In these practice problems, we will work on the **kinetics of radioactive reactions**. Most often, in chemistry at least, you will be asked to determine the * activity, quantity, the decay rate of radioactive isotopes, the time* required to drop the activity to a certain level, or apply those to techniques such as

**carbon dating**for calculating the age of ancient objects.

For all of these, there are **a few key concepts **and **main formulas** you will need to know or be able to use if they are provided.

**1)** Remember that **nuclear reactions follow the rules of first-order reactions**. You are going to need the integrated rate law of first-order reactions:

*where A _{t} is the current activity, A_{0} is the initial activity, k is the rate (decay) constant, and t is the time for which the decay is measured. *

It is important to remember that instead of activity (A), it can be the mass (m), moles, number of atoms (N), etc.

This formula is also used in carbon (or other elements) dating where we usually need to calculate *t* based on the initial activity (15.3 cpm/g C) and the half-life (5730 years) of ^{14}C.

**2)** Remember, the activity is the number of disintegrations per given time, and this, in turn, can be calculated using the differential rate law for first-order reactions:

*N here is the number of atoms (nuclei) and you will need to calculate in order to determine the activity of the nuclei.*

Essentially, activity (A) is the rate of radioactive processes.

**3)** Another key component is the half-life, which, remember is constant for first-order reactions, and is correlated to the rate constant of the process:

When solving half-life, or kinetics problems on nuclear reactions, there is often this initial lag when you don’t know where to start. A **great strategy** is to **write down the quantities given** in the problem, **and the unknown** next to them. After this, remember, **most questions are going to be around these equations**, so try to use them to find the link between the known and unknown in the problem.

#### Practice

Determine the half-life of a radioactive nuclide if its rate constant is 1.4 x 10^{-3} h^{-1}.

495 h

Radioactive reactions follow the first-order rate laws, and for first-order reactions, the half-life and the rate constant (decay constant) can be calculated using the following formula:

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{t_{{\rm{1/2}}}}}}\]

\[{t_{{\rm{1/2}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{\rm{k}}}\]

Therefore, for this example, the half-life would be:

\[{t_{{\rm{1/2}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{k}\, = \;\frac{{{\rm{0}}{\rm{.693}}}}{{1.4\, \times \,{{10}^{ – 3}}\,{{\rm{h}}^{ – 1}}}}\; = \,495\,{\rm{h}}\]

What is the rate (decay) constant for the beta decay of a radioactive nuclide if the half-life is 12.3 years?

0.0563 yrs^{-1}

Radioactive reactions follow the first-order rate laws, and for first-order reactions, the half-life and the rate constant (decay constant) can be calculated using the following formula:

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{t_{{\rm{1/2}}}}}}\]

Therefore,

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{12.3\,{\rm{years}}}}\; = \,0.0563\,{\rm{yr}}{{\rm{s}}^{ – 1}}\]

A radioactive isotope has a half-life of 14 days. How many grams of a 248-g sample remains after 56 days?

15.5 g

There are two ways of solving these types of problems: 1) by comparing the timeline with the half-life of the reaction, 2) by using the integrated rate law for first-order reactions.

Let’s start with the second one as it is a nice conceptual explanation.

**Approach 1)**

First, recall again that radioactivity follows the rules of **first-order reactions**, and for those, the **half-life does not depend on the concentration** (quantity for the radioactive process) of the starting material. In other words, the half-life is constant throughout the process, and during each half-life, the amount of the starting nuclide decreases by half.

It is always a good idea to **compare the time of the decay**, given in the problem, **to the half-life of the reaction**. Now, it may not be evident how many half-lives the given time takes, and therefore, it is always a good idea to **divide the time of the decay**, given in the problem, **by the half-life of the reaction**. If we divide the time (56 days) by the half-life (14 days), we will find out that the process takes 4 half-lives:

56 : 14 = 4

This means the mass of the nuclide is going to be halved 4 times:

Once you understand this concept, you do not have to draw the half-life segments, especially if the ratio of the time and the half-life is not a whole number. Instead, you can use the following formula to calculate the amount of the nuclide remaining (can also be the activity) after the given time:

The exponent indicates how many times the initial amount is halved. For example, if it was one time, we would simply multiply the initial amount by ½, if it is two times, then we multiply by ½ two times, and so forth.

