half life problem solving

31.5 Half-Life and Activity

Learning objectives.

By the end of this section, you will be able to:

• Define half-life.
• Define dating.
• Calculate age of old objects by radioactive dating.

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.

Why use a term like half-life rather than lifetime? The answer can be found by examining Figure 31.19 , 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 , t 1 / 2 t 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 N N to N / 2 N / 2 in one half-life, then to N / 4 N / 4 in the next, and to N / 8 N / 8 in the next, and so on. If N 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 t 1 / 2 t 1 / 2 . Thus, if N 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.

There is a tremendous range in the half-lives of various nuclides, from as short as 10 − 23 10 − 23 s for the most unstable, to more than 10 16 10 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 ( N 0 N 0 ) and the number ( N N ) at a later time t t :

where e = 2.71828 ... e = 2.71828 ... is the base of the natural logarithm, and λ λ is the decay constant for the nuclide. The shorter the half-life, the larger is the value of λ λ , and the faster the exponential e − λt e − λt decreases with time. The relationship between the decay constant λ λ and the half-life t 1 / 2 t 1 / 2 is

To see how the number of nuclei declines to half its original value in one half-life, let t = t 1 / 2 t = t 1 / 2 in the exponential in the equation N = N 0 e − λt N = N 0 e − λt . This gives N = N 0 e − λt = N 0 e −0.693 = 0.500 N 0 N = N 0 e − λt = N 0 e −0.693 = 0.500 N 0 . 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 N N by 2 ten times. This reduces it to N / 1024 N / 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 14 N 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 − 12 1 . 3 × 10 − 12 to find the number of 14 C 14 C nuclei in the 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 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 14 C 14 C nuclei in them is greater. Very old biological materials contain no 14 C 14 C 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 31.20 ). 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 14 C 14 C found in living tissues, allowing the shroud to be dated (see Example 31.4 ).

Example 31.4

How old is the shroud of turin.

Calculate the age of the Shroud of Turin given that the amount of 14 C 14 C found in it is 92% of that in living tissue.

Knowing that 92% of the 14 C 14 C remains means that N / N 0 = 0 . 92 N / N 0 = 0 . 92 . Therefore, the equation N = N 0 e − λt N = N 0 e − λt can be used to find λt λt . We also know that the half-life of 14 C 14 C is 5730 y, and so once λt λt is known, we can use the equation λ = 0 . 693 t 1 / 2 λ = 0 . 693 t 1 / 2 to find λ λ and then find t t as requested. Here, we postulate that the decrease in 14 C 14 C is solely due to nuclear decay.

Solving the equation N = N 0 e − λt N = N 0 e − λt for N / N 0 N / N 0 gives

Taking the natural logarithm of both sides of the equation yields

Rearranging to isolate t t gives

Now, the equation λ = 0 . 693 t 1 / 2 λ = 0 . 693 t 1 / 2 can be used to find λ λ for 14 C 14 C . Solving for λ λ and substituting the known half-life gives

We enter this value into the previous equation to find t t :

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 ± 60 1320 ± 60 . The uncertainty is typical of carbon-14 dating and is due to the small amount of 14 C 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 238 U 238 U . The decay series for 238 U 238 U ends with 206 Pb 206 Pb , 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 238 U 238 U has a half-life of 4 . 5 × 10 9 4 . 5 × 10 9 y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about 3 . 5 × 10 9 3 . 5 × 10 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 R R to be the rate of decay expressed in decays per unit time. In equation form, this is

where Δ N Δ N is the number of decays that occur in time Δ t Δ t . 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,

Activity R 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 226 Ra 226 Ra , in honor of Marie Curie’s work with radium. The definition of curie is

or 3 . 70 × 10 10 3 . 70 × 10 10 decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. 1 MBq = 100 microcuries ( μ Ci ) 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 R R should be proportional to the number of radioactive nuclei, N N , and inversely proportional to their half-life, t 1 / 2 t 1 / 2 . In fact, your intuition is correct. It can be shown that the activity of a source is

where N N is the number of radioactive nuclei present, having half-life t 1 / 2 t 1 / 2 . This relationship is useful in a variety of calculations, as the next two examples illustrate.

Example 31.5

How great is the 14 c 14 c activity in living tissue.

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.

To find the activity R R using the equation R = 0 . 693 N t 1 / 2 R = 0 . 693 N t 1 / 2 , we must know N N and t 1 / 2 t 1 / 2 . The half-life of 14 C 14 C can be found in Appendix B , and was stated above as 5730 y. To find N N , we first find the number of 12 C 12 C nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by 1 . 3 × 10 − 12 1 . 3 × 10 − 12 (the abundance of 14 C 14 C in a carbon sample from a living organism) to get the number of 14 C 14 C nuclei in a living organism.

One mole of carbon has a mass of 12.0 g, since it is nearly pure 12 C 12 C . (A mole has a mass in grams equal in magnitude to A A found in the periodic table.) Thus the number of carbon nuclei in a kilogram is

So the number of 14 C 14 C nuclei in 1 kg of carbon is

Now the activity R R is found using the equation R = 0 . 693 N t 1 / 2 R = 0 . 693 N t 1 / 2 .

Entering known values gives

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,

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

Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of 14 C 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 14 C 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 14 C 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 31.21 ). 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 131 I 131 I , 90 Sr 90 Sr , 137 Cs 137 Cs , 239 Pu 239 Pu , 238 U 238 U , and 235 U 235 U . Estimates are that the total amount of radiation released was about 100 million curies.

Human and Medical Applications

Example 31.6, what mass of 137 cs 137 cs escaped chernobyl.

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

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 N N released. Since the activity R R is given, and the half-life of 137 Cs 137 Cs is found in Appendix B to be 30.2 y, we can use the equation R = 0 . 693 N t 1 / 2 R = 0 . 693 N t 1 / 2 to find N N .

Solving the equation R = 0 . 693 N t 1 / 2 R = 0 . 693 N t 1 / 2 for N N gives

Entering the given values yields

Converting curies to becquerels and years to seconds, we get

One mole of a nuclide A X A X has a mass of A A grams, so that one mole of 137 Cs 137 Cs has a mass of 137 g. A mole has 6 . 02 × 10 23 6 . 02 × 10 23 nuclei. Thus the mass of 137 Cs 137 Cs released was

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 more recent reactors have a fundamentally safer design.

Activity R 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 . 693 N t 1 / 2 R = 0 . 693 N t 1 / 2 , the activity decreases as the number of radioactive nuclei decreases. The equation for R R as a function of time is found by combining the equations N = N 0 e − λt N = N 0 e − λt and R = 0 . 693 N t 1 / 2 R = 0 . 693 N t 1 / 2 , yielding

where R 0 R 0 is the activity at t = 0 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 = R 0 e − λt R = R 0 e − λt must be used to find R 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.

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How to Calculate Half Life

Last Updated: May 26, 2023 References

Understanding Half-Life

Learning the half-life equation, calculating from a graph, using a calculator, example problems, calculator, practice problems, and answers, expert q&a.

