Chapter 30, Problem 39E

(a)

To determine

The number of ${}^{14}\text{C}$ atoms present in $1\text{kg}$ wood.

Answer to Problem 39E

The number of ${}^{14}\text{C}$ atoms present in $1\text{kg}$ wood is $5.52\times {10}^{13}$.

Explanation of Solution

Write the expression for the number of ${}^{12}\text{C}$ atoms present in $1\text{Kg}$ wood.

    $n=\left(\frac{N}{M}\right)m$        (I)

Here, $N$ is the number of ${}^{12}\text{C}$ nuclei per mole, $M$ is the molar mass of ${}^{12}\text{C}$ and $m$ is the mass of the carbon.

Write the expression for the number of ${}^{14}\text{C}$ present.

    $n^{\prime }=n\left(\frac{p}{P}\right)$        (II)

Here, $n^{\prime }$ is the number of ${}^{14}\text{C}$ present, $p$ is the number of ${}^{14}\text{C}$ present for $P$ number of ${}^{12}\text{C}$ atoms.

Conclusion:

Substitute $6.02\times {10}^{23}$ for $N$, $12\text{gm}$ for $M$ and $1\text{kg}$ for $m$ in expression (I).

    $\begin{array}{c}n=\left(\frac{6.02\times {10}^{23}}{12\text{gm}}\right)\left(1\text{kg}\right)\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ =\left(\frac{6.623\times {10}^{23}}{12\text{gm}}\right)\left(1000\text{gm}\right)\\ =5.01\times {10}^{25}\end{array}$

Substitute $1$ for $p$, ${10}^{12}$ for $P$ and $5.52\times {10}^{22}$ for $n$ in expression (II).

    $\begin{array}{c}{n}^{\prime }=\left(5.01\times {10}^{25}\right)\left(\frac{1}{{10}^{12}}\right)\\ =5.01\times {10}^{13}\end{array}$

Thus, the number of ${}^{14}\text{C}$ atoms present in $1\text{kg}$ wood is $5.52\times {10}^{13}$.

(b)

To determine

The number of the atoms that will decay in $1\text{hr}$.

Answer to Problem 39E

The number of the atoms that will decay in $1\text{hr}$ is $5.519\times {10}^{13}$.

Explanation of Solution

Write the expression for the number of atoms that will decay.

    $N=N_0{e}^{-\lambda t}$        (II)

Here, $N$ is the number of atom present after time $t$, $N_0$ is the initial number of atom present and $\lambda$ is the decay constant.

Conclusion:

Substitute $5.52\times {10}^{10}$ for $N_0$, $3.84\times {10}^{-12}{\text{s}}^{-1}$ for $\lambda$ and $1\text{hr}$ for $t$ in expression (II).

    $\begin{array}{l}N=\left(5.52\times {10}^{13}\right)e^{-\left(\left(3.84\times {10}^{-12}{\text{s}}^{-1}\right)\left(1\text{hr}\right)\right)}\\ =\left(5.52\times {10}^{13}\right)e^{-\left(\left(3.84\times {10}^{-12}{\text{s}}^{-1}\right)\left(3600\text{s}\right)\right)}\\ \approx 5.519\times {10}^{13}\end{array}$

Thus, the number of atoms that will decay in $1\text{hr}$ is $5.519\times {10}^{13}$.