# 10.3: Radioactive Decay

In 1896, Antoine **Becquerel** discovered that a uranium-rich rock emits invisible rays that can darken a photographic plate in an enclosed container. Scientists offer three arguments for the nuclear origin of these rays. First, the effects of the radiation do not vary with chemical state; that is, whether the emitting material is in the form of an element or compound. Second, the radiation does not vary with changes in temperature or pressure—both factors that in sufficient degree can affect electrons in an atom. Third, the very large energy of the invisible rays (up to hundreds of eV) is not consistent with atomic electron transitions (only a few eV). Today, this radiation is explained by the conversion of mass into energy deep within the nucleus of an atom. The spontaneous emission of radiation from nuclei is called nuclear **radioactivity** (Figure).

**Figure 10.3.1.** The international ionizing radiation symbol is universally recognized as the warning symbol for nuclear radiation.

# Radioactive Decay Law

When an individual nucleus transforms into another with the emission of radiation, the nucleus is said to **decay**. Radioactive decay occurs for all nuclei with \(Z > 82\), and also for some unstable isotopes with \(Z < 83\). The decay rate is proportional to the number of original (undecayed) nuclei *N* in a substance. The number of nuclei lost to decay, \(-dN\) in time interval *dt*, is written

\[-\frac{dN}{dt} = \lambda N\]

where \(\lambda\) is called the decay constant. (The minus sign indicates the number of original nuclei decreases over time.) In other words, the more nuclei available to decay, the more that do decay (in time *dt*). This equation can be rewritten as

\[\frac{dN}{N} = -\lambda dt.\]

Integrating both sides of the equation, and defining \(N_0\) to be the number of nuclei at \(t = 0\), we obtain

\[\int_{N_0}^N \frac{dN'}{N} = - \int_0^t \lambda dt'.\]

This gives us

\[ln\frac{N}{N_0} = -\lambda t.\]

Taking the left and right sides of the equation as a power of *e*, we have the radioactive **decay law**.

Note: Radioactive Decay Law

The total number *N* of radioactive nuclei remaining after time *t* is

\[N = N_0e^{-\lambda t}\]

where \(\lambda\) is the decay constant for the particular nucleus.

The total number of nuclei drops very rapidly at first, and then more slowly (Figure).

**Figure 10.3.2.** A plot of the radioactive decay law demonstrates that the number of nuclei remaining in a decay sample drops dramatically during the first moments of decay.

The **half-life** \((T_{1/2})\) of a radioactive substance is defined as the time for half of the original nuclei to decay (or the time at which half of the original nuclei remain). The half-lives of unstable isotopes are shown in the chart of nuclides in [link]. The number of radioactive nuclei remaining after an integer (*n*) number of half-lives is therefore

\[N = \frac{N_0}{2^n}\]

If the decay constant \((\lambda)\) is large, the half-life is small, and vice versa. To determine the relationship between these quantities, note that when \(t = T_{1/2}\), then \(N = N_0/2\).

Thus, Equation can be rewritten as

\[\frac{N_0}{2} = N_0e^{-\lambda T_{1/2}}.\]

Dividing both sides by \(N_0\) and taking the natural logarithm yields

\[ln \frac{1}{2} = ln \space e^{-\lambda T_{1/2}}\]

which reduces to

\[\lambda = \frac{0.693}{T_{1/2}}.\]

Thus, if we know the half-life *T*_{1/2} of a radioactive substance, we can find its decay constant. The **lifetime** \(\overline{T}\) of a radioactive substance is defined as the average amount of time that a nucleus exists before decaying. The lifetime of a substance is just the reciprocal of the decay constant, written as

\[\overline{T} = \frac{1}{\lambda}.\]

The activity *A* is defined as the magnitude of the decay rate, or

\[A = -\frac{dN}{dt} = \lambda N = \lambda N_0 e^{-\lambda t}.\]

The infinitesimal change *dN* in the time interval *dt* is negative because the number of parent (undecayed) particles is decreasing, so the activity (*A*) is positive. Defining the initial activity as \(A_0 = \lambda N_0\), we have

\[A = A_0 e^{-\lambda t}.\]

Thus, the activity *A* of a radioactive substance decreases exponentially with time (Figure).

**Figure 10.3.3.** (a) A plot of the activity as a function of time (b) If we measure the activity at different times, we can plot ln A versus t, and obtain a straight line.

