10.3 Radioactive Decay

In 1896, Antoine Henri 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 10.8).

A yellow triangle with a black outline, enclosing a fan shaped structure is shown. The “fan” is black and has three blades. — **Figure 10.8** 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 $Z greater than 82,$ and also for some unstable isotopes with $Z less than 83.$ The decay rate is proportional to the number of original (undecayed) nuclei N in a substance. The number of nuclei lost to decay, $− d N$ in time interval dt, is written

$- \frac{d N}{d t} = λ N$

where $λ$ 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

$\frac{d N}{N} = − λ d t .$

Integrating both sides of the equation, and defining $N sub 0$ to be the number of nuclei at $t equals 0$ , we obtain

$\int_{N_{0}}^{N} \frac{d N^{'}}{N} = − \int_{0}^{t} λ d t^{'} .$

This gives us

$ln \frac{N}{N_{0}} = − λ t .$

(10.10)

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

The total number of nuclei drops very rapidly at first, and then more slowly (Figure 10.9).

A graph of N versus t is shown. It is labeled N equal to N subscript 0 e to the power minus lambda t. The value of N is maximum, N subscript 0, at t =0 and it reduces with time till it reaches 0. At t = T subscript half, N = N subscript 0 by 2 and at t = 2T subscript half, N = N subscript 0 by 4. — **Figure 10.9** 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 $open parenthesis T sub 1 / 2 close parenthesis$ 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 Figure 10.4. The number of radioactive nuclei remaining after an integer (n) number of half-lives is therefore

Function graph showing y = 100*exp(-ln(2)*t/Thalf) on t in [0, 170], y = 100 - 100*exp(-ln(2)*t/Thalf) on t in [0, 170], y = 50 and the parametric curve (Thalf, t) for t in [0, 50]. Adjustable parameter: Half-life T½ (strontium-90 = 28.8 y) (Thalf) = 28.8 y. Viewing window: x from -16.17 to 168.17, y from -9 to 105. — The decay law N = N₀e^(−λt) with the decay constant written the way this section derives it, λ = ln 2/T½, so the curve is 100·2^(−t/T½) percent of the original nuclei still undecayed. The rising dashed curve is the percentage that has decayed; the two always add to 100. The slider is the half-life, defaulting to the 28.8 y of strontium-90 from this section's example, and the vertical marker sits at exactly one half-life — where the falling curve crosses 50%, which is the definition rather than a coincidence. Two half-lives leave a quarter and three leave an eighth, so 86.4 years of ⁹⁰Sr leaves 12.5%. Drag the half-life and notice that the shape never changes, only the horizontal scale: one number fixes the whole decay, which is what makes ¹⁴C at 5730 y a usable clock for objects thousands of years old and ⁹⁰Sr a hazard for a human lifetime.

$N = \frac{N_{0}}{2^{n}}$

If the decay constant ( $λ$ ) is large, the half-life is small, and vice versa. To determine the relationship between these quantities, note that when $t equals T sub 1 / 2$ , then $N equals N sub 0 / 2$ . Thus, Equation 10.10 can be rewritten as

$\frac{N_{0}}{2} = N_{0} e^{− λ T_{1 / 2}} .$

Dividing both sides by $N sub 0$ and taking the natural logarithm yields

$ln \frac{1}{2} = ln e^{− λ T_{1 / 2}}$

which reduces to

$λ = \frac{0.693}{T_{1 / 2}} .$

(10.15)

Thus, if we know the half-life T_1/2 of a radioactive substance, we can find its decay constant. The lifetime $\bar{T}$ 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

$\bar{T} = \frac{1}{λ} .$

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

$A = - \frac{d N}{d t} = λ N = λ N_{0} e^{− λ t} .$

(10.17)

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 $A sub 0 equals λ N sub 0$ , we have

$A = A_{0} e^{− λ t} .$

Thus, the activity A of a radioactive substance decreases exponentially with time (Figure 10.10).

