College Physics

# 31.5Half-Life and Activity

College Physics31.5 Half-Life and Activity

Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by the Curies, decay faster than uranium. This means they have shorter lifetimes, producing a greater rate of decay. In this section we explore half-life and activity, the quantitative terms for lifetime and rate of decay.

### Half-Life

Why use a term like half-life rather than lifetime? The answer can be found by examining Figure 31.21, which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life, $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$. Half of the remaining nuclei decay in the next half-life. Further, half of that amount decays in the following half-life. Therefore, the number of radioactive nuclei decreases from $NN size 12{N} {}$ to $N/2N/2 size 12{N/2} {}$ in one half-life, then to $N/4N/4 size 12{N/4} {}$ in the next, and to $N/8N/8 size 12{N/8} {}$ in the next, and so on. If $NN size 12{N} {}$ is a large number, then many half-lives (not just two) pass before all of the nuclei decay. Nuclear decay is an example of a purely statistical process. A more precise definition of half-life is that each nucleus has a 50% chance of living for a time equal to one half-life $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$. Thus, if $NN size 12{N} {}$ is reasonably large, half of the original nuclei decay in a time of one half-life. If an individual nucleus makes it through that time, it still has a 50% chance of surviving through another half-life. Even if it happens to make it through hundreds of half-lives, it still has a 50% chance of surviving through one more. The probability of decay is the same no matter when you start counting. This is like random coin flipping. The chance of heads is 50%, no matter what has happened before.

Figure 31.21 Radioactive decay reduces the number of radioactive nuclei over time. In one half-life $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$, the number decreases to half of its original value. Half of what remains decay in the next half-life, and half of those in the next, and so on. This is an exponential decay, as seen in the graph of the number of nuclei present as a function of time.

There is a tremendous range in the half-lives of various nuclides, from as short as $10−2310−23 size 12{"10" rSup { size 8{ - "23"} } } {}$ s for the most unstable, to more than $10161016 size 12{"10" rSup { size 8{"16"} } } {}$ y for the least unstable, or about 46 orders of magnitude. Nuclides with the shortest half-lives are those for which the nuclear forces are least attractive, an indication of the extent to which the nuclear force can depend on the particular combination of neutrons and protons. The concept of half-life is applicable to other subatomic particles, as will be discussed in Particle Physics. It is also applicable to the decay of excited states in atoms and nuclei. The following equation gives the quantitative relationship between the original number of nuclei present at time zero ($N0N0 size 12{N rSub { size 8{0} } } {}$) and the number ($NN size 12{N} {}$) at a later time $tt size 12{t} {}$:

$N=N0e−λt,N=N0e−λt, size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {}$
31.36

where $e=2.71828...e=2.71828... size 12{e=2 "." "71828" "." "." "." } {}$ is the base of the natural logarithm, and $λλ size 12{λ} {}$ is the decay constant for the nuclide. The shorter the half-life, the larger is the value of $λλ size 12{λ} {}$, and the faster the exponential $e−λte−λt size 12{e rSup { size 8{ - λt} } } {}$ decreases with time. The relationship between the decay constant $λλ size 12{λ} {}$ and the half-life $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$ is

$λ= ln(2) t1/2 ≈ 0.693t1/2.λ= ln(2) t1/2 ≈ 0.693t1/2. size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } } {}$
31.37

To see how the number of nuclei declines to half its original value in one half-life, let $t=t1/2t=t1/2 size 12{t=t rSub { size 8{1/2} } } {}$ in the exponential in the equation $N=N0e−λtN=N0e−λt$. This gives $N=N0 e−λt =N0e−0.693 =0.500N0N=N0 e−λt =N0e−0.693 =0.500N0$. For integral numbers of half-lives, you can just divide the original number by 2 over and over again, rather than using the exponential relationship. For example, if ten half-lives have passed, we divide $NN size 12{N} {}$ by 2 ten times. This reduces it to $N/1024N/1024 size 12{ {N} slash {"1024"} } {}$. For an arbitrary time, not just a multiple of the half-life, the exponential relationship must be used.

Radioactive dating is a clever use of naturally occurring radioactivity. Its most famous application is carbon-14 dating. Carbon-14 has a half-life of 5730 years and is produced in a nuclear reaction induced when solar neutrinos strike $14N14N size 12{"" lSup { size 8{"14"} } N} {}$ in the atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the ecosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Thus, if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you multiply that number by $1.3×10−121.3×10−12$ to find the number of $14C14C$ nuclei in the object. When an organism dies, carbon exchange with the environment ceases, and $14C14C$ is not replenished as it decays. By comparing the abundance of $14C14C$ in an artifact, such as mummy wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of $14C14C$ nuclei in them is greater. Very old biological materials contain no $14C14C$ at all. There are instances in which the date of an artifact can be determined by other means, such as historical knowledge or tree-ring counting. These cross-references have confirmed the validity of carbon-14 dating and permitted us to calibrate the technique as well. Carbon-14 dating revolutionized parts of archaeology and is of such importance that it earned the 1960 Nobel Prize in chemistry for its developer, the American chemist Willard Libby (1908–1980).

