Chapter 20, Problem 88AP

Interpretation Introduction

Interpretation:

The amount, activity and the probability of ${}^{27}\text{Mg}$ after a certain time period are to be calculated.

Concept introduction:

The half-life of the reaction is the period of time required for the concentration of a reactant to decrease to one-half of its initial value. It is written as t_1/2. Thus, the half-life of a reaction is the time required for the reactant concentration to decrease from [N]₀ to [N]_0/2.

The amount of a particular radioactive isotope left after time t is given as:

$\text{ln}\frac{N_t}{N_o}\text{=}-kt$

Here, $k$

is the rate constant for the radioactive decay, $N_t$ is the amount of it left after time t, and $N_0$ is the initial amount of the sample.

Nearly all radioactive decays are of first order and the rate constant is given as:

$k\text{ = }\frac{0.693}{t_{1/2}}$

Here, $t_{1/2}$

is the half -life of the radioactive substance.

The rate of decay for a particular isotope is given by the rate law as:

$\text{Rate = }k\text{N}$

Here, $k$

is the rate constant for the radioactive decay and $\text{N}$ is the number of atoms in the sample. Rate is also called as activity of the isotope.

Answer to Problem 88AP

Solution:

$\text{4}\text{.71 }\times {\text{ 10}}^{\text{11}}\text{ Mg nuclei remains}$

$\begin{array}{l}\text{At, t}=0,\text{ activity = 0}\text{.138 Ci}\\ \text{At, t}=30\text{ min, activity = 0}\text{.0155 Ci }\end{array}$.

Probability is $\text{1}\text{.22 }\times {\text{ 10}}^{-3}$

Explanation of Solution

a) Number of ${}^{27}\text{Mg}$

nuclei left after 30 min.

Half -life of ${}^{27}\text{Mg}$

$=\text{ 9}\text{.50 min}$

Initial amount of ${}^{27}\text{Mg}$ present $=\text{ 4}\text{.20 }\times {\text{ 10}}^{\text{12}}$

${}^{27}\text{Mg}$

nuclei

Time (t) $=\text{ 30 min}$

It is also known that the rate constant for the radioactive decay is given as follows:

$k\text{ = }\frac{0.693}{t_{1/2}}$

Substitute the value of half-life in the above expression, we get rate constant,

$\begin{array}{c}k\text{ = }\left(\frac{0.693}{\text{9}\text{.50 min}}\right)\\ =0.0729{\text{ min}}^{-\text{1}}\end{array}$

Now, the number of Mg present after $30\text{ min}$

is calculated by the expression as follows:

$\text{ln}\frac{N_t}{N_o}=-kt$

Here, $N_o$ is initial rate of disintegration or original amount of radioactive substance while $N_t$

is the final rate of disintegration or the amount of radioactive substance left after time t.

Substitute the values of $N_o$, $k$, and $t$ in the above expression,

$\begin{array}{c}\text{ln}\frac{N_t}{\text{4}\text{.20 }\times {\text{10}}^{\text{12}}}=-\text{ (0}{\text{.0729 min}}^{-\text{1}})\times (30\text{ min)}\\ \left(\frac{N_t}{\text{4}\text{.20 }\times {\text{10}}^{\text{12}}}\right)={\text{e}}^{-\text{ (0}{\text{.0729 min}}^{-\text{1}})\times (30\text{ min)}}\\ \left(\frac{N_t}{\text{4}\text{.20 }\times {\text{10}}^{\text{12}}}\right)={\text{e}}^{-2.187}\\ \left(\frac{N_t}{\text{4}\text{.20 }\times {\text{10}}^{\text{12}}}\right)=\text{0}\text{.1225}\end{array}$

On solving further,

$N_t=\text{4}\text{.71 }\times {\text{ 10}}^{\text{11}}\text{ Mg nuclei}$

Thus, the number of ${}^{27}\text{Mg}$ nuclei remaining after $\text{30 min}$

is $\text{4}\text{.71 }\times {\text{ 10}}^{\text{11}}$.

b) The ${}^{27}\text{Mg}$ activities at $t=\text{0 min}$ and $t=30\text{ min}$.

