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 10.8).
Figure 10.8 The international ionizing radiation symbol is universally recognized as the warning symbol for nuclear radiation.
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 , Z > 82 , and also for some unstable isotopes with Z < 83 . 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, − d N − d N in time interval dt, is written
− d N d t = λ N − d N d t = λ Nwhere λ 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
d N N = − λ d t . d N N = − λ d t .Integrating both sides of the equation, and defining N 0 N 0 to be the number of nuclei at t = 0 t = 0 , we obtain
∫ N 0 N d N ′ N = − ∫ 0 t λ d t ′ . ∫ N 0 N d N ′ N = − ∫ 0 t λ d t ′ . ln N N 0 = − λ t . ln N N 0 = − λ t .Taking the left and right sides of the equation as a power of e, we have the radioactive decay law .
The total number N of radioactive nuclei remaining after time t is
N = N 0 e − λ t N = N 0 e − λ twhere λ λ is the decay constant for the particular nucleus.
The total number of nuclei drops very rapidly at first, and then more slowly (Figure 10.9).
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 ( T 1 / 2 ) ( 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 Figure 10.4. The number of radioactive nuclei remaining after an integer (n) number of half-lives is therefore
N = N 0 2 n N = N 0 2 nIf the decay constant ( λ λ ) is large, the half-life is small, and vice versa. To determine the relationship between these quantities, note that when t = T 1 / 2 t = T 1 / 2 , then N = N 0 / 2 N = N 0 / 2 . Thus, Equation 10.10 can be rewritten as
N 0 2 = N 0 e − λ T 1 / 2 . N 0 2 = N 0 e − λ T 1 / 2 .Dividing both sides by N 0 N 0 and taking the natural logarithm yields
ln 1 2 = ln e − λ T 1 / 2 ln 1 2 = ln e − λ T 1 / 2which reduces to
λ = 0.693 T 1 / 2 . λ = 0.693 T 1 / 2 .Thus, if we know the half-life T1/2 of a radioactive substance, we can find its decay constant. The lifetime T – 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
T – = 1 λ . T – = 1 λ .The activity A is defined as the magnitude of the decay rate, or
A = − d N d t = λ N = λ N 0 e − λ t . A = − d N d t = λ N = λ N 0 e − λ 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 = λ N 0 A 0 = λ N 0 , we have
A = A 0 e − λ t . A = A 0 e − λ t .Thus, the activity A of a radioactive substance decreases exponentially with time (Figure 10.10).
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.
The half-life of strontium-90, 38 90 Sr 38 90 Sr , is 28.8 y. Find (a) its decay constant and (b) the initial activity of 1.00 g of the material.
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.
λ = 0.693 T 1 / 2 = ( 0.693 T 1 / 2 ) ( 1 yr 3.16 × 10 7 s ) = 7.61 × 10 −10 s − 1 . λ = 0.693 T 1 / 2 = ( 0.693 T 1 / 2 ) ( 1 yr 3.16 × 10 7 s ) = 7.61 × 10 −10 s − 1 .
N 0 = 1.00 g 89.91 g ( N A ) = 6.70 × 10 21 nuclei . N 0 = 1.00 g 89.91 g ( N A ) = 6.70 × 10 21 nuclei .
From this, we find that the activity A 0 A 0 at t = 0 t = 0 for 1.00 g of strontium-90 isA 0 = λ N 0 = ( 7.61 × 10 −10 s −1 ) ( 6.70 × 10 21 nuclei ) = 5.10 × 10 12 decays/s . A 0 = λ N 0 = ( 7.61 × 10 −10 s −1 ) ( 6.70 × 10 21 nuclei ) = 5.10 × 10 12 decays/s .
Expressing λ 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 . 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 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 . ln A = − λ t + ln A 0 .This equation follows the linear form y = m x + b y = m x + b . If we plot ln A versus t, we expect a straight line with slope − λ − λ and y-intercept ln A 0 ln A 0 (Figure 10.10(b)). Activity A is expressed in units of becquerels (Bq), where one 1 Bq = 1 decay per second 1 Bq = 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 226 Ra 226 Ra . The relationship between the Bq and Ci is
1 Ci = 3.70 × 10 10 Bq . 1 Ci = 3.70 × 10 10 Bq .Approximately 20 % 20 % of the human body by mass is carbon. Calculate the activity due to 14 C 14 C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.