Therefore, in this case, the exponent is 4 because the time corresponds to 4 half-lives.

\({\rm{Remaining}}\,{\rm{nuclide}}\,{\rm{ = }}\,{\rm{initial}}\, \times \,{\left( {\frac{1}{2}} \right)^4}\)

\({\rm{Remaining}}\,{\rm{nuclide}}\,{\rm{ = }}\,{\rm{248}}\, \times \,{\left( {\frac{1}{2}} \right)^4}\; = \;15.5\,{\rm{g}}\)

**Approach 2)**

The other option is to use the integrated rate law for first-order reactions:

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

The activity can be replaced by moles, number of atoms, or mass as long as we are talking about the same species. Therefore, we can write that:

\[\ln \;\frac{m}{{ {m_0}}}\; = \; – kt\]

Instead of the rate constant k, we will use the formula discussed in the previous expression to express it with the half life since it is given in the problem:

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{t_{{\rm{1/2}}}}}}\]

\[\ln \;\frac{m}{{ 248\,g}}\; = \; – \frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{14}}\,{\rm{days}}}}\,{\rm{ \times }}\,{\rm{56}}\,{\rm{days}}\]

\[\ln \;\frac{m}{{ 248\,g}}\; = \; – {\rm{2}}{\rm{.772}}\]

Next, we simplify the equation by raising to power which gets rid of the logarithm:

\[{e^{\ln \;\frac{m}{{ 248\,g}}}}\; = \;{e^{ – {\rm{2}}{\rm{.772}}}}\]

\[\frac{m}{{ 248\,g}}\; = \;0.0625368\]

m = 0.0625368 x 248 g = 15.5 g

A 2.34 g sample of radioactive isotope with a half-life of 4.6 days was obtained from a mine. How many grams of the isotope remains after 8.5 days?

0.65 g

The time does not add up as whole-number series of half-lives, so we will use the integrated law:

\[\ln \;\frac{m}{{ {m_0}}}\; = \; – kt\]

Instead of the rate constant k, we will use the formula discussed in the previous expression to express it with the half-life since it is given in the problem:

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{t_{{\rm{1/2}}}}}}\]

\[\ln \;\frac{m}{{ 2.34\,g}}\; = \; – \frac{{0.693}}{{4.6\,\cancel{{\rm{d}}}}} \times \,8.5\,\cancel{{\rm{d}}}\]

\[\ln \;\frac{m}{{ 2.34\,g}}\; = \; – 1.2805\]

\[\frac{m}{{ 2.34\,g}}\; = \;{e^{ – 1.2805}}\]

\[\frac{m}{{ 2.34\,g}}\; = \;0.2779\]

m = 0.2779 x 2.34 g = 0.65 g

**To confirm** the answer, we can always **use the shortcut** (see problem 3 for the explanation). 8.5 : 4.6 = 1.8478 half-lives, so this will be the exponent of ½:

Therefore, the remaining amount would be:

\({\rm{Remaining}}\,{\rm{nuclide}}\,{\rm{ = }}\,{\rm{2}}{\rm{.34}}\,{\rm{g}}\, \times \,{\left( {\frac{1}{2}} \right)^{1.8478}}\; = \;0.65\,{\rm{g}}\)

The initial activity of a radioactive sample was found to be 1564 disintegrations per minute. After 7.50 h the activity dropped to 895 disintegrations per minute. What is the half-life of the nuclide?

9.31 h

This is a similar problem where we need to use the integrated rate law of first-order reactions:

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

Once we know the rate constant, we can then determine the half-life of the reaction.

\[\ln \;\frac{{895}}{{ 1564}}\; = \; – k\, \times \,7.5\,{\rm{h}}\]

-0.558178 = – *k* x 7.5 h

*k* = 0.07442376 h^{-1}

^{ }^{ }

\[{t_{{\rm{1/2}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{\rm{k}}}\, = \,\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{0}}{\rm{.07442376}}\,{{\rm{h}}^{{\rm{ – 1}}}}}}\; = \,9.31\,{\rm{h}}\]

A sample of a radioactive isotope has an initial activity of 35 dis/min. 2.00 hours later the activity is measured to be 16 dis/min. Determine how many atoms of the radioisotope were in the sample initially.