Example

**Decay Constant and Activity of Strontium-90**

The half-life of strontium-90, \(_{38}^{90}Sr\), is 28.8 y. Find (a) its decay constant and (b) the initial activity of 1.00 g of the material.

**Strategy**

We can find the decay constant directly from Equation. To determine the activity, we first need to find the number of nuclei present.

**Solution**

a. The decay constant is found to be

\[\lambda = \frac{0.693}{T_{1/2}} = \left(\frac{0.693}{T_{1/2}}\right)\left(\frac{1 \space yr}{3.16 \times 10^7 \space s}\right) = 7.61 \times 10^{-10} \space s^{-1}.\]

b. The atomic mass of \(_{38}^{90}Sr\) is 89.91 g. Using Avogadro’s number \(N_A = 6.022 \times 10^{23}\) atoms/mol, we find the initial number of nuclei in 1.00 g of the material:

\[N_0 = \frac{1.00 \space g}{89.91 \space g} (N_A) = 6.70 \times 10^{21} \space nuclei.\]

From this, we find that the activity \(A_0\) at \(t = 0\) for 1.00 g of strontium-90 is

\[A_0 = \lambda N_0 = (7.61 \times 10^{-10} s^{-1})(6.70 \times 10^{21} \space nuclei) = 5.10 \times 10^{12} \space decays/s.\]

Expressing \(\lambda\) in terms of the half-life of the substance, we get

\[A = A_0 e^{-(0.693/T_{1/2})T_{1/2}} = A_0 e^{-0.693} = A_0/2.\]

Therefore, the activity is halved after one half-life. We can determine the decay constant \(\lambda\) by measuring the activity as a function of time. Taking the natural logarithm of the left and right sides of Equation, we get

\[ln \space A = - \lambda t + ln \space A_0.\]

This equation follows the linear form \(y = mx + b\). If we plot ln *A* versus *t*, we expect a straight line with slope \(-\lambda\) and *y*-intercept \(ln \space A_0\) (Figure(b)). Activity *A* is expressed in units of **becquerels** (Bq), where one \(1 \space Bq = 1 \space decay \space per \space second\). This quantity can also be expressed in decays per minute or decays per year. One of the most common units for activity is the **curie** (Ci), defined to be the activity of 1 g of \(^{226}Ra\). The relationship between the Bq and Ci is

\[1 \space Ci = 3.70 \times 10^{10}Bq.\]

Example

**What is**\(^{14}C\) **Activity in Living Tissue?**

Approximately \(20\%\) of the human body by mass is carbon. Calculate the activity due to \(^{14}C\) in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

**Strategy**

The activity of \(^{14}C\) is determined using the equation \(A_0 = \lambda N_0\), where *λ* is the decay constant and \(N_0\) is the number of radioactive nuclei. The number of \(^{14}C\) nuclei in a 1.00-kg sample is determined in two steps. First, we determine the number of \(^{12}C\) nuclei using the concept of a mole. Second, we multiply this value by \(1.3 \times 10^{-12}\) (the known abundance of \(^{14}C\) in a carbon sample from a living organism) to determine the number of \(^{14}C\) nuclei in a living organism. The decay constant is determined from the known half-life of \(^{14}C\) (available from [link]).

**Solution**

One mole of carbon has a mass of 12.0 g, since it is nearly pure \(^{12}C\). Thus, the number of carbon nuclei in a kilogram is

\[N(^{12}C) = \frac{6.02 \times 10^{23} mol^{-1}}{12.0 \space g/mol} \times (1000 \space g) = 5.02 \times 10^{25}.\]

The number of \(^{14}C\) nuclei in 1 kg of carbon is therefore

\[N(^{14}C) = (5.02 \times 10^{25})(1.3 \times 10^{_12}) = 6.52 \times 10^{13}.\]

Now we can find the activity *A* by using the equation \(A = \frac{0.693 \space N}{t_{1/2}}\). Entering known values gives us

\[A = \frac{0.693 (6.52 \times 10^{13})}{5730 \space y} = 7.89 \times 10^9 \space y^{-1}\]

or \(7.89 \times 10^9\) decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,

\[A = (7.89 \times 10^9 \space y^{-1}) \frac{1.00 \space y}{3.16 \times 10^7 \space s} = 250 \space Bq,\]

or 250 decays per second. To express *A* in curies, we use the definition of a curie,

\[A = \frac{250 \space Bq}{3.7 \times 10^{10} \space Bq/Ci} = 6.76 \times 10^{-9} Ci.\] Thus,

\[A = 6.76 \space nCi.\]

Significance Approximately \(20\%\) of the human body by weight is carbon. Hundreds of \(^{14}C\) decays take place in the human body every second. Carbon-14 and other naturally occurring radioactive substances in the body compose a person’s background exposure to nuclear radiation. As we will see later in this chapter, this activity level is well below the maximum recommended dosages.

