr}\right)\left(-1.20\right)\\ =9.95\times {10}^3\text{yr}\end{array}$

Therefore, the lifetime of specimen is $9.95\times {10}^3\text{yr}$.