Figure a shows a graph of A versus t. It starts at point A subscript 0 and reduces with time. The rate of reduction decreases slowly till A is very close to 0, making a curved plot on the graph. The plot is labeled A = A subscript 0 e to the power minus lambda t. Figure b shows a graph of ln A versus t. It starts at ln A subscript 0 and slopes downwards in a straight line. The slope of the line is labeled minus lambda t. — **Figure 10.10** (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 1 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 10.15. To determine the activity, we first need to find the number of nuclei present.

Solution

The decay constant is found to be
$λ = \frac{0.693}{T_{1 / 2}} = (\frac{0.693}{T_{1 / 2}}) (\frac{1 yr}{3.16 \times 10^{7} s}) = 7.61 \times 10^{−10} s^{− 1} .$
The atomic mass of $38 90 S r$ 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:
$N_{0} = \frac{1.00 g}{89.91 g} (N_{A}) = 6.70 \times 10^{21} nuclei .$
From this, we find that the activity $A sub 0$ at $t equals 0$ for 1.00 g of strontium-90 is
$\begin{matrix} A_{0} & = λ N_{0} \\ = (7.61 \times 10^{−10} s^{−1}) (6.70 \times 10^{21} nuclei) \\ = 5.10 \times 10^{12} decays/s . \end{matrix}$

Expressing $λ$ 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 $λ$ by measuring the activity as a function of time. Taking the natural logarithm of the left and right sides of Equation 10.17, we get

$ln A = − λ t + ln A_{0} .$

This equation follows the linear form $y = m x + b$ . If we plot ln A versus t, we expect a straight line with slope $− λ$ and y-intercept $ln A sub 0$ (Figure 10.10(b)). Activity A is expressed in units of becquerels (Bq), where one $1 Bq equals 1 decay per second$ . 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 $to the power 226 Ra$ . The relationship between the Bq and Ci is

$1 Ci = 3.70 \times 10^{10} Bq .$

Example 2 What is $to the power 14 C$ Activity in Living Tissue?

Approximately $20 %$ of the human body by mass is carbon. Calculate the activity due to $to the power 14 C$ in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

Strategy The activity of $to the power 14 C$ is determined using the equation $A sub 0 equals λ N sub 0$ , where λ is the decay constant and $N sub 0$ is the number of radioactive nuclei. The number of $to the power 14 C$ nuclei in a 1.00-kg sample is determined in two steps. First, we determine the number of $to the power 12 C$ nuclei using the concept of a mole. Second, we multiply this value by $1.3 \times 10^{−12}$ (the known abundance of $to the power 14 C$ in a carbon sample from a living organism) to determine the number of $to the power 14 C$ nuclei in a living organism. The decay constant is determined from the known half-life of $to the power 14 C$ (available from Figure 10.4).

Solution One mole of carbon has a mass of 12.0 g, since it is nearly pure $to the power 12 C$ . Thus, the number of carbon nuclei in a kilogram is

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

The number of $to the power 14 C$ 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 $A = \frac{0.693 N}{t_{1 / 2}} .$ Entering known values gives us

$A = \frac{0.693 (6.52 \times 10^{13})}{5730 y} = 7.89 \times 10^{9} y^{− 1}$

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

$A = (7.89 \times 10^{9} y^{− 1}) \frac{1.00 y}{3.16 \times 10^{7} s} = 250 Bq,$

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

$A = \frac{250 Bq}{3.7 \times 10^{10} Bq/Ci} = 6.76 \times 10^{−9} Ci .$

Thus,

$A equals 6.76 nCi.$

Significance Approximately $20 %$ of the human body by weight is carbon. Hundreds of $to the power 14 C$ 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 $λ$ and Equation 10.10 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 $to the power 14 N$ 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 $to the power 14 C$ nuclei in that object. When an organism dies, carbon exchange with the environment ceases, and $to the power 14 C$ is not replenished as it decays.

By comparing the abundance of $to the power 14 C$ 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 $to the power 14 C$ nuclei in them is greater. Very old biological materials contain no $to the power 14 C$ at all. The validity of carbon dating can be checked by other means, such as by historical knowledge or by tree-ring counting.