This article was co-authored by Meredith Juncker, PhD and by wikiHow staff writer, Hannah Madden . Meredith Juncker is a PhD candidate in Biochemistry and Molecular Biology at Louisiana State University Health Sciences Center. Her studies are focused on proteins and neurodegenerative diseases. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been viewed 1,159,315 times.

The half-life of a substance undergoing decay is the time it takes for the amount of the substance to decrease by half. It was originally used to describe the decay of radioactive elements like uranium or plutonium, but it can be used for any substance which undergoes decay along a set, or exponential, rate. You can calculate the half-life of any substance, given the rate of decay, which is the initial quantity of the substance and the quantity remaining after a measured period of time. [1] X Research source

• Elements like uranium and plutonium are most often studied with half-life in mind.

• Therefore, you can calculate the half-life for a particular element and know for certain how quickly it will break down no matter what.

• Technically, there are 2 types of carbon: carbon-14, which decays, and carbon-12, which stays constant.

• Simply replacing the variable doesn't tell us everything, though. We still have to account for the actual half-life, which is, for our purposes, a constant.

• On half-life graphs, the x-axis will usually show the timeline, while the y-axis usually shows the rate of decay.

• For example, if the starting point is 1,640, divide 1,640 / 2 to get 820.
• If you are working with a semi log  plot, meaning the count rate is not evenly spaced, you’ll have to take the logarithm of any number from the vertical axis. [11] X Research source

• If you know the half-life but you don’t know the initial quantity, you can input the half-life, the quantity that remains, and the time that has passed. As long as you know 3 of the 4 values, you’ll be able to use a half-life calculator.

• If you don’t know the half-life but you do know the decay constant and the mean lifetime, you can input those instead. Just like the initial equation, you only need to know 2 of the 3 values to get the third one.

• This is a helpful visual, and it can be useful if you don’t want to do all of the equation work.

• Check to see if the solution makes sense. Since 112 g is less than half of 300 g, at least one half-life must have elapsed. Our answer checks out.

• Substitute and evaluate.

• Remember to check your solution intuitively to see if it makes sense.

• For this particular equation, the actual length of the half-life did not play a role.

You Might Also Like

• ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Half-lives_and_Pharmacokinetics
• ↑ https://chem.libretexts.org/Courses/Furman_University/CHM101%3A_Chemistry_and_Global_Awareness_(Gordon)/05%3A_Basics_of_Nuclear_Science/5.07%3A_Calculating_Half-Life
• ↑ https://atomic.lindahall.org/what-is-meant-by-half-life.html
• ↑ http://faculty.bard.edu/belk/math213/ExponentialDecay.pdf
• ↑ https://www.ausetute.com.au/halflife.html
• ↑ https://socratic.org/chemistry/nuclear-chemistry/nuclear-half-life-calculations
• ↑ https://www.khanacademy.org/test-prep/mcat/physical-processes/atomic-nucleus/a/decay-graphs-and-half-lives-article
• ↑ https://www.gcsescience.com/prad17-measuring-half-life.htm
• ↑ https://www.calculator.net/half-life-calculator.html
• ↑ https://www.youtube.com/watch?v=5mKrIv1lo1E&feature=youtu.be&t=163
• ↑ https://www.chemteam.info/Radioactivity/Radioactivity-Half-Life-probs1-10.html

About This Article

To find the half life of a substance, or the time it takes for a substance to decrease by half, you’ll be using a variation of the exponential decay formula. Plug in ½ for a, use the time for x, and multiply the left side by the initial quantity of the substance. Rearrange the equation so that you’re solving for what the problem asks for, whether that’s half life, mass, or another value. Plug in the values you have and solve, writing the answer in seconds, days, or years. To see the half life equation and look at examples, read on! Did this summary help you? Yes No

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(1/2) number of half-lives = decimal amount remaining

(1/2) 0 = 1

(1/2) 1 = 0.5

(1/2) 2 = 0.25 (1/2) 3 = 0.125 (1/2) 4 = 0.0625

1/2 2 = 1/4

half-lives exponent fraction decimal one 1/2 1 1/2 0.5 two 1/2 2 1/4 0.25 three 1/2 3 1/8 0.125 four 1/2 4 1/16 0.0625 five 1/2 5 1/32 0.03125

(1/2) 2.45 = 0.18301 (1/2) 0.5882 = 0.66517

total time elasped ÷ length of half-life = number of half-lives elapsed

17190 yr / 5730 yr = 3 half-lives

12300 / 5730 = 2.1466 half-lives

(1/2) 2.1466 = 0.225844

starting amount times decimal amount remaining = remaining amount

(N o ) (decimal amount remaining) = N

decimal amount remaining = (1/2) total time / half-life time

(N o ) (1/2) total time / half-life time = N

(88.0) (1/2) x / 12.3 = 11.0

11.0 g / 88.0 g = 0.125

(1/2) n = 0.125 n log 0.5 = log 0.125 n = log 0.125 / log 0.5 n = 3

12.3 yr times 3 = 36.9 yr

(N o ) (1/2) total time / half-life time = N (88.0) (1/2) x/12.3 = 11.0 (1/2) x/12.3 = 0.125 log (1/2) x/12.3 = log 0.125 (x/12.3) (−0.30103) = -0.90309 x/12.3 = 3 x = 36.9 yr

23.5 g / 84.0 g = 0.279762 Note that I rounded off, but not too much.

(1/2) n = 0.279762 n log 0.5 = log 0.279762 n = 1.837728

12.3 yr times 1.837728 = 22.6 yr

13.0 g / 208 g = 0.0625

recognize 0.0625 as being associated with 4 half-lives or (1/2) n = 0.0625 n log 0.5 = log 0.0625 n = 4

60.0 hr / 4 = 15.0 hr

(208) (1/2) 60/n = 13

0.0625 = (1/2) 60/n Note: the 0.0625 comes from 13 divided by 208

log 0.0625 = log (1/2) (60/n) log 0.0625 = (60/n) log 0.5 −1.2041 = (60/n) (−0.3010)

4 = 60/n 4n = 60 n = 15 hr

0.598 / 27.5 = 0.02174545 (1/2) n = 0.02174545 n = 5.52314 574 yr / 5.52314 = 104 yr (to three sig figs)

(1/2) n = 0.40 n log 0.5 = log 0.40 n = 1.32193

5730 yr times 1.32193 = 7575 yr

100 yr / 25 yr = 4 (1/2) 4 = 0.0625 20 g / 0.0625 = 320 g Here's the 320 g decaying over four half-lives: 320 ---> 160 160 ---> 80 80 ---> 40 40 ---> 20

4 days / 14.3 days = 0.27972

(1/2) 0.27972 = 0.82375

73.3 is to 0.82375 as x is to 1 x = 88.9833 g rounded to three sig figs, 89.0 g

30 s / 7.5 s = 4 (1/2) 4 = 0.0625 x is to 1 as 144 is to 0.0625 144 / 0.0625 = 2304

100% minus 3.402% = 96.598% We will use it as 0.96598

(1/2) n = 0.96598 n log 0.5 = log 0.96598 n = 0.049935

1 day is to 0.049935 as x is to 1 x = 20 days

(1/2) n = 0.4324 n log 0.5 = log 0.4324 n = log 0.4324 / log 0.5 n = 1.20956 1.20956 times 0.007685 yr = 0.00930 yr 0.00930 yr times (365 day / yr) times (24 hr / day) = 81.468 hr To three sig figs, this is 81.5 hours

(1/2) n = 0.795 n log 0.5 = log 0.795 n = 0.330973 The problem does not provide the half-life of C-14. We look it up and find it to be 5730 year. 0.330973 times 5730 yr = 1896.47529 Rounded off to three significant figures, the answer would be 1.90 x 10 3 years. (Using 1900 would be wrong, as that shows only two sig figs. Using 1900. would also be incorrect, as that shows four SF.)

