Chapter 4, Problem 4.43P

(a)

To determine

The radioactivity of lead.

Answer to Problem 4.43P

The activity of lead is $1.96\times {10}^{-11}\text{mCi}$ .

Explanation of Solution

Given:

The half- life of lead is $22\text{yr}$ and the original quantity of lead is $100\text{mCi}$ .

Concept used:

When the nucleus of a radioactive substance decays in other nuclei, then this decay is known as the radioactive decay of nucleus. The nucleus that is going to decay is known as the parent nuclei and a newly formed nucleus is known as the daughter nuclei.

Write the expression for the radioactivity.

  $N=N_0{e}^{-\lambda T}$ ....... (1)

Here, $N$ is the number of newly formed nuclei, $N_0$ is the original nucleus, $\lambda$ is the decay constant and $T$ is the time of decay.

When the nuclei of the radioactive substance decay to half of its initial value, the time required in decaying is known as half-life.

Write the expression for the half-life.

  $T_{\frac{1}{2}}=\frac{0.693}{\lambda }$

Rearrange the above expression for the decay constant.

  $\lambda =\frac{0.693}{T_{\frac{1}{2}}}$ ........ (2)

Here, $T_{\frac{1}{2}}$ is the half-life of radioactive element.

Write the expression for the activity of the nucleus.

  $R=\lambda N$ ....... (3)

Here, $R$ is the activity of the nucleus.

Calculation:

Substitute $22\text{yr}$ for $T_{\frac{1}{2}}$ in equation (2).

  $\begin{array}{c}\lambda =\frac{0.693}{22\text{yr}}\\ =\frac{0.693}{22\text{yr}\left(\frac{31536000\text{s}}{1\text{yr}}\right)}\\ =9.98\times {10}^{-10}\text{decay/s}\end{array}$

Substitute $10\text{yr}$ for $T$ , $100\text{mCi}$ for $N_0$ and $9.98\times {10}^{-10}\text{decay/s}$ for $\lambda$ in equation (1).

  $\begin{array}{c}N=\left(100\text{mCi}\right)e^{-\left(9.98\times {10}^{-10}\text{decay/s}\right)\left(10\text{yr}\right)}\\ =\left(100\text{mCi}\right)e^{-\left(9.98\times {10}^{-10}\text{decay/s}\right)\left(10\text{yr}\left(\frac{31536000\text{s}}{1\text{yr}}\right)\right)}\\ =72.99\text{mCi}\end{array}$

Substitute $72.99\text{mCi}$ for $N$ and $9.98\times {10}^{-10}\text{decay/s}$ for $\lambda$ in equation (3).

  $\begin{array}{c}R=\left(9.98\times {10}^{-10}\text{decay/s}\right)\left(72.99\text{mCi}\right)\\ =\left(9.98\times {10}^{-10}\text{decay/s}\left(\frac{0.27\times {10}^{-7}\text{mCi}}{1\text{decay/s}}\right)\right)\left(72.99\text{mCi}\right)\\ =1.96\times {10}^{-11}\text{mCi}\end{array}$

Conclusion:

Thus, the activity of lead is $1.96\times {10}^{-11}\text{mCi}$ .

(b)

To determine

The radioactivity of polonium.

Answer to Problem 4.43P

The activity of polonium is $2.94\times {10}^{-14}\text{mCi}$ .

Explanation of Solution

Given:

The half- life of polonium is $138\text{days}$ and the original quantity of lead is $100\text{mCi}$ .

Calculation:

Substitute $138\text{days}$ for $T_{\frac{1}{2}}$ in equation (2).

  $\begin{array}{c}\lambda =\frac{0.693}{138\text{days}}\\ =\frac{0.693}{138\text{days}\left(\frac{86400\text{s}}{1\text{days}}\right)}\\ =5.81\times {10}^{-8}\text{decay/s}\end{array}$

Substitute $10\text{yr}$ for $T$ , $100\text{mCi}$ for $N_0$ and $5.81\times {10}^{-8}\text{decay/s}$ for $\lambda$ in equation (1).