# Radioactive Dating

**Radioactive dating** is a technique that uses naturally occurring radioactivity to determine the age of a material, such as a rock or an ancient artifact. The basic approach is to estimate the original number of nuclei in a material and the present number of nuclei in the material (after decay), and then use the known value of the decay constant \(\lambda\) and Equation to calculate the total time of the decay, *t*.

An important method of **radioactive dating** is **carbon-14 dating**. Carbon-14 nuclei are produced when high-energy solar radiation strikes \(^{14}N\) nuclei in the upper atmosphere and subsequently decay with a half-life of 5730 years. Radioactive carbon has the same chemistry as stable carbon, so it combines with the ecosphere and eventually becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Therefore, if you know the number of carbon nuclei in an object, you multiply that number by \(1.3 \times 10^{-12}\) to find the number of \(^{14}C\) nuclei in that object. When an organism dies, carbon exchange with the environment ceases, and \(^{14}C\) is not replenished as it decays.

By comparing the abundance of \(^{14}C\) in an artifact, such as mummy wrappings, with the normal abundance in living tissue, it is possible to determine the mummy’s age (or the time since the person’s death). Carbon-14 dating can be used for biological tissues as old as 50,000 years, but is generally most accurate for younger samples, since the abundance of \(^{14}C\) nuclei in them is greater. Very old biological materials contain no \(^{14}C\) at all. The validity of carbon dating can be checked by other means, such as by historical knowledge or by tree-ring counting.

Example

An Ancient Burial Cave In an ancient burial cave, your team of archaeologists discovers ancient wood furniture. Only \(80\%\) of the original \(^{14}C\) remains in the wood. How old is the furniture?

Strategy

The problem statement implies that \(N/N_0 = 0.80\). Therefore, the equation \(N = N_0e^{-\lambda t}\) can be used to find the product, \(\lambda t\). We know the half-life of \(^{14}C\) is 5730 y, so we also know the decay constant, and therefore the total decay time *t*.

**Solution**

Solving the equation \(N = N_0 e^{-\lambda t}\) for \(N/N_0\) gives us

\[\frac{N}{N_0} = e^{-\lambda t}.\] Thus \[0.80 = e^{-\lambda t}.\]

Taking the natural logarithm of both sides of the equation yields

\[ln \space 0.80 = - \lambda t,\] so that

\[-0.223 = -\lambda t.\]

Rearranging the equation to isolate *t* gives us

\[t = \frac{0.223}{\left(\frac{0.693}{5730 \space y}\right)} = 1844 \space y.\]

**Significance**

The furniture is almost 2000 years old—an impressive discovery. The typical uncertainty on carbon-14 dating is about \(5\%\), so the furniture is anywhere between 1750 and 1950 years old. This date range must be confirmed by other evidence, such as historical records.

**Check Your Understanding** A radioactive nuclide has a high decay rate. What does this mean for its half-life and activity?

Half-life is inversely related to decay rate, so the half-life is short. Activity depends on both the number of decaying particles and the decay rate, so the activity can be great or small.

Visit the Radioactive Dating Game to learn about the types of radiometric dating and try your hand at dating some ancient objects.

# Summary

- In the decay of a radioactive substance, if the decay constant \((\lambda)\) is large, the half-life is small, and vice versa.
- The radioactive decay law, \(N = N_0 e^{-\lambda t}\), uses the properties of radioactive substances to estimate the age of a substance.
- Radioactive carbon has the same chemistry as stable carbon, so it mixes into the ecosphere and eventually becomes part of every living organism. By comparing the abundance of \(^{14}C\) in an artifact with the normal abundance in living tissue, it is possible to determine the artifact’s age.

# Conceptual Questions

Exercise

How is the initial activity rate of a radioactive substance related to its half-life?

Exercise

For the carbon dating described in this chapter, what important assumption is made about the time variation in the intensity of cosmic rays?

That it is constant.

# Problems

Exercise

A sample of radioactive material is obtained from a very old rock. A plot ln*A* verses *t* yields a slope value of \(-10^{-9} s^{-1}\) (see Figure(b)). What is the half-life of this material?