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3.1: Half-Life

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Learning Objectives

• To define half-life .
• To solve word problems involving half-life, time, and mass of a radioactive substance.

Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life . The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

Consider the following example. Suppose we have 100.0 g of 3 H (tritium, a radioactive isotope of hydrogen). It has a half-life of 12.3 y. After 12.3 y, half of the sample will have decayed to 3 He by emitting a beta particle, so that only 50.0 g of the original 3 H remains. After another 12.3 y—making a total of 24.6 y—another half of the remaining 3 H will have decayed, leaving 25.0 g of 3 H. After another 12.3 y—now a total of 36.9 y—another half of the remaining 3 H will have decayed, leaving 12.5 g of 3 H. This sequence of events is illustrated in Figure \(\PageIndex{1}\).

We can determine the amount of a radioactive isotope remaining after a given number half-lives by using the following expression:

\[\mathrm{amount\: remaining=initial\: amount\times \left ( \dfrac{1}{2} \right )^n} \label{Eq1}\]

where n is the number of half-lives. This expression works even if the number of half-lives is not a whole number.

Example \(\PageIndex{1}\): Fluorine-20

The half-life of 20 F is 11.0 s. If a sample initially contains 5.00 g of 20 F, how much 20 F remains after 44.0 s?

If we compare the time that has passed to the isotope’s half-life, we note that 44.0 s is exactly 4 half-lives, so we can use Equation \ref{Eq1} with \(n = 4\). Substituting and solving results in the following:

\[ \begin{align*} \mathrm{amount\: remaining} &=5.00\: g \times \left(\dfrac{1}{2}\right)^4 \\[4pt] &=5.00\: g \times \dfrac{1}{16} \\[4pt] &=0.313\: g \end{align*}\]

Less than one-third of a gram of 20 F remains.

Exercise \(\PageIndex{1}\): Titanium-44

The half-life of 44 Ti is 60.0 y. A sample initially contains 0.600 g of 44 Ti. How much 44 Ti remains after 180.0 y?

Half-lives of isotopes range from fractions of a microsecond to billions of years. Table \(\PageIndex{1}\) lists the half-lives of some isotopes.

Example \(\PageIndex{2}\): Iodine-131

The isotope \(\ce{I}\)-131 is used in treatment for thyroid cancer and has a half-life of 8.05 days. If a clinic initially has 400 mg, how long will it take for the sample to decay to only 25 mg of \(\ce{I}\)-131?

We begin by determining how many half-lives it would take for the sample to decay from 400 mg to 25 mg of \(\ce{I}\)-131. There are several ways to do this! We know that for each half-life the mass of \(\ce{I}\)-131 is reduced to one half of the previous value so we could solve this conceptually: 

\[ \begin{align*} \dfrac{400 \: \text{mg}}{2} = 200 \: \text{mg} \\ \dfrac{200 \: \text{mg}}{2} = 100 \: \text{mg} \\ \dfrac{100 \: \text{mg}}{2} = 50 \: \text{mg} \\ \dfrac{50 \: \text{mg}}{2} = 25 \: \text{mg} \end{align*}\]

We had to divide the original mass by 2 four times to get the final mass. Therefore, it took four half-lives to get from 400 mg to 25 mg of \(\ce{I}\)-131. 

Once the number of half-lives has been determined, we can calculate the total time. 

\[ \begin{align*} \text{total time} = \text{half-life}\times \text{number of half-lives}  \\ = 8.05 \: \text{days} \times 4 \: \text{half-lives} = 32.2 \: \text{days} \end{align*} \]

A more sophisticated way to solve the problem is to make use of percentages or fractions. The graph below shows the percentage of \(\ce{I}\)-131 remaining after each half-life. These percentages are always the same, regardless of the initial mass of \(\ce{I}\)-131. 

Typical radioactive decay curve.

The graph above illustrates a typical decay curve. Since the half-life for \(\ce{I}-131\) is 8.05 days, the x-axis has units of days. The y-axis indicates the percentage of sample remaining. Initially, (at elapsed time = 0), 100 % of the sample remains. After one half-life (8.05 days for \(\ce{I}-131\)), 50 % remains. After two half-lives, 25 % of the initial sample remains. We can use this pattern as an alternate way of calculating the number of half-lives when the initial and final masses are known. 

Alternate method of solving: 

\[\dfrac{\text{final mass}}{\text{initial mass}}\times 100 = \text{percentage remaining} \]

\[\dfrac{25 \: \text{mg}}{400 \: \text{mg}}\times 100 = 6.25 \text{%} \nonumber \]

This result, 6.25 % remaining, corresponds to four half-lives. Careful! Don't count 100 % as the first half-life. At time = 0, nothing has happened yet. 50 % remaining = 1 half-life, 25 % remaining = 2 half-lives, 12.5 % remaining = 3 half-lives, and 6.25 % remaining = 4 half-lives. These percentages are the same for all isotopes so you could calculate them once and then use them for multiple problems. 

To finish this problem, multiply the number of half-lives by the length of each half-life to get the total time. 

Exercise \(\PageIndex{2}\)

A sample of \(\ce{Ac}\)-225 originally contained 80 grams and after 50 days only 2.5 grams of the original \(\ce{Ac}\)-225 remain. What is the half life of \(\ce{Ac}\)-225?

10 days. Notice that the phrase "what is the half-life?" means "how long is the half-life?".

Looking Closer: Half-Lives of Radioactive Elements

Many people think that the half-life of a radioactive element represents the amount of time an element is radioactive. In fact, it is the time required for half—not all—of the element to decay radioactively. Occasionally, however, the daughter element is also radioactive, so its radioactivity must also be considered.

The expected working life of an ionization-type smoke detector (described in the opening essay) is about 10 years. In that time, americium -241, which has a half-life of about 432 y, loses less than 4% of its radioactivity. A half-life of 432 y may seem long to us, but it is not very long as half-lives go. Uranium-238, the most common isotope of uranium, has a half-life of about 4.5 × 10 9 y, while thorium-232 has a half-life of 14 × 10 9 y.