5.37 x 10^{3} atoms

The number of atoms and activity are correlated by the following formula:

A = *k* · N or A_{0} = *k* · N_{0}

_{ }

Therefore, the plan would be to determine the rate constant and use it to calculate the initial amount of atoms.

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

One thing to address before we start the calculations is to make sure the units of the time match to those of the disintegrations. Let’s convert the two hours to minutes. Because 1 hour is 60 min, the duration for the decay will be 120 min.

\[\ln \;\frac{{16}}{{ 35}}\; = \; – k\; \times \,120\,{\rm{min}}\]

-0.782759 = – *k* x 120 min

*k* = 0.006522994 min^{-1}

^{ }

Next, we use this to determine the A_{0}.

A_{0} = *k* · N_{0}

_{ }

\[{N_0}\, = \;\frac{{{A_0}}}{{ k}}\; = \;\frac{{35}}{{ 0.006522994}}\, = \,5.37\, \times \,{10^3}\,{\rm{atoms}}\]

An archaeological sample gives 4.80 disintegrations of ^{14}C per minute per gram of total carbon. How old is the sample if the initial decay rate of ^{14}C is 15.3 cpm/g C, and its half-life is 5730 years?

9.59 x 10^{3} years

We are going to use the integrated rate law for first-order reactions to determine the age of the sample (*t*).

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

\[{\rm{ln}}\;\frac{{{\rm{4}}{\rm{.80}}}}{{{\rm{ 15}}{\rm{.3}}}}\;{\rm{ = }}\;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{5730}}\;{\rm{yrs}}}}\, \times \,t\]

-1.159237 = -0.000120942 *t*

* **t *= 9.59 x 10^{3} years

How old is an ancient painting with a ^{14}C activity of 7.60 cpm/g if the initial decay rate of ^{14}C is 15.3 cpm/g C, and its half-life is 5730 years?

5.79 x 10^{3} years

We are going to use the integrated rate law for first-order reactions to determine the age of the sample (*t*).

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

\[{\rm{ln}}\;\frac{{{\rm{7}}{\rm{.60}}}}{{{\rm{ 15}}{\rm{.3}}}}\;{\rm{ = }}\;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{5730}}\;{\rm{yrs}}}}\, \times \,t\]

-0.699704 = -0.000120942 *t*

* **t *= 5.79 x 10^{3} years

A wooden tray discovered by a group of archeologists has a ^{14}C activity that is 73% of the current ^{14}C activity. How old is the tray?

2.60 x 10^{3} years

73% of the current activity means A = 0.73 A_{0}, and this is what we will use in the formula:

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

\[\ln \;\frac{{0.73\,{A_0}}}{{ {A_0}}}\; = \; – kt\]

\[{\rm{ln}}\;{\rm{0}}{\rm{.73}}\;{\rm{ = }}\;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{5730}}\;{\rm{yrs}}}}\, \times \,t\]

-0.3147107 = -0.000120942 *t*

* **t *= 2.60 x 10^{3} years

The activity of 400. mg sample of ^{14}C carbon collected from an ancient cloth is 145 disintegrations per hour. How old is the cloth if the activity of a current 1.00-g sample of carbon is 921 disintegrations per hour?