  $\begin{array}{c}N=\left(100\text{mCi}\right)e^{-\left(5.81\times {10}^{-8}\text{decay/s}\right)\left(10\text{yr}\right)}\\ =\left(100\text{mCi}\right)e^{-\left(5.81\times {10}^{-8}\text{decay/s}\right)\left(10\text{yr}\left(\frac{31536000\text{s}}{1\text{yr}}\right)\right)}\\ =1.09\times {10}^{-6}\text{mCi}\end{array}$

Substitute $1.09\times {10}^{-6}\text{mCi}$ for $N$ and $5.81\times {10}^{-8}\text{decay/s}$ for $\lambda$ in equation (3).

  $\begin{array}{c}R=\left(5.81\times {10}^{-8}\text{decay/s}\right)\left(1.09\times {10}^{-6}\text{mCi}\right)\\ =\left(5.81\times {10}^{-8}\text{decay/s}\left(\frac{0.27\times {10}^{-7}\text{mCi}}{1\text{decay/s}}\right)\right)\left(1.09\times {10}^{-6}\text{mCi}\right)\\ =2.94\times {10}^{-14}\text{mCi}\end{array}$

Conclusion:

Thus, the activity of polonium is $2.94\times {10}^{-14}\text{mCi}$ .

(c)

To determine

Weight of the polonium.

Answer to Problem 4.43P

The weight of the polonium is $3.8\times {10}^{-31}\text{gm}$ .

Explanation of Solution

Given:

The activity of polonium is $1.09\times {10}^{-6}\text{mCi}$ , decay constant is $5.81\times {10}^{-8}\text{decay/s}$ and the half- life of polonium is $138\text{days}$ .

Concept used:

Write the expression for the number of nuclei in terms of weight.

  $N=\frac{m}{M}{N}_A$

Rearrange the above expression for the given mass of the element.

  $m=\frac{NM}{N_A}$ ........ (4)

Here, $m$ is the given mass of an element, $M$ is the molar mass and $N_A$ is the Avogadro number.

Calculation:

Substitute $1.09\times {10}^{-6}\text{mCi}$ for $N$ , $210$ for $M$ and $6.022\times {10}^{23}$ for $N_A$ in equation (4).

  $\begin{array}{c}m=\frac{\left(1.09\times {10}^{-6}\text{mCi}\right)\left(210\right)}{\left(6.022\times {10}^{23}\right)}\\ =3.8\times {10}^{-31}\text{gm}\end{array}$

Conclusion:

Thus, the weight of the polonium is $3.8\times {10}^{-31}\text{gm}$ .

(d)

To determine

The weight of lead at the time of sealing and the weight of lead after decay.

Answer to Problem 4.43P

1. The weight of the lead at the time of sealing is $3487.21\times {10}^{-20}\text{gm}$ .
2. The weight of the lead after decay is $6.83\times {10}^{-36}\text{gm}$ .

Explanation of Solution

Calculation:

Substitute $100\text{mCi}$ for $N$ , $210$ for $M$ and $6.022\times {10}^{23}$ for $N_A$ in equation (4).

  $\begin{array}{c}m=\frac{\left(100\text{mCi}\right)\left(210\right)}{\left(6.022\times {10}^{23}\right)}\\ =3487.21\times {10}^{-20}\text{gm}\end{array}$

Substitute $1.96\times {10}^{-11}\text{mCi}$ for $N$ , $210$ for $M$ and $6.022\times {10}^{23}$ for $N_A$ in equation (4).

  $\begin{array}{c}m=\frac{\left(1.96\times {10}^{-11}\text{mCi}\right)\left(210\right)}{\left(6.022\times {10}^{23}\right)}\\ =6.83\times {10}^{-36}\text{gm}\end{array}$