On the other hand, some nuclei have extremely short half-lives, presenting challenges to the scientists who study them. The longest-lived isotope of lawrencium , 262 Lr, has a half-life of 3.6 h, while the shortest-lived isotope of lawrencium, 252 Lr, has a half-life of 0.36 s. As of this writing, the largest atom ever detected has atomic number 118, mass number 293, and a half-life of 120 ns. Can you imagine how quickly an experiment must be done to determine the properties of elements that exist for so short a time?

Key Takeaways

• Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively.
• The amount of material left over after a certain number of half-lives can be easily calculated.

Concept Review Exercises

• Define half-life .
• Describe a way to determine the amount of radioactive isotope remaining after a given number of half-lives.
• Half-life is the amount of time needed for half of a radioactive material to decay.
• take half of the initial amount for each half-life of time elapsed

10 half life problems and answers examples

Expert Answer(s) - 1

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Chemistry Tutor

Still stuck with a Chemistry question

1. A radioisotope decays from 150 mg to 120.2 mg in 5 days. Calculate the half-life of this isotope.

          From N t = 1 2 t t 1 / 2 N o                                                                                                  

we rearrange this equation to take the form

              t 1 / 2 =  t   log 2 log N o N t

                     = 5 log 2 log 150 120 . 2

                     = 15.6482 days

2. Iodine-131 has a half-life of 8 days. What is the mass of this nuclide that remaines when 40 g of it decays in 48 hours?

We use the relalation  N t =  1 2 t t 1 / 2 N o

                                     =  0 . 5 48 8 X 40

                                     = 0.6250 g

3. A sample of Pd-100 decayed to a mass of 30 mg in 16 days. Given that the half-life of Pd is 4 days, calculate the initial mass of the sample

We are required to find  N o , when we have  N t   and t. rearranging the equation used in example 2, we obtain;

              N o =  2 t t 1 / 2   N t

                   = 2 16 4 X  30

                   = 480 g

4. Given that 50.0 g of C-14 decayed in 4 half-lives, how much of this isotope is left? C-14 Half-Life = 5730 Years.

We utilize the equation that relate amount remaining, initial mass and number of half-lives,n.

              N t =  1 2 n   X  N o

                   =  ( 1 2 ) 4   X 50

                   = 3.125 g

5. What is the half-life of an isotope that is 80 % remained after 16 days?

      % remaining= 80 100

 Therefore  N t = 80,  N o = 100, Now using the half-life equation in example 1, we have

                  t 1 / 2 =  16   log 2 log   100 80

                         = 49.7 days 

6. A certain radioisotope has a half-life of 9 days. What percentage of an initial mass of this isotope remains after 25 days?

This question requires that we calculate N t , when we have t and  t 1 / 2 ,  

                  N t =  0 . 5 t t 1 / 2 N o ,

             percentage = N t N o   x   100

                              =  0 . 5 t t 1 / 2 x 100

                              = 14.58 %

7.A radioactive source has a half-life of 80 s. How long will it take for 5/6 of the source to decay?

Fraction remaining =  1 6 , in this case  N t = 1,  N o = 6, we are required to determine t, when we have  t 1 / 2 , now from N t =  N o x  0 . 5 t t 1 / 2 ,we obtain the relation for calculating t as;

                        t =  t 1 / 2   log   N o N t log   2

                          =  80   log   1 6 log   2

                          = 206.797 seconds

8. Achaelogist dated Holy shroud using radiocarbon ( C 14 ), he found out that the amount of C 14 found in Holy shroud is 92 % of that in the living tossue (half-life of carbon-14=5730 years)

a) Calculate the decay constant for C 14  assuming that it decreases solely due to nuclear decay 

  Solution

         λ = 0 . 693 t 1 / 2

          = 0.0001 y e a r - 1

b) Calculate the age of Shroud of Turin

            N t =  N o e - λ t

              t =  1 λ ln N o N t

                = 5730 0 . 693 ln   100 72

                = 690 years

9. The activity of a radioactive source decreased from 6000  per minute to 500 counts per minute in 10 hours. Calculate the half-life of the radioactive source.

We use the equation  A t =  ( 1 2 ) t t 1 / 2   A o , where  A t is the activity in time t,  A o is the original activity

                                    500 = ( 1 2 ) 10 t 1 / 2 ×   6000

                                    t 1 / 2 =  10   log   2 log   12

                                          =  2 . 789   h o u r s

We can also use the relation  A t =  ( 1 2 ) n A o , where n is the number of half-lives

                                            A t  =  A o 2 n

                                            n =  t t 1 / 2 =  l og   A o A t log   2  

                                          t 1 / 2 =    10   log   2 log   6000 500  

                                                  = 2.789 hours

10. Calculate the percentage of the original radioisotope that remained after three half-lives elapsed

   %   r e m a i n e d =

                     = 0 . 5 3   ×   100

                     =  12 . 5   %

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How to Solve Half-Life Problems | Step-by-Step Guide and Examples

By Happy Sharer

Introduction

Half-life problems involve calculations that determine how long it takes for a given amount of a substance to decrease by half. They are commonly used in fields such as chemistry, physics, and biology, and can be used to model a variety of phenomena, including the decay of radioactive elements and the rate of chemical reactions. This article provides a step-by-step guide on how to solve half-life problems, along with different types of problems and examples of how to tackle them.

Step-by-Step Guide on How to Solve Half-Life Problems

Half-life problems can be solved using a five-step process. First, you need to identify the type of problem. Then, you need to calculate the initial amount of substance. Next, you need to calculate the decay constant. After that, you must apply the equation for half-life. Finally, you should check your answer.

Step 1: Identify the Type of Problem

The first step in solving half-life problems is to identify the type of problem. There are three main types of half-life problems: exponential decay, radioactive decay, and first-order kinetics. Each type of problem requires a slightly different approach, so it’s important to know which one you’re dealing with.

Step 2: Calculate the Initial Amount of Substance

Once you’ve identified the type of problem, you need to calculate the initial amount of substance. This can be done by finding the total amount of substance at the beginning of the problem and subtracting any amount that has already been lost. For example, if you’re given a problem involving the decay of a radioactive element, you would need to subtract the amount that has already decayed from the initial amount.

Step 3: Calculate the Decay Constant

The next step is to calculate the decay constant. The decay constant is a measure of how quickly the substance is decaying. It can be calculated by dividing the rate of decay by the initial amount of substance. For example, if the rate of decay is 0.5 grams per second and the initial amount of substance is 10 grams, then the decay constant is 0.05 grams per second.

Step 4: Apply the Equation for Half-Life

Once you’ve calculated the decay constant, you can use the equation for half-life to calculate how long it will take for the amount of substance to decrease by half. The equation is: t = ln(2) / λ , where t is the time it takes for the amount of substance to decrease by half, ln is the natural logarithm, and λ is the decay constant. For example, if the decay constant is 0.05, then the half-life is 14.2 seconds.