7.71 x 10^{3} years

We need to follow the same steps in this problem as we have been doing so far. The only difference is that the masses of samples are not equal, and therefore, we need to calibrate them. Let’s adjust the activity of the collected sample to 1.00 g. 400. mg is 0.400 g, so, to find the activity of the old sample per gram of caron, we need to divide 145 by 0.400:

A = 145 ÷ 0.400 = 362.5 dis/min

And now we can use the integrated rate law to determine the age of the sample:

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

\[{\mathop{\rm l}\nolimits} {\rm{n}}\;\frac{{{\rm{362}}{\rm{.5}}}}{{{\rm{ 921}}}}\;{\rm{ = }}\;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{5730}}\,{\rm{yrs}}}}t\]

-0.932436 = -0.000120942 *t** *

*t *= 7.71 x 10^{3} years

The bones of a camel were found to have ^{14}C activity of 3.40 dis/min · g carbon. Determine, approximately, how long ago the animal lived given the initial decay rate of ^{14}C is 15.3 cpm/g carbon, and its half-life is 5730 years?

1.24 x 10^{4} years

We are going to use the integrated rate law for first-order reactions to determine the age of the animal (*t*).

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

\[{\rm{ln}}\;\frac{{{\rm{3}}{\rm{.40}}}}{{{\rm{ 15}}{\rm{.3}}}}\;{\rm{ = }}\;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{5730}}\;{\rm{yrs}}}}\, \times \,t\]

-1.504077 = -0.000120942 *t*

* **t *= 1.24 x 10^{4} years

What is the activity of a 1.50-g sample of Molybdenum-93 given the half-life is 4000. years? Express your answer in becquerel (*Bq*) which is the activity equivalent to 1 disintegration per second.

5.33 x 10^{10} *Bq*

To determine the activity, we need the rate constant because it is calculated using the following formula:

A = *k* · N

N is the number of atoms, and we can determine it using the molar mass of ^{93}Mo.

For the *k*, we need to use its correlation with the half-life, making sure the units are converted from years to seconds because of the units *Bq*.

Let’s start with the number of ^{93}Mo. You can review this article on converting the mass to number of atoms.

\[{\rm{N}}\;{{\rm{(}}^{93}}{\rm{Mo)}}\;{\rm{ = }}\,{\rm{1}}{\rm{.50}}\,\cancel{{{\rm{g}}\,}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}{{{\rm{93}}\;\cancel{{{\rm{g}}\,}}}}\; \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{atoms}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\,{\rm{ = }}\;{\rm{9}}{\rm{.71}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{21}}}}{\,^{93}}{\rm{Mo}}\,{\rm{atoms}}\,\]

Next, we calculate the rate constant and convert the years to seconds as well. It will be a good practice on dimensional analysis as anyways:

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{t_{{\rm{1/2}}}}}}\]

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{4000}}{\rm{.}}\,\cancel{{{\rm{yrs}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{{\rm{yr}}}}}}{{{\rm{365}}\,\cancel{{\rm{d}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{\rm{d}}}}}{{{\rm{24}}\,\cancel{{\rm{h}}}}}{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{\rm{h}}}}}{{{\rm{60}}\,\cancel{{{\rm{min}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{{\rm{min}}}}}}{{{\rm{60}}\,{\rm{sec}}}}\, = \,5.494\, \times \,{10^{ – 12}}{\sec ^{ – 1}}\]

And now, we can calculate the activity:

A = *k* · N = (5.494 x 10^{-12} sec^{-1}) x (9.71 x 10^{21} atoms) = 5.33 x 10^{10} *Bq *

The half -life of ^{60}Co is 5.3 years. What will be the activity of a ^{60}Co sample in 47 years if the current activity is 7864 dis/min?

16.9 dis/min

We are going to use the integrated rate law for first-order reactions to determine the activity in 47 years (A).

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

As always, we will use the half-life to determine the *k*.

\[\ln \;\frac{A}{{ 7864}}\; = \;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{5}}{\rm{.3}}\,{\rm{yrs}}}}\; \times \,{\rm{47}}\,{\rm{yrs}}\]

\[\ln \;\frac{A}{{ 7864}}\; = \;{\rm{ – }}6.14547\]

\[\frac{A}{{ 7864}}\; = \;{e^{{\rm{ – }}6.14547}}\; = \;0.002143\]

A = 0.002143 x 7864 = 16.9 dis/min

**To confirm** the answer, we can always **use the shortcut** (see problem 3 for the explanation). 47 : 5.3 = 8.8679 half-lives, so this will be the exponent of ½:

Therefore, the remaining amount would be:

\({\rm{Current}}\;{\rm{A}}\,{\rm{ = }}\,{\rm{7864}}\,{\rm{dis/min}}\,{\rm{ \times }}\,{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{{\rm{8}}{\rm{.8679}}}}\;{\rm{ = }}\;{\rm{16}}{\rm{.9}}\;{\rm{dis/min}}\)

A sample of a radioactive nuclide has an activity of 1850 disintegrations per minute. After 5.0 h the activity dropped to 1065 disintegrations per minute. Determine the half-life of the radioisotope.

6.3 h

We are going to use the integrated rate law for first-order reactions to determine the half-life of the nuclide.

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – kt\]

Let’s abbreviate half-life as HL to use in the formula. Remember, HL = 0.693/*k*.

\[\ln \;\frac{{{A_t}}}{{ {A_0}}}\; = \; – \frac{{0.693}}{{HL}}t\]

\[\ln \;\frac{{{\rm{1065}}}}{{{\rm{ 1850}}}}\;{\rm{ = }}\;{\rm{ – }}\frac{{{\rm{0}}{\rm{.693}}}}{{HL}}\,{\rm{ \times }}\;{\rm{5}}{\rm{.0}}\;{\rm{h}}\]

\[ – 0.552211\; = {\rm{ – }}\frac{{{\rm{3}}{\rm{.465}}\,{\rm{h}}}}{{HL}}\,\]

\[HL\; = {\rm{ – }}\frac{{{\rm{3}}{\rm{.465}}\,{\rm{h}}}}{{ – 0.552211}}\; = \,6.3\;{\rm{h}}\,\]

The answer is rounded to two significant figures.

The half-life of Np-239 is 2.40 days. How many decay events will a 1.80-g sample of this nuclide produce in the first second assuming the atomic mass is 239 u?

1.52 x 10^{16} decays/sec

The decay events in the first sentence are given by A_{0} which is calculated using the following formula:

A_{0} = *k* · N_{0}

_{ }

N_{0} is the initial number of the atoms which we can calculate using the mass to the number of atoms conversion. You can review this article on converting the mass to number of atoms.

\[{{\rm{N}}_{\rm{0}}}\;{{\rm{(}}^{239}}{\rm{Np)}}\;{\rm{ = }}\,{\rm{1}}{\rm{.80}}\,\cancel{{{\rm{g}}\,}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}{{{\rm{239}}\;\cancel{{{\rm{g}}\,}}}}\; \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{atoms}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\,{\rm{ = }}\;{\rm{4}}{\rm{.5339}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{21}}}}{\,^{239}}{\rm{Np}}\,{\rm{atoms}}\,\]

For the *k*, we need to use its correlation with the half-life, making sure the units are converted from days to seconds.

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{t_{{\rm{1/2}}}}}}\]

\[k\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{2}}{\rm{.4}}\;\cancel{{\rm{d}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{\rm{d}}}}}{{{\rm{24}}\,\cancel{{\rm{h}}}}}{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{\rm{h}}}}}{{{\rm{60}}\,\cancel{{{\rm{min}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\,\cancel{{{\rm{min}}}}}}{{{\rm{60}}\,{\rm{sec}}}}\, = \,3.3420\; \times \,{10^{ – 6}}{\sec ^{ – 1}}\]

And now, we can calculate the initial activity:

*A*_{0} = *k* · N_{0} = (3.3420 x 10^{-6} sec^{-1}) x (4.5339 x 10^{21} atoms) = 1.52 x 10^{16} decays/sec

The half-life of ^{90}Sr is 20.0 years. How long does it take for a sample of ^{90}Sr to reach 25% of the initial amount?

40.0 years

Notice that 25% is reached after 2 half-lives. And therefore, it will take 40 years. See problem 3 for more explanations.

Let’s also do this by using the formula as we are going to need it in the next problem which is similar but more difficult.

