Chapter 45, Problem 65AP

(a)

To determine

The activity in curies due to each of the isotopes.

Answer to Problem 65AP

The activity in curies due to each of the isotopes is $330\mu \text{Ci}$, $16\mu \text{Ci}$ and $310\mu \text{Ci}$.

Explanation of Solution

Write the expression to find the mass of ${}^{\text{238}}\text{U}$ in the sample.

  $m_{{}^{\text{238}}\text{U}}=M-m_{{}^{\text{235}}\text{U}}-m_{{}^{\text{234}}\text{U}}$

Here, $m_{{}^{\text{238}}\text{U}}$ is the mass of ${}^{\text{238}}\text{U}$, $M$ is the total mass of the sample, $m_{{}^{\text{235}}\text{U}}$ is the mass of ${}^{\text{235}}\text{U}$, $m_{{}^{\text{234}}\text{U}}$ is the mass of ${}^{\text{234}}\text{U}$.

Substitute $1.00\text{kg}$ for $M$, $\left(1.00kg\right)\left(0.00720\right)$ for $m_{{}^{\text{235}}\text{U}}$ and $\left(1.00kg\right)\left(0.0000500\right)$ for ${}^{\text{234}}\text{U}$ to find the mass of ${}^{\text{238}}\text{U}$ in the sample.

  $\begin{array}{c}{m}_{{}^{\text{238}}\text{U}}=1.00\text{kg}-\left(1.00kg\right)\left(0.00720\right)-\left(1.00kg\right)\left(0.0000500\right)\\ =0.993\text{kg}\end{array}$

Conclusion:

Write the expression to find the number of nuclei in ${}^{\text{238}}\text{U}$.

  $N=m_{{}^{\text{238}}\text{U}}\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{ mol}}\right)\left(\frac{\text{1 mol}}{0.238\text{ kg}}\right)$

Substitute $0.993\text{kg}$ for $m_{{}^{\text{238}}\text{U}}$ to find the number of nuclei in ${}^{\text{238}}\text{U}$.

  $\begin{array}{c}N=0.993\text{kg}\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{ mol}}\right)\left(\frac{\text{1 mol}}{0.238\text{ kg}}\right)\\ =2.51\times {10}^{24}\text{nuclei}\end{array}$

Write the expression to find the activity of ${}^{\text{238}}\text{U}$

  $\begin{array}{c}R=\lambda N\\ =\frac{\ln 2}{T_{1/2}}N\end{array}$

Here, $T_{1/2}$ is the half-life, $\lambda$ is the decay constant.

Substitute $2.51\times {10}^{24}\text{nuclei}$ for $N$ and $4.47\times {10}^9\text{yr}$ for $T_{1/2}$ to find the activity of ${}^{\text{238}}\text{U}$.

  $\begin{array}{c}R=\frac{\ln 2}{4.47\times {10}^9\text{yr}}\left(2.51\times {10}^{24}\text{nuclei}\right)\left(\frac{1\text{ yr}}{3.16\times {10}^7\text{s}}\right)\left(\frac{1\text{ Ci}}{3.70\times {10}^{10}{\text{s}}^{-1}}\right)\\ =3.3\times {10}^{-4}\text{Ci}\frac{{10}^6\mu \text{Ci}}{1.0\text{Ci}}\\ =330\mu \text{Ci}\end{array}$

Write the expression to find the number of nuclei in ${}^{\text{235}}\text{U}$.

  $N=m_{{}^{\text{235}}\text{U}}\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{ mol}}\right)\left(\frac{\text{1 mol}}{0.238\text{ kg}}\right)$

Substitute $\left(1.00kg\right)\left(0.00720\right)$ for $m_{{}^{\text{235}}\text{U}}$ to find the number of nuclei in ${}^{\text{235}}\text{U}$.

  $\begin{array}{c}N=\left(1.00kg\right)\left(0.00720\right)\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{ mol}}\right)\left(\frac{\text{1 mol}}{0.238\text{ kg}}\right)\\ =1.84\times {10}^{22}\text{nuclei}\end{array}$

Write the expression to find the activity of ${}^{\text{235}}\text{U}$

  $\begin{array}{c}R=\lambda N\\ =\frac{\ln 2}{T_{1/2}}N\end{array}$

Here, $T_{1/2}$ is the half-life, $\lambda$ is the decay constant.

Substitute $1.84\times {10}^{22}\text{nuclei}$ for $N$ and $4.47\times {10}^9\text{yr}$ for $T_{1/2}$ to find the activity of ${}^{\text{235}}\text{U}$.

  $\begin{array}{c}R=\frac{\ln 2}{7.04\times {10}^8\text{ yr}}\left(1.84\times {10}^{22}\text{nuclei}\right)\left(\frac{1\text{ yr}}{3.16\times {10}^7\text{s}}\right)\left(\frac{1\text{ Ci}}{3.70\times {10}^{10}{\text{s}}^{-1}}\right)\\ =3.6\times {10}^{-5}\text{Ci}\frac{{10}^6\mu \text{Ci}}{1.0\text{Ci}}\\ =16\mu \text{Ci}\end{array}$

Write the expression to find the number of nuclei in ${}^{\text{234}}\text{U}$.

  $N=m_{{}^{\text{234}}\text{U}}\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{ mol}}\right)\left(\frac{\text{1 mol}}{0.238\text{ kg}}\right)$

Substitute $\left(1.00kg\right)\left(0.0000500\right)$ for $m_{{}^{\text{234}}\text{U}}$ to find the number of nuclei in ${}^{\text{234}}\text{U}$.