Step 5: Check Your Answer

The final step is to check your answer. You can do this by comparing your result to known values or by performing a simple calculation to confirm your answer. For example, if you were given a problem involving the decay of a radioactive element, you could compare your answer to the known half-life of the element. If your answer is within a reasonable range, then you can be confident that you have solved the problem correctly.

Different Types of Half-Life Problems and How to Tackle Them

In addition to the steps outlined above, there are several different types of half-life problems that you may encounter. Each type of problem requires a slightly different approach, so it’s important to understand the differences between them.

Exponential Decay

Exponential decay occurs when the rate of decay is proportional to the amount of substance present. This type of problem is often encountered when dealing with radioactive elements. To solve an exponential decay problem, you need to calculate the initial amount of substance, the decay constant, and the half-life.

Radioactive Decay

Radioactive decay occurs when an unstable nucleus breaks down into two or more smaller nuclei. This type of problem is often encountered when dealing with radioactive elements. To solve a radioactive decay problem, you need to calculate the initial amount of substance, the decay constant, and the half-life.

First-Order Kinetics

First-order kinetics occurs when the rate of a reaction is determined by the concentration of reactants. This type of problem is often encountered when dealing with chemical reactions. To solve a first-order kinetics problem, you need to calculate the initial amount of reactant, the rate constant, and the half-life.

Examples of Half-Life Problems and How to Solve Them

To help illustrate the process of solving half-life problems, here are three examples of how to tackle them. In each example, we walk through the steps of identifying the type of problem, calculating the initial amount of substance, calculating the decay constant, applying the equation for half-life, and checking the answer.

You are given the following information about a radioactive element: The initial amount of the element is 20 grams, and the rate of decay is 0.5 grams per second. What is the half-life of the element?

In this example, we are dealing with a radioactive decay problem. The first step is to calculate the initial amount of substance, which is 20 grams. Next, we need to calculate the decay constant. This can be done by dividing the rate of decay (0.5 grams per second) by the initial amount of substance (20 grams), giving us a decay constant of 0.025 grams per second. Finally, we can use the equation for half-life ( t = ln(2) / λ ) to calculate the half-life. Plugging in the decay constant of 0.025, we get a half-life of 27.8 seconds.

You are given the following information about a chemical reaction: The initial concentration of reactant A is 0.5 moles per liter, and the rate constant is 0.1 moles per liter per second. What is the half-life of the reaction?

In this example, we are dealing with a first-order kinetics problem. The first step is to calculate the initial amount of reactant, which is 0.5 moles per liter. Next, we need to calculate the rate constant. This can be done by dividing the rate of reaction (0.1 moles per liter per second) by the initial concentration of reactant (0.5 moles per liter), giving us a rate constant of 0.2 moles per liter per second. Finally, we can use the equation for half-life ( t = ln(2) / λ ) to calculate the half-life. Plugging in the rate constant of 0.2, we get a half-life of 3.5 seconds.

You are given the following information about a radioactive element: The initial amount of the element is 40 grams, and the rate of decay is 0.25 grams per second. What is the half-life of the element?

In this example, we are dealing with a radioactive decay problem. The first step is to calculate the initial amount of substance, which is 40 grams. Next, we need to calculate the decay constant. This can be done by dividing the rate of decay (0.25 grams per second) by the initial amount of substance (40 grams), giving us a decay constant of 0.00625 grams per second. Finally, we can use the equation for half-life ( t = ln(2) / λ ) to calculate the half-life. Plugging in the decay constant of 0.00625, we get a half-life of 112 seconds.

Mathematics Behind Half-Life Problems and How to Utilize It

Half-life problems require an understanding of basic mathematics, including the formula for half-life, logarithms, and the calculation of decay constants. Here, we provide an overview of these topics and explain how they can be used to solve half-life problems.

The Formula for Half-Life

The formula for half-life is t = ln(2) / λ , where t is the time it takes for the amount of substance to decrease by half, ln is the natural logarithm, and λ is the decay constant. This formula can be used to calculate the half-life of any given substance or reaction.

Using Logarithms to Solve Problems

Logarithms can be used to simplify the calculation of half-life. For example, instead of calculating the half-life directly from the formula, you can use a logarithmic equation to solve for the decay constant, which can then be plugged into the formula for half-life. This simplifies the calculation and makes it easier to check your answer.

Calculating the Decay Constant

The decay constant is a measure of how quickly the substance is decaying. It can be calculated by dividing the rate of decay by the initial amount of substance. For example, if the rate of decay is 0.5 grams per second and the initial amount of substance is 10 grams, then the decay constant is 0.05 grams per second.

Strategies for Working Through Half-Life Problems Quickly and Accurately

Half-life problems can be challenging, but there are several strategies that can help make them easier to solve. These include estimating solutions, developing a systematic approach, and double-checking your answers.

Estimating Solutions

One strategy for working through half-life problems is to estimate the solution before attempting to solve the problem. This can be done by examining the data and making educated guesses about the answer. Once you have an estimate, you can use it to check your work and make sure that your answer is reasonable.

Developing a Systematic Approach

Another strategy for working through half-life problems is to develop a systematic approach. This involves breaking the problem down into smaller steps and focusing on one step at a time. By taking a systematic approach, you can ensure that you don’t miss any important details and that your answer is accurate.

Double-Checking Your Answers

Finally, it’s important to double-check your answers. This can be done by comparing your result to known values or by performing a simple calculation to confirm your answer. For example, if you were given a problem involving the decay of a radioactive element, you could compare your answer to the known half-life of the element. If your answer is within a reasonable range, then you can be confident that you have solved the problem correctly.

Tips and Resources for Learning and Understanding Half-Life Problems

Half-life problems can be tricky, but there are several tips and resources that can help you learn and understand them. These include practice problems, online tutorials, and textbooks.

Practice Problems

One way to learn how to solve half-life problems is to practice them. There are many websites that offer free practice problems, and some textbooks even include practice problems at the end of chapters. Practicing half-life problems can help you become familiar with the process and improve your problem-solving skills.

Online Tutorials

Another way to learn about half-life problems is to watch online tutorials. There are many videos and tutorials available on the internet that provide step-by-step instructions for solving half-life problems. Watching these tutorials can help you gain a better understanding of the concepts and methods involved.

Finally, textbooks can be a great resource for learning about half-life problems. Many textbooks include detailed explanations of the concepts behind half-life problems and provide examples of how to solve them. Reading textbooks can help you gain a deeper understanding of the material and give you the knowledge you need to solve more complex problems.

Half-life problems can be solved using a step-by-step approach. To solve a half-life problem, you need to identify the type of problem, calculate the initial amount of substance, calculate the decay constant, apply the equation for half-life, and check your answer. Different types of half-life problems require different approaches, and examples are provided to illustrate how to tackle them. Mathematics is also an important part of solving half-life problems, and strategies such as estimating solutions, developing a systematic approach, and double-checking your answers can help you work through them quickly and accurately. Finally, practice problems, online tutorials, and textbooks are all valuable resources for learning and understanding half-life problems.