We can set N_{0} as 100% and N as 25% and use these in the integrated rate law.

\[\ln \;\frac{N}{{ {N_0}}}\; = \; – kt\]

\[\ln \;\frac{{25\% }}{{ 100\% }}\; = \; – \frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{20}}\;{\rm{yrs}}}}t\]

ln 0.25 = -0.03465 *t*

* **t *= 40.0 years

Fluorine-18 (*t*_{1/2 }= 109 min) is commonly used in PET scans to locate certain cells. Can the PET scan be done 5 hours after the administration if the minimal amount of fluorine-18 is 20% of its initial value?

It will not be possible to do the scan if the sample is administered 5 hours prior to the scan.

What we need to do here is use the integrated rate law to find out how long it takes for the initial amount of ^{19}F to decrease to its 20%. If it takes less than 5 hours, then the solution containing the nuclide must be administered less than 5 hours before the scan.

Let’s set N_{0} as 100% and N as 20%.

\[\ln \;\frac{N}{{ {N_0}}}\; = \; – kt\]

\[\ln \;\frac{{20\% }}{{ 100\% }}\; = \; – \frac{{0.693}}{{109\,\min }}t\]

ln 0.2 = -0.0063578 *t** *

*t *= 253 min

So, the initial amount drops to 20% after 253 min which is 4.22 hours. Therefore, it will not be possible to do the scan if the sample is administered 5 hours prior to the scan. In 5 hours, the amount will be below 20%.

A patient was given 1.80-mg dose of Iodine-131 to treat a thyroid disease. How long will it take to decrease the amount of I-131 to 0.400 mg if the half-life of the nuclide is 8.0 days? Assume the nuclide is not removed from the body by other means.

17 days

We can use the integrated rate law to determine the time. Remember, in the logarithm expression, mass can be used as well.

\[\ln \;\frac{m}{{ {m_0}}}\; = \; – kt\]

\[\ln \;\frac{{0.400}}{{ 1.80}}\; = \; – \frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{8}}{\rm{.0}}\;{\rm{days}}}}t\]

-1.5041 = -0.086625 *t*

* **t *= 17 days

The answer is reasonable since 17 days is a little more than 2 half-lives which would decrease the amount to 0.45 mg (1.8 : 2 : 2 = 0.45).

In a certain experiment, 1.60 g of ^{18}F needed to be used for imaging specific sells. The source for this nuclide is NaF which takes 3 hours to be delivered. Given the half-life of ^{18}F is 109 min, what mass of NaF must be ordered from the supplier?

5.03 g

The m is 1.60 g, and we need to find the m_{0}. We can use the integrated rate law to determine the time. 3 hours is 180 min, and this is what we need to use to match the units of the half-life.

\[\ln \;\frac{m}{{ {m_0}}}\; = \; – kt\]

\[\ln \;\frac{{1.60}}{{ {m_0}}}\; = \; – \frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{109}}\,{\rm{min}}}} \times \,180\,\min \]

\[\frac{{1.60}}{{ {m_0}}}\; = \;{e^{ – 1.1444}}\]

1.60 = 0.3184 *m*_{0}

*m*_{0 }= 5.03 g

Thallium-201 is in medical imaging for nuclear cardiology and certain cells. How many grams of the ^{201}Tl nuclide can be delivered to the hospital from a nuclear facility that is 350 miles away if a 30.0-g sample is transported in a truck that drives at 65 mi/h. The half-life of ^{201}Tl is 73.1 hours.

28.5 g

The time it will take to deliver the sample is 350 : 65 = 5.385 hours. So, we **need to determine the m** given m

_{0}= 30.0 g.

\[\ln \;\frac{m}{{ {m_0}}}\; = \; – kt\]

\[\ln \;\frac{m}{{ 30.0}}\; = \; – \frac{{{\rm{0}}{\rm{.693}}}}{{{\rm{73}}{\rm{.1}}\,{\rm{h}}}} \times \,5.385\,{\rm{h}}\]

\[\frac{m}{{ 30.0}}\; = \;{e^{ – 0.0511}}\]

\[\frac{m}{{ 30.0}}\; = \;0.95023\]

m = 28.5 g