Chapter 29, Problem 84P

(a)

To determine

The mass of ${}^{40}\text{K}$ found in a broccoli stalk containing $300\text{ mg}$ of potassium.

Answer to Problem 84P

The mass of ${}^{40}\text{K}$ found in a broccoli stalk containing $300\text{ mg}$ of potassium is $3.6\times {10}^{-8}\text{ kg}$.

Explanation of Solution

In natural occurring of potassium, $0.0117\%$ are radioactive ${}^{40}\text{K}$. The atomic mass of potassium is $39.0983\text{ u}$ and ${}^{40}\text{K}$ is $39.9639985\text{ u}$. The broccoli stalk contains a mass of $m=300\text{ mg}$, what will be the mass of radioactive ${}^{40}\text{K}$.

Write the formula for the number of nuclei

    $N=\frac{m{N}_A}{M}$                                                                                           (I)

Here, $N$ is the number of nuclei, $m$ is the mass, $N_A$ is the Avogadro’s number and $M$ is the molar mass.

Conclusion:

Divide the mass of potassium and ${}^{40}\text{K}$, and substitute in equation (I) to find the value of $m_{40}$

$\begin{array}{l}\frac{m_{40}}{m}=\frac{\raisebox{1ex}{$N_{40}{M}_{40}$}\!\left/ \!\raisebox{-1ex}{$N_{\text{A}}$}\right.}{\raisebox{1ex}{$NM$}\!\left/ \!\raisebox{-1ex}{$N_{\text{A}}$}\right.}=\frac{N_{40}}{N}\times \frac{M_{40}}{M}\\ m_{40}=\frac{N_{40}}{N}\times \frac{M_{40}}{M}\times m\end{array}$

Substitute $0.0117\%$ for $\frac{N_{40}}{N}$, $300\text{ mg}$ for $m$  $39.0983\text{ u}$ for $M$  and $39.9639985\text{ u}$ for $M_{40}$ in above equation to find the value of $m_{40}$

$\begin{array}{l}{m}_{40}=\left(0.0117\right)\times \frac{39.9639985\text{ u}}{39.0983\text{ u}}\times 300\times {10}^{-3}\text{ kg}\\ m_{40}=3.6\times {10}^{-8}\text{ kg}\end{array}$

Thus, the mass of ${}^{40}\text{K}$ found in a broccoli stalk containing $300\text{ mg}$ of potassium is $3.6\times {10}^{-8}\text{ kg}$.

(b)

To determine

The activity of broccoli stalk due to ${}^{40}\text{K}$.

Answer to Problem 84P

The activity of broccoli stalk due to ${}^{40}\text{K}$ is $9.5\text{ Bq}$.

Explanation of Solution

The mass of ${}^{40}\text{K}$ found in a broccoli stalk containing $300\text{ mg}$ of potassium is $m=3.6\times {10}^{-8}\text{ kg}$,  and its molar mass of is $M=39.9639985\text{g}/\text{mol}$ and its half-life is $T_{1/2}=1.248\times {10}^9\text{ yr}$ . The Avogadro’s number is $N_A=6.022\times {10}^{23}\text{nuclei}/\text{mol}$.

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 activity

    $R=\frac{N}{\tau }$                                                                                                (III)

Here, $R$ is the activity.

Conclusion:

Substitute equation (I) and (II) in (III) to solve for $R_0$

$\begin{array}{l}R=\frac{N}{\tau }\\ R=\frac{m{N}_A\ln 2}{T_{1/2}M}\end{array}$

Substitute $3.6\times {10}^{-8}\text{ kg}$ for $m$, $6.022\times {10}^{23}\text{nuclei}/\text{mol}$ for $N_A$, $1.248\times {10}^9\text{ yr}$ for $T_{1/2}$ and $39.9639985\text{g}/\text{mol}$ for $M$ in the above equation to find the value of $R$ . [Note: $1\text{ yr}=3.156\times {10}^7\text{ s}$]

$\begin{array}{l}R=\frac{3.6\times {10}^{-8}\text{ kg}\times 6.022\times {10}^{23}\text{nuclei}/\text{mol}\times \ln 2}{39.9639985\text{g}/\text{mol}\times 1.248\times {10}^9\text{ yr}\times 3.156\times {10}^7\text{s}/\text{yr}}\\ R=9.5\text{ Bq}\end{array}$

Thus, the activity of broccoli stalk due to ${}^{40}\text{K}$ is $9.5\text{ Bq}$.