Chapter 31, Problem 49P

(a)

To determine

Daughter nucleus.

Answer to Problem 49P

The daughter nucleus is $\text{P}\text{o}$

Explanation of Solution

Given:

Radon gas, $\text{R}\text{n}$

Calculation:

The decay reaction is given by:

  $\text{R}\text{n}\to \text{P}\text{o}+\mathrm{\alpha }\left(\text{H}\text{e}\right)$

Hence, the daughter nucleus is $\text{P}\text{o}$

Conclusion:

Hence, the daughter nucleus is $\text{P}\text{o}$

(b)

To determine

The half-life of daughter nucleus.

Answer to Problem 49P

The half life of the daughter nucleus $\text{P}\text{o}$ is 3.1 min.

Explanation of Solution

Given:

Radon gas, $\text{R}\text{n}$

Calculation:

The daughter nucleus further undergoes a reaction as follows:

  $\text{P}\text{o}\to \text{P}\text{b}+\text{H}\text{e}$

So, the daughter nucleus is radioactive.

And the half life of the daughter nucleus $\text{P}\text{o}$ is 3.1 min.

Conclusion:

Hence, the half life of the daughter nucleus $\text{P}\text{o}$ is 3.1 min.

(c)

To determine

Whether the daughter nucleus is a noble gas or chemically reacting.

Answer to Problem 49P

The daughter nucleus is chemically reactive.

Explanation of Solution

Given:

Radon gas, $\text{R}\text{n}$

Calculation:

The decay reaction of the daughter nucleus is given by:

  $\text{P}\text{b}\to \text{B}\text{i}+\mathrm{e}+\overline{{\mathrm{\nu }}_{\mathrm{e}}}$

So, the daughter nucleus is chemically reactive.

Conclusion:

The daughter nucleus is chemically reactive.

(d)

To determine

The activity after 1 month later.

Answer to Problem 49P

The activity after 1 month is $\mathrm{R}=2.4\times {10}^4\text{Bq}$

Explanation of Solution

Given:

Radon gas, $\text{R}\text{n}$

Calculation:

The half-life of radon can be given as:

  $\begin{array}{l}{\mathrm{t}}_{\frac{1}{2}}=3.8\text{ days}\\ {\mathrm{t}}_{\frac{1}{2}}=3.8\times 24\times 3600\text{ s}\\ {\mathrm{t}}_{\frac{1}{2}}=3.28\times {10}^5\text{s}\end{array}$

The number of radon atoms in 1.0 ng is given as:

  $\begin{array}{l}{\mathrm{N}}_0=\frac{\left(1.0\times {10}^{-9}\text{g}\right)\left(6.022\times {10}^{23}\right)}{\left(222\text{g}\right)}\\ {\mathrm{N}}_0=2.71\times {10}^{12}\end{array}$

Now, the activity equation is given by:

  $\begin{array}{l}{\mathrm{R}}_0=\mathrm{\lambda }\mathrm{N}=\left(\frac{0.693}{{\mathrm{t}}_{\frac{1}{2}}}\right)\left(2.71\times {10}^{12}\right)\\ {\mathrm{R}}_0=\left(\frac{0.693}{3.28\times {10}^5\text{s}}\right)\left(2.71\times {10}^{12}\right)\\ {\mathrm{R}}_0=5.73\times {10}^6\text{Bq}\end{array}$

After one month, the activity is given by:

  $\begin{array}{l}\mathrm{R}={\mathrm{R}}_0{\mathrm{e}}^{-\left(\frac{0.693\left(30\right)}{3.8}\right)}\\ \mathrm{R}=\left(5.73\times {10}^6\text{Bq}\right){\mathrm{e}}^{-\left(\frac{0.693\left(30\right)}{3.8}\right)}\\ \mathrm{R}=2.4\times {10}^4\text{Bq}\end{array}$

Conclusion:

Hence, the activity after 1 month is $\mathrm{R}=2.4\times {10}^4\text{Bq}$