Example 3 An Ancient Burial Cave

In an ancient burial cave, your team of archaeologists discovers ancient wood furniture. Only $80 %$ of the original $to the power 14 C$ remains in the wood. How old is the furniture?

Strategy The problem statement implies that $N / N sub 0 equals 0.80.$ Therefore, the equation $N = N_{0} e^{− λ t}$ can be used to find the product, $λ t$ . We know the half-life of $to the power 14 C$ is 5730 y, so we also know the decay constant, and therefore the total decay time t.

Solution Solving the equation $N = N_{0} e^{− λ t}$ for $N / N sub 0$ gives us

$\frac{N}{N_{0}} = e^{− λ t} .$

Thus,

$0.80 = e^{− λ t} .$

Taking the natural logarithm of both sides of the equation yields

$ln 0.80 = - λ t,$

so that

$−0.223 equals − λ t.$

Rearranging the equation to isolate t gives us

$t = \frac{0.223}{λ},$

where

$λ = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{5730 y} .$

Combining this information yields

$t = \frac{0.223}{(\frac{0.693}{5730 y})} = 1844 y .$

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.

Summary

In the decay of a radioactive substance, if the decay constant ( $λ$ ) is large, the half-life is small, and vice versa.
The radioactive decay law, $N = N_{0} e^{− λ t},$ 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 $to the power 14 C$ in an artifact with the normal abundance in living tissue, it is possible to determine the artifact’s age.

Conceptual Questions

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

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

A sample of radioactive material is obtained from a very old rock. A plot lnA verses t yields a slope value of $− 10^{−9} s^{−1}$ (see Figure 10.10(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 $T sub 1 / 2 equals 693 million years.$

Show that: $\bar{T} = \frac{1}{λ}$ .

The half-life of strontium-91, $38 91 S r$ is 9.70 h. Find (a) its decay constant and (b) for an initial 1.00-g sample, the activity after 15 hours.

a. The decay constant is $λ = 1.99 \times 10^{−5} s^{− 1}$ . b. Since strontium-91 has an atomic mass of 90.90 g, the number of nuclei in a 1.00-g sample is initially
$N_{0} = 6.63 \times 10^{21} nuclei .$
The initial activity for strontium-91 is
$\begin{matrix} A_{0} & = λ N_{0} \\ = 1.32 \times 10^{17} decays/s \end{matrix}$
The activity at $t = 15.0 h = 5.40 \times 10^{4} s$ is
$A = 4.51 \times 10^{16} decays/s .$

A sample of pure carbon-14 $open parenthesis T sub 1 / 2 equals 5730 y close parenthesis$ has an activity of $1.0 μ Ci.$ What is the mass of the sample?

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?

$1.20 \times 10^{− 2} mol$ ; $6.00 \times 10^{− 3} mol$ ; $3.75 \times 10^{− 4} mol$

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

Calculate the activity $R$ , in curies of 1.00 g of $to the power 226 R a.$ (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 $to the power 226 R a$ is more precisely known than it was when the Ci unit was established.

Natural uranium consists of $to the power 235 U$ $open parenthesis percent abundance equals 0.7200 %$ , $λ = 3.12 \times 10^{−17} / s)$ and $to the power 238 U$ $open parenthesis percent abundance equals 99.27 %$ , $λ = 4.92 \times 10^{−18} / s) .$ What were the values for percent abundance of $to the power 235 U$ and $to the power 238 U$ when Earth formed $4.5 \times 10^{9}$ years ago?

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 $to the power 226 R a$ 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?

a. $2.73 μ g$ ; b. $9.76 \times 10^{4} Bq$

The $to the power 210 P o$ source used in a physics laboratory is labeled as having an activity of $1.0 μCi$ 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?

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 $to the power 235 U$ removed for reactor use and is nearly pure $to the power 238 U$ . Depleted uranium has been erroneously called nonradioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure $to the power 238 U$ . (b) Calculate the activity of 60.0 g of natural uranium, neglecting the $to the power 234 U$ and all daughter nuclides.

a. $7.46 \times 10^{5} Bq$ ; b. $7.75 \times 10^{5} Bq$