One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 31.22). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus, and so the shroud was never disregarded completely and remained controversial over the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92% of the $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ found in living tissues, allowing the shroud to be dated (see Example 31.4).

Figure 31.22 Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material. (credit: Butko, Wikimedia Commons)

### Example 31.4

#### How Old Is the Shroud of Turin?

Calculate the age of the Shroud of Turin given that the amount of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ found in it is 92% of that in living tissue.

#### Strategy

Knowing that 92% of the $14C14C$ remains means that $N/N0=0.92N/N0=0.92 size 12{N/N rSub { size 8{0} } =0 "." "92"} {}$. Therefore, the equation $N=N0e−λtN=N0e−λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {}$ can be used to find $λtλt size 12{λt} {}$. We also know that the half-life of $14C14C$ is 5730 y, and so once $λtλt size 12{λt} {}$ is known, we can use the equation $λ=0.693t1/2λ=0.693t1/2 size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } } {}$ to find $λλ size 12{λ} {}$ and then find $tt size 12{t} {}$ as requested. Here, we postulate that the decrease in $14C14C$ is solely due to nuclear decay.

#### Solution

Solving the equation $N=N0e−λtN=N0e−λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {}$ for $N/N0N/N0 size 12{N/N rSub { size 8{0} } } {}$ gives

$NN0=e−λt.NN0=e−λt. size 12{ { {N} over {N rSub { size 8{0} } } } =e rSup { size 8{-λt} } } {}$
31.38

Thus,

$0.92=e−λt.0.92=e−λt. size 12{0 "." "92"=e rSup { size 8{ - λt} } } {}$
31.39

Taking the natural logarithm of both sides of the equation yields

$ln0.92=–λtln0.92=–λt size 12{"ln "0 "." "92""=-"λt} {}$
31.40

so that

$−0.0834=−λt.−0.0834=−λt. size 12{ - 0 "." "0834"= - λt} {}$
31.41

Rearranging to isolate $tt size 12{t} {}$ gives

$t=0.0834λ.t=0.0834λ. size 12{t= { {0 "." "0834"} over {λ} } } {}$
31.42

Now, the equation $λ=0.693t1/2λ=0.693t1/2 size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } } {}$ can be used to find $λλ size 12{λ} {}$ for $14C14C size 12{"" lSup { size 8{"14"} } C} {}$. Solving for $λλ size 12{λ} {}$ and substituting the known half-life gives

$λ=0.693t1/2=0.6935730 y.λ=0.693t1/2=0.6935730 y. size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } = { {0 "." "693"} over {"5730"" y"} } } {}$
31.43

We enter this value into the previous equation to find $tt size 12{t} {}$:

$t=0.08340.6935730 y=690 y.t=0.08340.6935730 y=690 y. size 12{t= { {0 "." "0834"} over { { {0 "." "693"} over {"5730"" y"} } } } ="690"" y"} {}$
31.44

#### Discussion

This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. $1320±601320±60 size 12{"1320" +- "60"} {}$. The uncertainty is typical of carbon-14 dating and is due to the small amount of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived.

There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of $238U238U$. The decay series for $238U238U$ ends with $206Pb206Pb$, so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon-14 dating, the technique can be verified by a consistent body of knowledge. Since $238U238U$ has a half-life of $4.5×1094.5×109$ y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about $3.5×1093.5×109 size 12{3 "." 5 times "10" rSup { size 8{9} } } {}$ years ago.

### Activity, the Rate of Decay

What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define activity $RR size 12{R} {}$ to be the rate of decay expressed in decays per unit time. In equation form, this is

$R=ΔNΔtR=ΔNΔt size 12{R= { {ΔN} over {Δt} } } {}$
31.45

where $ΔNΔN size 12{ΔN} {}$ is the number of decays that occur in time $ΔtΔt size 12{Δt} {}$. The SI unit for activity is one decay per second and is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is,

$1Bq=1 decay/s.1Bq=1 decay/s. size 12{1" Bq"="1 decay/s"} {}$
31.46

Activity $RR size 12{R} {}$ is often expressed in other units, such as 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 $226Ra226Ra$, in honor of Marie Curie’s work with radium. The definition of curie is

$1 Ci=3.70×1010 Bq, 1 Ci=3.70×1010 Bq, size 12{1" Ci"=3 "." "70" times "10" rSup { size 8{"10"} } " Bq"} {}$
31.47

or $3.70×10103.70×1010 size 12{3 "." "70" times "10" rSup { size 8{"10"} } } {}$ decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. $1 MBq=100 microcuries (μCi)1 MBq=100 microcuries (μCi) size 12{"1 MBq"="100 microcuries " $$μ"Ci"$$ } {}$. In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq).

Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity $RR size 12{R} {}$ should be proportional to the number of radioactive nuclei, $NN size 12{N} {}$, and inversely proportional to their half-life, $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$. In fact, your intuition is correct. It can be shown that the activity of a source is

$R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$
31.48

where $NN size 12{N} {}$ is the number of radioactive nuclei present, having half-life $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$. This relationship is useful in a variety of calculations, as the next two examples illustrate.

### Example 31.5

#### How Great Is the $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ Activity in Living Tissue?

Calculate the activity due to $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

#### Strategy

To find the activity $RR size 12{R} {}$ using the equation $R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$, we must know $NN size 12{N} {}$ and $t1/2t1/2 size 12{t rSub { size 8{1/2} } } {}$. The half-life of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ can be found in Appendix B, and was stated above as 5730 y. To find $NN size 12{N} {}$, we first find the number of $12C12C size 12{"" lSup { size 8{"12"} } C} {}$ nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by $1.3×10−121.3×10−12 size 12{1 "." 3×"10" rSup { size 8{ +- "12"} } } {}$ (the abundance of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ in a carbon sample from a living organism) to get the number of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ nuclei in a living organism.

#### Solution

One mole of carbon has a mass of 12.0 g, since it is nearly pure $12C12C size 12{"" lSup { size 8{"12"} } C} {}$. (A mole has a mass in grams equal in magnitude to $AA size 12{A} {}$ found in the periodic table.) Thus the number of carbon nuclei in a kilogram is

$N(12C)= 6.02 × 1023 mol–1 12.0 g/mol ×(1000 g) =5.02×1025. N(12C)= 6.02 × 1023 mol–1 12.0 g/mol ×(1000 g) =5.02×1025.$
31.49

So the number of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ nuclei in 1 kg of carbon is

$N(14C)= (5.02×1025 )(1.3×10−12)=6.52×1013.N(14C)= (5.02×1025 )(1.3×10−12)=6.52×1013. size 12{N $$rSup { size 8{"14"} } C$$ = $$5 "." "02" times "10" rSup { size 8{"25"} }$$ $$1 "." 3 times "10" rSup { size 8{ - "12"} }$$ =6 "." "52" times "10" rSup { size 8{"13"} } } {}$
31.50

Now the activity $RR size 12{R} {}$ is found using the equation $R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$.

Entering known values gives

$R=0.693(6.52×1013)5730 y=7.89×109y–1,R=0.693(6.52×1013)5730 y=7.89×109y–1, size 12{R= { {0 "." "693" $$6 "." "52"´"10" rSup { size 8{"13"} }$$ } over {"5730"" y"} } =7 "." "89"´"10" rSup { size 8{9} } /y} {}$
31.51

or $7.89×1097.89×109 size 12{7 "." "89" times "10" rSup { size 8{9} } } {}$ decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,

$R = (7.89 × 10 9 y –1 ) 1.00 y 3. 16× 107 s=250 Bq, R = (7.89 × 10 9 y –1 ) 1.00 y 3. 16× 107 s=250 Bq, size 12{R=7 "." "89"´"10" rSup { size 8{9} } /y cdot { {1 "." "00"" y"} over {3 "." "16"´"10" rSup { size 8{7} } " s"} } ="250"" Bq"} {}$
31.52

or 250 decays per second. To express $RR size 12{R} {}$ in curies, we use the definition of a curie,

$R=250 Bq3.7×1010 Bq/Ci=6.76×10−9 Ci.R=250 Bq3.7×1010 Bq/Ci=6.76×10−9 Ci. size 12{R= { {"250"" Bq"} over {3 "." 7´"10" rSup { size 8{"10"} } " Bq/Ci"} } =6 "." "75"´"10" rSup { size 8{-9} } " Ci"} {}$
31.53

Thus,

$R=6.76nCi.R=6.76nCi. size 12{R=6 "." "75" "nCi"} {}$
31.54

#### Discussion

Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect $14C14C size 12{"" lSup { size 8{"14"} } C} {}$ in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less $14C14C size 12{"" lSup { size 8{"14"} } C} {}$, and for samples more than 50 thousand years old, it is impossible.

Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see Figure 31.23). Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of $131I131I$, $90Sr90Sr$, $137Cs137Cs$, $239Pu239Pu$, $238U238U$, and $235U235U$. Estimates are that the total amount of radiation released was about 100 million curies.

### Human and Medical Applications

Figure 31.23 The Chernobyl reactor. More than 100 people died soon after its meltdown, and there will be thousands of deaths from radiation-induced cancer in the future. While the accident was due to a series of human errors, the cleanup efforts were heroic. Most of the immediate fatalities were firefighters and reactor personnel. (credit: Elena Filatova)

### Example 31.6

#### What Mass of $137Cs137Cs$ Escaped Chernobyl?

It is estimated that the Chernobyl disaster released 6.0 MCi of $137Cs137Cs$ into the environment. Calculate the mass of $137Cs137Cs$ released.

#### Strategy

We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei $NN size 12{N} {}$ released. Since the activity $RR size 12{R} {}$ is given, and the half-life of $137Cs137Cs size 12{"" lSup { size 8{"137"} } "Cs"} {}$ is found in Appendix B to be 30.2 y, we can use the equation $R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$ to find $NN size 12{N} {}$.

#### Solution

Solving the equation $R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$ for $NN size 12{N} {}$ gives

$N=Rt1/20.693.N=Rt1/20.693. size 12{N= { { ital "Rt""" lSub { size 8{1/2} } } over {0 "." "693"} } } {}$
31.55

Entering the given values yields

$N=(6.0 MCi)(30.2 y)0.693.N=(6.0 MCi)(30.2 y)0.693. size 12{N= { { $$6 "." 0" MCi"$$ $$"30" "." 2" y"$$ } over {0 "." "693"} } } {}$
31.56

Converting curies to becquerels and years to seconds, we get

N = ( 6 . 0 × 10 6 Ci ) ( 3 . 7 × 10 10 Bq/Ci ) ( 30.2 y ) ( 3 . 16 × 10 7 s/y ) 0.693 = 3 . 1 × 10 26 . N = ( 6 . 0 × 10 6 Ci ) ( 3 . 7 × 10 10 Bq/Ci ) ( 30.2 y ) ( 3 . 16 × 10 7 s/y ) 0.693 = 3 . 1 × 10 26 . alignl { stack { size 12{N= { { $$6 "." 0´"10" rSup { size 8{6} } " Ci"$$ $$3 "." 7´"10" rSup { size 8{"10"} } " Bq/Ci"$$ $$"30" "." 2" y"$$ $$3 "." "16"´"10" rSup { size 8{7} } " s/y"$$ } over {0 "." "693"} } } {} # " "=3 "." 1´"10" rSup { size 8{"26"} } "." {} } } {}
31.57

One mole of a nuclide $AXAX size 12{"" lSup { size 8{A} } X} {}$ has a mass of $AA size 12{A} {}$ grams, so that one mole of $137Cs137Cs size 12{"" lSup { size 8{"137"} } "Cs"} {}$ has a mass of 137 g. A mole has $6.02 ×10236.02 ×1023 size 12{6 "." "02 " times "10" rSup { size 8{"23"} } } {}$ nuclei. Thus the mass of $137Cs137Cs size 12{"" lSup { size 8{"137"} } "Cs"} {}$ released was

m = 137 g 6.02 × 10 23 ( 3 . 1 × 10 26 ) = 70 × 10 3 g = 70 kg . m = 137 g 6.02 × 10 23 ( 3 . 1 × 10 26 ) = 70 × 10 3 g = 70 kg . alignl { stack { size 12{m= left ( { {"137"" g"} over {6 "." "02 "´"10" rSup { size 8{"23"} } } } right ) $$3 "." 1´"10" rSup { size 8{"26"} }$$ ="70"´"10" rSup { size 8{3} } " g"} {} # " "="70 kg" "." {} } } {}
31.58

#### Discussion

While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that Western reactors have a fundamentally safer design.

Activity $RR size 12{R} {}$ decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since $R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$, the activity decreases as the number of radioactive nuclei decreases. The equation for $RR size 12{R} {}$ as a function of time is found by combining the equations $N=N0e−λtN=N0e−λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {}$ and $R=0.693Nt1/2R=0.693Nt1/2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}$, yielding

$R=R0e−λt,R=R0e−λt, size 12{R=R rSub { size 8{0} } e rSup { size 8{ - λt} } } {}$
31.59

where $R0R0 size 12{R rSub { size 8{0} } } {}$ is the activity at $t=0t=0 size 12{t=0} {}$. This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation $R=R0e−λtR=R0e−λt size 12{R=R rSub { size 8{0} } e rSup { size 8{ - λt} } } {}$ must be used to find $RR size 12{R} {}$.

### PhET Explorations

#### Alpha Decay

Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.