  $\begin{array}{c}N=\left(1.00kg\right)\left(0.0000500\right)\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{ mol}}\right)\left(\frac{\text{1 mol}}{0.238\text{ kg}}\right)\\ =1.29\times {10}^{20}\text{nuclei}\end{array}$

Write the expression to find the activity of ${}^{\text{234}}\text{U}$

  $\begin{array}{c}R=\lambda N\\ =\frac{\ln 2}{T_{1/2}}N\end{array}$

Here, $T_{1/2}$ is the half-life, $\lambda$ is the decay constant.

Substitute $1.29\times {10}^{20}\text{nuclei}$ for $N$ and $4.47\times {10}^9\text{yr}$ for $T_{1/2}$ to find the activity of ${}^{\text{234}}\text{U}$.

  $\begin{array}{c}R=\frac{\ln 2}{2.44\times {10}^5\text{ yr}}\left(1.29\times {10}^{20}\text{nuclei}\right)\left(\frac{1\text{ yr}}{3.16\times {10}^7\text{s}}\right)\left(\frac{1\text{ Ci}}{3.70\times {10}^{10}{\text{s}}^{-1}}\right)\\ =3.1\times {10}^{-4}\text{Ci}\frac{{10}^6\mu \text{Ci}}{1.0\text{Ci}}\\ =310\mu \text{Ci}\end{array}$

Thus, The activity in curies due to each of the isotopes is $330\mu \text{Ci}$, $16\mu \text{Ci}$ and $310\mu \text{Ci}$.

(b)

To determine

The fraction of the total activity due to isotopes.

Answer to Problem 65AP

The fraction of the total activity due to isotopes are $50\%$, $2.4\%$, $47\%$.

Explanation of Solution

the activity of ${}^{\text{238}}\text{U}$ is $330\mu \text{Ci}$, the activity of ${}^{\text{235}}\text{U}$ is $16\mu \text{Ci}$, the activity of ${}^{\text{234}}\text{U}$ is $310\mu \text{Ci}$.

Write the expression to find the total activity of the sample.

  $R_t=R_{{}^{\text{238}}\text{U}}+R_{{}^{\text{235}}\text{U}}+R_{{}^{\text{234}}\text{U}}$

Here, $R_t$ is the total activity of the sample, $R_{{}^{\text{238}}\text{U}}$ is the activity of ${}^{\text{238}}\text{U}$, $R_{{}^{\text{235}}\text{U}}$ is the activity of ${}^{\text{235}}\text{U}$ and $R_{{}^{\text{234}}\text{U}}$ is the activity of ${}^{\text{234}}\text{U}$.

Substitute $330\mu \text{Ci}$ for $R_{{}^{\text{238}}\text{U}}$, $16\mu \text{Ci}$ for $R_{{}^{\text{235}}\text{U}}$ and $310\mu \text{Ci}$ for $R_{{}^{\text{234}}\text{U}}$ to find $R_t$.

  $\begin{array}{c}{R}_t=330\mu \text{Ci}+16\mu \text{Ci}+310\mu \text{Ci}\\ =656\mu \text{Ci}\end{array}$

Conclusion:

Thus, the fraction of activity due to ${}^{\text{238}}\text{U}$ is $\frac{330\mu \text{Ci}}{656\mu \text{Ci}}\times 100\%=50\%$, the fraction of activity due to ${}^{\text{235}}\text{U}$ is $\frac{16\mu \text{Ci}}{656\mu \text{Ci}}\times 100\%=2.4\%$, the fraction of activity due to ${}^{\text{234}}\text{U}$ is $\frac{310\mu \text{Ci}}{656\mu \text{Ci}}\times 100\%=47\%$.

(c)

To determine

Whether activity of the sample is dangerous.

Answer to Problem 65AP

Yes. The activity of the sample is dangerous.

Explanation of Solution

The activity of ${}^{\text{238}}\text{U}$ is $330\mu \text{Ci}$, the activity of ${}^{\text{235}}\text{U}$ is $16\mu \text{Ci}$, the activity of ${}^{\text{234}}\text{U}$ is $310\mu \text{Ci}$.

Write the expression to find the total activity of the sample.

  $R_t=R_{{}^{\text{238}}\text{U}}+R_{{}^{\text{235}}\text{U}}+R_{{}^{\text{234}}\text{U}}$

Here, $R_t$ is the total activity of the sample, $R_{{}^{\text{238}}\text{U}}$ is the activity of ${}^{\text{238}}\text{U}$, $R_{{}^{\text{235}}\text{U}}$ is the activity of ${}^{\text{235}}\text{U}$ and $R_{{}^{\text{234}}\text{U}}$ is the activity of ${}^{\text{234}}\text{U}$.

Substitute $330\mu \text{Ci}$ for $R_{{}^{\text{238}}\text{U}}$, $16\mu \text{Ci}$ for $R_{{}^{\text{235}}\text{U}}$ and $310\mu \text{Ci}$ for $R_{{}^{\text{234}}\text{U}}$ to find $R_t$.

  $\begin{array}{c}{R}_t=330\mu \text{Ci}+16\mu \text{Ci}+310\mu \text{Ci}\\ =656\mu \text{Ci}\end{array}$

Conclusion:

Since, the activity of the sample is $656\mu \text{Ci}$; it is dangerous to human. But in the laboratories, it is used with taking enough precautions to reduce the human contact with the sample.