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Hi, I'm Happy Sharer and I love sharing interesting and useful knowledge with others. I have a passion for learning and enjoy explaining complex concepts in a simple way.

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Chemistry Steps

General Chemistry

Nuclear chemistry.

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. 

• Alpha Particles, Beta Particles, and Gamma Rays
• Balancing Nuclear Reactions
• Carbon Dating Practice Problems
• The energy  of Nuclear Reactions
• Nuclear Binding Energy
• Nuclear Fission and Fusion

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

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

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

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?

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?

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.

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?

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?

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?

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?

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?

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.

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?

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.

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?

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?

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?

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.

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?

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.

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Half-life Problems And Answers Examples

• by Abnurlion
• April 13, 2023 November 11, 2023

Half-life Problems And Answers Examples: Here are 18 half-life problems and answers to help you understand how to solve a question about half-life:

In 24 days, a radioactive isotope decreased in mass from 64 grams to 2 grams. What is the half-life of the radioactive material?

The half-life of the radioactive material is 4.8 days.

Explanation

We are going to apply two methods to arrive at our answer

Method 1: Conventional method

64 g to 32 g = 1 half-life

32 g to 16 g = 2 half-life

16 g to 8 g = 3 half-life

8 g to 4 g = 4 half-life

4 g to 2 g = 5 half-life

If 5 half-life is equal to 24 days. Then 1 half-life will be T

Therefore, T x 5 half-life = 1 half-life x 24 days

T = (1 half-life x 24 days) / 5 half-life = 4.8 days

Method 2: Zhepwo Method

Decay time , t = 24 days

Half-life, T = ?

Initial mass, N 1 = 64 g

Final mass, N 2 = 2 g

We will use the formula

R = N 1 / N 2 = 64 / 2 = 32

T = t / (log 2 R) = 24 / (log 2 32) = 24 / 5 = 4.8 days

Therefore, the half-life of the radioactive material is 4.8 days

The half-life of radioactive material is 6 hours. What quantity of 1 kg of the material would decay in 24 hours?

The final answer to the above question is (15/16) 0.9375 kg

We will also apply two methods to solve the above question

Method 1: Conventional Method

After 6 hours, 1/2 kg decays, and 1/2 kg will remain

Another 6 hours, 1/2 of 1/2 kg decays, 1/4 kg remains

6 hours after, 1/2 of 1/4 kg decays, 1/8 kg remains

After another 6 hours, 1/2 of 1/8 kg decays, 1/16 kg remains

Therefore, the material decayed would be:

1 kg – 1/16 kg = (15 / 16) kg = 0.9375 kg

Half-life, T = 6 hrs

Initial mass, N 1 = 1 kg

Mass of the decayed material, N d = ?

This implies that

n = t / T = 24 / 6 = 4

R = 2 n = 2 4 = 16

N d = N 1 (R – 1 / R) = 1 (16-1 / 16) = (15/16) kg

Therefore, the quantity of 1 kg of the material that would decay in 24 hours is 15/16 kilograms or 0.9375 kilograms.

A radioactive sample initially contains N atoms. After three (3) half-lives, what is the number of atoms that have disintegrated?

The final answer to the above question is (7/8) N

We will use two methods to solve the above question

First step: After 1 half-life, N/2 decay, N/2 remain

Second Step: After 2 half-life, 1/2 of N/2 decay, N/4 remain

Third step: After 3 half-life, 1/2 of N/4 decay, N/8 remains

The number of atoms disintegrated = Sum of the disintegrated atoms or fractions

This is now equal to

= N/2 + (1/2 x N/2) + (1/2 x N/4) = N/2 + N/4 + N/8 = (7/8)N

Alternatively, we can also apply the following steps to solve the problem using Zhepwo method:

Initial number of atoms, N 1 = N

Number of half-lives, n = 3

Also, the number of atoms decayed, N d = ?

R = 2 n = 2 3 = 8

N d = N 1 (R – 1 / R) = N ( 8 – 1 / 8) = N x 7/8 = (7/8)N

Therefore, the number of atoms that have disintegrated is (7/8)N

After three half-lives, the fraction of a radioactive material that has decayed is

The final answer to the above question is 7/8

We will use two methods

1 half-life implies 1/2 decays, ,1/2 remains

2 half-life shows that 1/4 decays, 1/4 remains

3 half-life, 1/8 decays, 1/8 remains

Hence, the fraction decayed = original fraction – remaining fraction = (1 – 1/8) = 7/8

Note that the original fraction is 1/1 which is equal to 1

The fraction decayed f d = ?

Disintegration ratio, R = 2 n = 2 3 = 8

f d = R – 1 / R = 8 – 1 / 8 = 7/8

Therefore, the fraction of radioactive material that has decayed is 7/8

The half-life of a radioactive element is 24 hours. Calculate the fraction of the original element that would have disintegrated in 96 hours.

The final answer to the above question is 15/16

I will apply two methods to solve the problem

After 24 hrs, 1/2 disintegrate, 1/2 remain

Another 24 hrs, 1/4 decay, 1/4 remain

Next 24 hrs, 1/8 decay, 1/8 remain

The next 24 hrs, 1/16 decay, 1/16 remain

Fraction disintegrate = sum of decayed fractions = 1/2 + 1/4 + 1/8 + 1/16 = 15/16

Half-life, T = 24 hrs

Decay time, t = 96 hrs

Fraction disintegrated, f d = ?

n = t/T = 96/24 = 4

f d = (R – 1) / R = (16 – 1) / 16 = 15/16

Therefore, the fraction of the original element that would have disintegrated in 96 hours is 15/16.

A radioactive isotope has a half-life of 20 hours. What fraction of the original radioactive nuclei will remain after 80 hours?

The final answer to the above question is 1/16

We will still use the two methods to find the answer to the problem

Let n be the original number of nuclei

After 20 hours, n/2 disintegrates and n/2 remains

The next 20 hours, 1/2 of n/2 disintegrates and n/4 remains

Next 20 hrs, 1/2 of n/4 decayed, and n/8 remains

After another 20 hrs, 1/2 of n/8 decayed, and n/16 remains

Half-life, T = 20 hrs

Decay time, t = 80 hrs

Number of half-lives, n = t/T = 80/20 = 4

Disintegration ratio, R = 2 n = 2 4 = 16

Fraction remaining, f r = 1/R = 1/16

We can also use an alternative method

T = 20, t = 80

Fraction remaining, f r = 1 / (2 t/T ) = 1 / (2 80/20 ) = 1 / 2 4 = 1 / 16

Therefore, after 80 hours. The fraction of the original number remaining would be 1/16

Two radioactive elements A and B have half-lives of 100 and 50 years respectively. Samples of A and B initially contain equal number of atoms. What is the ratio of remaining atoms of A to that of B after 200 years?