The decay constant is equal to the negative value of the slope or \(- 10^{-9} s^{-1}\). The half-life of the nuclei, and thus the material, is \(T_{1/2} = 663 \space million \space years.

Exercise

Show that: \(\overline{T} = \frac{1}{\lambda}\).

Exercise

The half-life of strontium-91, \(_{38}^{91}Sr\) is 9.70 h. Find (a) its decay constant and (b) for an initial 1.00-g sample, the activity after 15 hours.

[Hide Solution]

a. The decay constant is \(\lambda = 1.99 \times 10^{-5}s^{-1}\). b. Since strontium-91 has an atomic mass of 90.90 g, the number of nuclei in a 1.00-g sample is initially \(N_0 = 6.63 \times 10^{21} \space nuclei\). The initial activity for strontium-91 is

\(A_0 = \lambda N_0\)

\(= 1.32 \times 10^{17} \space decays/s\)

The activity at \(t = 15.0 \space h = 5.40 \times 10^4 s\) is \(A = 4.51 \times 10^{16} \space decays/s\).

Exercise

A sample of pure carbon-14 \((T_{1/2} = 5730 \space y)\) has an activity of \(1.0 \space \mu Ci\). What is the mass of the sample?

Exercise

A radioactive sample initially contains \(2.40 \times 10^{-2}\) mol of a radioactive material whose half-life is 6.00 h. How many moles of the radioactive material remain after 6.00 h? After 12.0 h? After 36.0 h?

[Hide Solution]

\(1.20 \times 10^{-2} mol\); \(6.00 \times 10^{-3} mol\); \(3.75 \times 10^{-4}mol\)

Exercise

An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of \(^{14}C\). Estimate the minimum age of the charcoal, noting that \(2^{10} = 1024\).

Exercise

Calculate the activity \(R\), in curies of 1.00 g of \(^{226}Ra\). (b) Explain why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.

a. 0.988 Ci; b. The half-life of \(^{226}Ra\) is more precisely known than it was when the Ci unit was established.

Exercise

Natural uranium consists of \(^{235}U\) (percent abundance = \(0.7200 \%\), \(\lambda = 3.12 \times 10^{-17} /s\)) and \(^{238}U\) (percent abundance = \(99.27 \%\), \(\lambda = 4.92 \times 10^{-18} /s\)). What were the values for percent abundance of \(^{235}U\) and \(^{238}U\) when Earth formed \(4.5 \times 10^9\) years ago?

Exercise

World War II aircraft had instruments with glowing radium-painted dials. The activity of one such instrument was \(1.0 \times 10^5\) Bq when new. (a) What mass of \(^{226}Ra\) was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made?

[Hide Solution]

a. \(2.73 \mu g\); b. \(9.76 \times 10^4 \space Bq\)

Exercise

The \(^{210}Po\) source used in a physics laboratory is labeled as having an activity of \(1 \space \mu Ci\) on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus?

Exercise

Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its \(^{235}U\) removed for reactor use and is nearly pure \(^{238}U\). Depleted uranium has been erroneously called nonradioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure \(^{238}U\). (b) Calculate the activity of 60.0 g of natural uranium, neglecting the \(^{234}U\) and all daughter nuclides.

[Hide Solution]

a. \(7.46 \times 10^5 \space Bq\); b. \(7.75 \times 10^5 \space Bq\)

## Glossary

- activity
- magnitude of the decay rate for radioactive nuclides

- becquerel (Bq)
- SI unit for the decay rate of a radioactive material, equal to 1 decay/second

- carbon-14 dating
- method to determine the age of formerly living tissue using the ratio \(^{14}C/^{12}C\)
- curie (Ci)
- unit of decay rate, or the activity of 1 g of \(^{226}Ra\), equal to \(3.70 \times 10^{10} \space Bq\)
- decay
- process by which an individual atomic nucleus of an unstable atom loses mass and energy by emitting ionizing particles
- decay constant
- quantity that is inversely proportional to the half-life and that is used in equation for number of nuclei as a function of time
- half-life
- time for half of the original nuclei to decay (or half of the original nuclei remain)
- lifetime
- average time that a nucleus exists before decaying
- radioactive dating
- application of radioactive decay in which the age of a material is determined by the amount of radioactivity of a particular type that occurs
- radioactive decay law
- describes the exponential decrease of parent nuclei in a radioactive sample
- radioactivity
- spontaneous emission of radiation from nuclei