At first, convert the rate constant unit into ${\sec }^{-1}$

It is known that one minute is equal to $60\text{ sec}$.

Thus, the conversion factor is $\frac{\text{1 min}}{60\text{ sec}}$.

The above conversion factor of the rate constant can be written as follows:

$\left(\text{0}\text{.0729 }\frac{1}{\cancel{\min }}\times \frac{1\cancel{\text{min}}}{\text{60 sec}}\right)=1.22\times {\text{10}}^{-3}{\text{sec}}^{-\text{1}}$

Now, at $\text{t = 0 min}$,

The activity of $\text{Mg nuclei}$ is given as follows:

$\text{Activity = }k\text{N}$

Substitute the values of $k$ and $\text{N}$ in the above expression,

$\begin{array}{c}\text{Activity}=\text{(1}\text{.22 }\times {\text{ 10}}^{\text{-3}}{\sec }^{-1}\text{)}\times \text{(4}\text{.20 }\times {\text{ 10}}^{\text{12}}\text{ nuclei)}\\ \text{Activity}=\text{5}\text{.12}\times {\text{10}}^9\text{ decays/sec }\end{array}$

Now, by the definition of 1 curie, $3.7\times {\text{ 10}}^{\text{10}}$

disintegration occurs per second.

Thus, the conversion factor is $\frac{\text{1 Ci}}{\text{3}\text{.7 }\times {\text{ 10}}^{\text{10}}\text{ disintegration/sec}}$

Hence, by using the above conversion factor, the rate can be converted to curie as follows:

$\left(\text{5}\text{.12 }\times {\text{ 10}}^{\text{9}}\cancel{\text{decays/sec}}\times \frac{1\text{ Ci}}{\text{3}\text{.70 }\times {\text{ 10}}^{\text{10}}\cancel{\text{decays/sec}}}\right)=0.138\text{ Ci}$

Thus, the radioactivity of $\text{Mg nuclei}$

at $t=\text{0 min}$

is $0.138\text{ Ci}$.

Similarly, at $t=\text{30 min}$

The activity of $\text{Mg nuclei}$ is given as follows:

$\text{Activity = }k\text{N}$

Substitute the values $k$ and $\text{N}$ in the above expression,

$\begin{array}{c}\text{Activity}=\text{(1}\text{.22}\times {\text{10}}^{-\text{3}}{\sec }^{-1}\text{)}\times \text{(4}\text{.71}\times {\text{10}}^{\text{11}}\text{ nuclei)}\\ \text{Activity}=\text{5}\text{.75}\times {\text{10}}^8\text{ decays/sec }\end{array}$

Now, by the definition of 1 curie, $3.7\times {\text{ 10}}^{\text{10}}$

disintegration occurs per second.

Thus, the conversion factor is $\frac{\text{1 Ci}}{\text{3}\text{.7 }\times {\text{ 10}}^{\text{10}}\text{ disintegration/sec}}$.

Hence, by using the above conversion factor, the rate can be converted to curie as:

$\text{5}\text{.75 }\times {\text{ 10}}^8\cancel{\text{decays/sec}}\times \frac{1\text{ Ci}}{\text{3}\text{.70 }\times {\text{ 10}}^{\text{10}}\cancel{\text{decays/sec}}}=0.0155\text{ Ci}$

Thus, the radioactivity of $\text{Mg nuclei}$

at $\text{t = 30 min}$

is $0.0155\text{ Ci}$.

c) The probability that ${}^{\text{27}}\text{Mg}$ nucleus decays during a 1 sec.

From part (b) results, it is known that the rate constant of $\text{Mg nuclei}$

is $1.22\times {\text{10}}^{-3}{\text{ sec}}^{-\text{1}}$ and its half-life is $=\text{9}\text{.50 min}$. As it can be seen that the half-life is much greater than $\text{1 sec}$.

Thus, it can be said that $1.22\times {\text{ 10}}^{-3}\text{ decays}$

are occurring per second.

Hence, the probability of $\text{Mg nuclei}$

that decays in $\text{1 sec}$ is $1.22\times {\text{ 10}}^{-3}$.