The ratio of remaining atoms of A to that of B after 200 years is 4:1

You can employ any of the two methods below to arrive at your answer:

1. Conventional Method:

When we allow n to be as the original number of the nuclei

Sample A: Half-life = 100 years

After 100 years, n/2 disintegrates and n/2 remains

Additionally, after 100 years, n/4 disintegrates and n/4 remains

Thus, after 200 years, the fraction of the original number that will remain is 1/4

Sample B: Half-life = 50 years

After 50 years, n/2 disintegrate, n/2 remain

Another 50 years, n/4 disintegrate and n/4 will remain

50 years more, n/8 will decay, n/8 remain

After 50 years, n/16 decay, n/16 remain

Therefore, after 200 years, fraction of atoms that will remain is 1/16

Ratio of A:B = 1/4 : 1/16

We will now multiply both sides by 16 to obtain:

The ratio of A:B = 4 : 1

2. Zhepwo Method

T is the Half-life for A and B which is 100 and 50 respectively

And t is the decay time for A and B which is 200 and 200 respectively

The number of half-lives n = t/T which is A = 200/100 = 2, and B = 200/50 = 4

Disintegrating ratio R = 2 n and it will give us 2 2 = 4 and 2 4 = 16

The initial number of atoms, N 1 = 1

Thus, we have A = 16 (1/4) = 4, and B = 16 (1/16) = 1

Hence we have a ratio of A:B = 4:1

A radioactive substance has a half-life of 80 days. If the initial number of atoms in the sample is 6 x 10 10 , how many atoms would remain at the end of 320 days?

The number of atoms that would remain at the end of 320 days is 3.8 x 10 9 atoms

T = 80; t = 320; and N 1 = 6 x 10 10

We will also use the formula N 2 = (N 1 / 2 t/T )

Therefore, by substituting our data into the above formula, we will have:

N 2 = (N 1 / 2 t/T ) = (6 x 10 10 ) / 2 320/80 = 6 x 10 10 / 2 4 = 6 x 10 10 / 16 = 3.8 x 10 9 atoms

A percentage of the original nuclei of a sample of a radioactive substance left after 5 half-lives is?

The percentage left is 3%

Number of hlaf-lives (n) = 5

original amount or fraction (N 1 ) = 1

Disintegrating ratio, R = 2 n = 2 5 = 32

Fraction remaining, f r = 1/R

The formula we will apply is:

Percentage of the original left = [amount left (fraction remaining) / original amount] x 100

Percentage of the original left = (f r / N 1 ) x 100 = [(1/32) / 1] x 100 = [1/32] x 100 = 3.125 = 3%

A radioactive substance of mass 768 grams has a half-life of 3 years. After how many years does this substance leave only 6 grams undecayed?

The answer is 21 years

Half-life T = 3 years

Initial mass present, N 1 = 768 g

Final mass remaining, N 2 = 6 g

Decay time, t = ?

The disintegration ratio, R = N 1 / N 2 = 768 / 6 = 128

from the formula below:

T = t / (log 2 R)

Making t subject of the formula and substituting our values, we will have:

t = T x log 2 R = 3 x log 2 128 = 3 x 7 = 21 years

An element whose half-life is 10 days is of mass 12 grams. Calculate the time during which 11.25 grams of the element would have decayed.

The final answer to this question is 40 days

Half-life, T = 10 days

Initial mass present, N 1 = 12 g

The mass of the element decayed, N d = 11.25 grams

N d = N 1 – N 2

Thus, N 2 = N 1 – N d = 12 – 11.25 = 0.75 grams

R = N 1 / N 2 = 12 / 0.75 = 16

Applying the formula T = t / (log 2 R) and making t the subject of the formula, we will have:

t = T x log 2 R = 10 x log 2 16 = 10 x 4 = 40 years

A radioactive element decreases in mass from 100g to 15g in 6 days. What is the half-life of the radioactive material?

The half-life of the radioactive element is 2.2 days

Initial mass present, N 1 = 100g

Final mass present, N 2 = 15g

Decay time, t = 6 days

Disintegration ratio, R = N 1 / N 2 = 100 / 5 = 6.67

The half-life formula T = t / (log 2 R) = 6 / log 2 6.67

Therefore, we can now say that

T = (6 x log2) / log 2 6.67 = (6 x 0.30103) / 0.824 = 2.2 days

The time it will take a certain radioactive material with a half-life of 50 days to reduce 1/32 of its original number is

The final answer to the above question is 250 days

Half-life, T = 50 days

Disintegration ratio, R = 32

Fraction remaining, f r = 1/R = 1/32

Using the formula, we will obtain

t = T x log 2 R = 50 x log 2 32 = 50 x 5 = 250 days

A radioactive substance has a half-life of 3 minutes. After 9 minutes, the count rate was observed to be 200, what was the count rate at zero time?

Note that count rate at zero time is the same as the initial count rate

9 min ===> count rate 200

6 min ===> count rate (200 x 2) = 400

3 min ===> count rate (400 x 2) = 800

0 min ===> count rate (800 x 2) = 1600

Therefore, count rate at zero time is 1600

The count rate of a radioactive material is 800 count/min. If the half-life of the material is 4 days, what would the the count rate be 16 days later?

Initial count rate = 800 count/min

Half-life = 4 days

After 4 days ==> (1/2) x 800 = 400 count/min

8 days after ==> (1/2) x 400 = 200 count/min

12 days after ==> (1/2) x 200 = 100 count/min

4 days after ==> (1/2) x 100 = 50 count/min

Therefore, the count rate 16 days later is 50 count/min

The half-life of a radioactive source is 1 minute. If a rate meter connected to the source registers 200μA at a given time, what would be its reading after 3 minutes?

A rate meter measures the count rate of radioactive of radioactive substance

Halff-life = 1 minute

1 minutes after, rate meter reads (1/2) x 200μ = 100μ

2 minutes after, rate meter reads (1/2) x 100μ = 50μ

3 minutes after, rate meter reads (1/2) x 50μ = 25μ

Therefore, rate meter reading after 3min is 25μ.

In 90 seconds, the mass of a radioactive element reduces to 1/32 of its original values. Determine the half-life of the element.

Decay time, t = 90 seconds

Fraction of initial mass remaining, F r = 1/32

Additionally, F r = 1/R

Therefore, by equating the above formulas, we will have

This implies that R = 32

Hence, Half-life (T) = t / (log 2 R) = 90 / (log 2 32) = 90 / 5 = 18 seconds

A radioactive substance has a half-life of 3 days. If a mass of 1.55 g of this substance is left after decaying for 15 days, determine the original value of the mass.

Half-life (T) = 3 days2

Decay time, t = 15 days

Final mass remaining, N 2 = 1.55 g

Therefore, n = t/T = 15/3 = 5

R = N 1 / N 2

Hence, N 1 = RN 2 = 32 x 1.55

Thus, the original mass N 1 =49.6 grams

In 90 seconds, the mass of a radioactive element reduces to 1/16 of its original value. Determine the half-life of the element.