The activity of 14 C 14 C is determined using the equation A 0 = λ N 0 A 0 = λ N 0 , where λ is the decay constant and N 0 N 0 is the number of radioactive nuclei. The number of 14 C 14 C nuclei in a 1.00-kg sample is determined in two steps. First, we determine the number of 12 C 12 C nuclei using the concept of a mole. Second, we multiply this value by 1.3 × 10 −12 1.3 × 10 −12 (the known abundance of 14 C 14 C in a carbon sample from a living organism) to determine the number of 14 C 14 C nuclei in a living organism. The decay constant is determined from the known half-life of 14 C 14 C (available from Figure 10.4).
One mole of carbon has a mass of 12.0 g, since it is nearly pure 12 C 12 C . Thus, the number of carbon nuclei in a kilogram is
N ( 12 C ) = 6.02 × 10 23 mol −1 12.0 g/mol × ( 1000 g ) = 5.02 × 10 25 . N ( 12 C ) = 6.02 × 10 23 mol −1 12.0 g/mol × ( 1000 g ) = 5.02 × 10 25 .
The number of 14 C 14 C nuclei in 1 kg of carbon is therefore
N ( 14 C ) = ( 5.02 × 10 25 ) ( 1.3 × 10 −12 ) = 6.52 × 10 13 . N ( 14 C ) = ( 5.02 × 10 25 ) ( 1.3 × 10 −12 ) = 6.52 × 10 13 .
Now we can find the activity A by using the equation A = 0.693 N t 1 / 2 . A = 0.693 N t 1 / 2 . Entering known values gives us
A = 0.693 ( 6.52 × 10 13 ) 5730 y = 7.89 × 10 9 y − 1 A = 0.693 ( 6.52 × 10 13 ) 5730 y = 7.89 × 10 9 y − 1
or 7.89 × 10 9 7.89 × 10 9 decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,
A = ( 7.89 × 10 9 y − 1 ) 1.00 y 3.16 × 10 7 s = 250 Bq , A = ( 7.89 × 10 9 y − 1 ) 1.00 y 3.16 × 10 7 s = 250 Bq ,
or 250 decays per second. To express A in curies, we use the definition of a curie,
A = 250 Bq 3.7 × 10 10 Bq/Ci = 6.76 × 10 −9 Ci . A = 250 Bq 3.7 × 10 10 Bq/Ci = 6.76 × 10 −9 Ci . A = 6.76 nCi . A = 6.76 nCi .Approximately 20 % 20 % of the human body by weight is carbon. Hundreds of 14 C 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 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 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 14 N 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 × 10 −12 1.3 × 10 −12 to find the number of 14 C 14 C nuclei in that object. When an organism dies, carbon exchange with the environment ceases, and 14 C 14 C is not replenished as it decays.
By comparing the abundance of 14 C 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 14 C nuclei in them is greater. Very old biological materials contain no 14 C 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.
In an ancient burial cave, your team of archaeologists discovers ancient wood furniture. Only 80 % 80 % of the original 14 C 14 C remains in the wood. How old is the furniture?
The problem statement implies that N / N 0 = 0.80 . N / N 0 = 0.80 . Therefore, the equation N = N 0 e − λ t N = N 0 e − λ t can be used to find the product, λ t λ t . We know the half-life of 14 C 14 C is 5730 y, so we also know the decay constant, and therefore the total decay time t.
Taking the natural logarithm of both sides of the equation yields
ln 0.80 = − λ t , ln 0.80 = − λ t , −0.223 = − λ t . −0.223 = − λ t .Rearranging the equation to isolate t gives us
t = 0.223 λ , t = 0.223 λ , λ = 0.693 t 1 / 2 = 0.693 5730 y . λ = 0.693 t 1 / 2 = 0.693 5730 y .Combining this information yields
t = 0.223 ( 0.693 5730 y ) = 1844 y . t = 0.223 ( 0.693 5730 y ) = 1844 y .The furniture is almost 2000 years old—an impressive discovery. The typical uncertainty on carbon-14 dating is about 5 % 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.
A radioactive nuclide has a high decay rate. What does this mean for its half-life and activity?
Visit the Radioactive Dating Game to learn about the types of radiometric dating and try your hand at dating some ancient objects.
This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.
Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.
Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction
Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction
© Jul 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Give today and help us reach more students.