Decay time t = 90 seconds, Fraction of mass remaining, F r = 1/16

But F r = 1/R

Therefore, 1/R = 1/16, and the value of R = 16

Half-life (T) = t / (log 2 R) = t / (log 2 16) = 90/4 = 22.5 seconds

Picture of Half-Life Formula

You may also like to read:

What is Radioactivity

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• Chemistry Concept Questions and Answers
• Half Life Questions

Half-Life Questions

Half-Life or previously known as the Half-Life Period is one of the common terminologies used in Science to describe the radioactive decay of a particular sample or element within a certain period of time.

However, this concept is also widely used to describe various types of decay processes, especially exponential and non-exponential decay. Apart from science, the term is used in medical sciences to represent the biological half-life of certain chemicals in the human body or in drugs.

Half-Life Chemistry Questions with Solutions

Q1. An isotope of caesium (Cs-137) has a half-life of 30 years. If 1.0g of Cs-137 disintegrates over a period of 90 years, how many grams of Cs-137 would remain?

b.) 0.125 g

c.) 0.00125 g

Correct Answer- (b.) 0.125 g

Q2. Selenium-83 has a half-life of 25.0 minutes. How many minutes would it take for a 10.0 mg sample to decay and only have 1.25 mg of it remain?

a.) 75 minutes

b.) 75 days

c.) 75 seconds

d.) 75 hours

Correct Answer- (a.) 75 minutes

Q3. How long does it take a 100.00g sample of As-81, with a half-life of 33 seconds, to decay to 6.25g?

a.) 122 seconds

b.) 101 seconds

c.) 132 seconds

d.) 22 seconds

Correct Answer – (c.) 132 seconds

Q4. What is the half-life of a radioactive isotope if a 500.0g sample decays to 62.5g in 24.3 hours?

a.) 8.1 hours

b.) 6.1 hours

c.) 5 hours

d.) 24 hours

Correct Answer- (a.) 8.1 hours

Q5. What is the half-life of Polonium-214 if, after 820 seconds, a 1.0g sample decays to 0.03125g?

a.) 164 minutes

b.) 164 seconds

c.) 64 seconds

d.) 160 minutes

Correct Answer- (b.) 164 seconds

Q6. The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes have elapsed?

To begin, we’ll count the number of half-lives that have passed. This can be obtained by doing the following:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

Number of half-lives

n = 7.2/2.4 = 3

Thus, three half-lives have passed.

Finally, we will calculate the remaining amount. This can be obtained by doing the following:

N 0 (original amount) = 100 g

(n) = number of half-lives

Amount remaining (N) =?

N = 100 / 2 3

N = 100 / 8

As a result, the amount of Zn-71 remaining after 7.2 minutes is 12.5 g.

Q7. Pd-100 has a half-life of 3.6 days. If one had 6.02 x 10 23 atoms at the start, how many atoms would be present after 20.0 days?

Half-life = 3.6 days

Initial atoms = 6.02 ×10 23 atoms

Time = 20days

To calculate the atoms present after 20 days, we use the formula below.

Thus, the number of atoms available is 1.28 × 10 22 atoms.

Q8. Os-182 has a half-life of 21.5 hours. How many grams of a 10.0 gram sample would have decayed after exactly three half-lives?

Answer. The amount of the radioactive substance that will remain after 3- half- lives=(½) 3 × a,

where a = initial concentration of the radioactive element.

So, amount of the radioactive substance that remains aftet 3- half-lives=( ½)³x10 = 10/8= 1.25 g.

Therefore, the number of grams of the radioactive substance that decayed in 3 half-lives = (10 – 1.25) g

Q9. After 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

Answer. The remaining decimal fraction is:

2.00 mg / 128.0 mg = 0.015625

The half-lives that must have expired to get to 0.015625?

(½) n = 0.015625

n log 0.5 = 0.015625

n = log 0.5 / 0.015625 n = 6

Calculation of the half-life:

24 days divided by 6 half-lives equals 4.00 days

Q10. A radioactive isotope decayed to 17/32 of its original mass after 60 minutes. Find the half-life of this radioisotope.

Answer. The amount that remains

17/32 = 0.53125

(1/2) n = 0.53125

n log 0.5 = log 0.53125

n = 0.91254

Half-lives that have elapsed are therefore, n = 0.9125

60 minutes divided by 0.91254 equals 65.75 minutes.

Therefore, n = 66 minutes

Q11. How long will it take for a 40 gram sample of I-131 (half-life = 8.040 days) to decay to 1/100 of its original mass?

Answer. (1/2) n = 0.01

n log 0.5 = log 0.01

6.64 x 8.040 days = 53.4 days

Therefore, it will take 53.4 days to decay to 1/100 of its original mass.

Q12. At time zero, there are 10.0 grams of W-187. If the half-life is 23.9 hours, how much will be present at the end of one day? Two days? Seven days?

24.0 hr / 23.9 hr/half-life = 1.0042 half-lives

One day = one half-life; (1/2) 1.0042 = 0.4985465 remaining = 4.98 g

Two days = two half-lives; (1/2) 2.0084 = 0.2485486 remaining = 2.48 g

Seven days = 7 half-lives; (1/2) 7.0234 = 0.0076549 remaining = 0.0765 g

Q13. 100.0 grams of an isotope with a half-life of 36.0 hours is present at time zero. How much time will have elapsed when 5.00 grams remains?

The afraction amount remaining will be-

5.00 / 100.0 = 0.05

(1/2) n = 0.05

n log 0.5 = log 0.05

n = 4.32 half-lives

36.0 hours x 4.32 = 155.6 hours

Q14. How much time will be required for a sample of H-3 to lose 75% of its radioactivity? The half-life of tritium is 12.26 years.

If you lose 75%, then 25% remains.

(1/2) n = 0.25

n = 2 (Since, (1/2) 2 = 1/4 and 1/4 = 0.25)

12.26 x 2 = 24.52 years

Therefore, 24.52 years of time will be required for a sample of H-3 to lose 75% of its radioactivity

Q15. The half-life for the radioactive decay of 14 C is 5730 years. An archaeological artifact containing wood had only 80% of the 14 C found in a living tree. Estimate the age of the sample.

Answer. Decay constant, k = 0.693/t 1/2 = 0.693/5730 years = 1/209 × 10 –4 /year

= 1846 years (approx)

Practise Questions on Half-Life

Q1. A newly prepared radioactive nuclide has a decay constant λ of 10 –6 s –1 . What is the approximate half-life of the nuclide?

d.) 1 month

Q2. If the decay constant of a radioactive nuclide is 6.93 x 10 –3 sec –1 , its half-life in minutes is:

Q3. A first-order reaction takes 40 min for 30% decomposition. Calculate t 1/2 .

Q4. What will be the time for 50% completion of a first-order reaction if it takes 72 min for 75% completion?

Q5. How much time will it take for 90% completion of a reaction if 80% of a first-order reaction was completed in 70 min?

Click the PDF to check the answers for Practice Questions. Download PDF

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