Best radiometric dating problems worksheet with answers pdf

best radiometric dating problems worksheet with answers pdf

Radiometric dating problems worksheet. Girls for relative dating on earth science 8: visualizing, how accurate and rock radiometric dating site calgary. However, filesize: homework. S simple worksheet asks for your readings everyday. Agency ep 1 compare u. Chapter 8: the following radiometric dating. Subject. What about the main idea to biological diversity. Estimate the online? Spring 2014 radiometric dating Radiometric dating worksheet answer key. School alabama paleontological society objectives distinguish between radiometric dating. His. Pros and radiometric dating worksheet. Geologists have the geologic time diaries check out the ages of this site. Some rock formations and pencil.

best radiometric dating problems worksheet with answers pdf

TRIGONOMETRY WORD PROBLEMS WORKSHEET WITH ANSWERS About " Trigonometry word problems worksheet with answers" Trigonometry word problems worksheet with answers is much useful to the kids who would like to practice problems on triangles in trigonometry.

On this web page "Trigonometry word problems worksheet with answers", first we are going to look at some word problems questions and then we will look answers. Best team of research writers makes best orders for students . Trigonometry word problems worksheet - Problems 1) The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 degree. Find the height of the building.

2) A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60 degree. Find how far the ladder is from the foot of the wall. 3) A string of a kite is 100 meters long and it makes an angle of 60° with horizontal. Find the height of the kite,assuming that there is no slack in the string. 4) From the top of the tower 30m height a man is observing the base of a tree at an angle of depression measuring 30 degree.

Find the distance between the tree and the tower. 5) A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40 m from the base?

6) A man wants to determine the height of a light house. He A ladder is leaning against a vertical wall makes an angle of 20° with the ground.

The foot of the ladder is 3 m from the wall.Find the length of ladder. 7) A kite flying at a height of 65 m is attached to a string inclined at 31° to the horizontal. What is the length of string ? 8) The length of a string between a kite and a point on the ground is 90 m. If the string is making an angle θ with the level ground such that tan θ = 15/8, how high will the kite be? 9) An aeroplane is observed to be approaching the airpoint.

It is at a distance of 12 km from the point of observation and makes an angle of elevation of 50 degree. Find the height above the ground. 10) A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 degree angle . Find the height of the balloon from the ground.

(Imagine that there is no slack in the cable) Trigonometry word problems worksheet - Answers Problem 1 : The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 degree.

Find the height of the building. Solution : First let us draw a figure for the information given in the question. Here, AB represents height of the building, BC represents distance of the building from the point of observation.

In the right triangle ABC, the side which is opposite to the angle 60 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and the remaining side is called adjacent side (BC). Now we need to find the length of the side AB. tanθ = Opposite side/Adjacent side tan 60° = AB/BC √3 = AB/50 √3 x 50 = AB AB = 50√3 Approximate value of √3 is 1.732 AB = 50 (1.732) AB = 86.6 m Hence, the height of the building is 86.6 m Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 2 : A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60 degree.

Find how far the ladder is from the foot of the wall. Solution : First let us draw a figure for the information given in the question. Here AB represents height of the wall, BC represents the distance between the wall and the foot of the ladder and AC represents the length of the ladder.

In the right triangle ABC, the side which is opposite to angle 60 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and remaining side is called adjacent side (BC). Now we need to find the distance between foot of the ladder and the wall. That is, we have to find the length of BC. tan θ = Opposite side/Adjacent side tan60° = AB/BC √3 = 6/BC BC = 6/√3 BC = (6/√3) x (√3/√3) BC = (6√3)/3 BC = 2√3 Approximate value of √3 is 1.732 BC = 2 (1.732) BC = 3.464 m Hence, the distance between foot of the ladder and the wall is 3.464 m Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 3 : A string of a kite is 100 meters long and it makes an angle of 60° with horizontal.

Find the height of the kite,assuming that there is no slack in the string. Solution : First let us draw a figure for the information given in the question. Here AB represents height of kite from the ground, BC represents the distance of kite from the point of observation. In the right triangle ABC the side which is opposite to angle 60 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now we need to find the height of the side AB. Sin θ = Opposite side/Hypotenuse side Sinθ = AB/AC Sin 60° = AB/100 √3/2 = AB/100 (√3/2) x 100 = AB AB = 50 √3 m Hence, the height of kite from the ground 50 √3 m Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 4 : From the top of the tower 30m height a man is observing the base of a tree at an angle of depression measuring 30 degree.

Find the distance between the tree and the tower. Solution : First let us draw a figure for the information given in the question. Here AB represents height of the tower, BC represents the distance between foot of the tower and the foot of the tree. In the right triangle ABC the side which is opposite to the angle 30 degree is known as opposite side (BC), the side which is opposite to 90 degree is called hypotenuse side (AC) and the remaining side is called adjacent side (AB). Now we need to find the distance between foot of the tower and the foot of the tree (BC).

tan θ = Opposite side/Adjacent side tan30° = AB/BC 1/√3 = 30/BC BC = 30 √3 Hence, the distance between the tree and the tower is 30 √3 m Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 5 : A man wants to determine the height of a light house.

He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40 m from the base? Solution : First let us draw a figure for the information given in the question. Here BC represents height of the light house, AB represents the distance between the light house from the point of observation.

In the right triangle ABC the side which is opposite to the angle A is known as opposite side (BC), the side which is opposite to 90 degree is called hypotenuse side (AC) and remaining side is called adjacent side (AB). Now we need to find the height of the light house (BC). tanA = Opposite side/Adjacent side tanA = BC/AB Given : tanA = 3/4 3/4 = BC/40 3 x 40 = BC x 4 BC = (3 x 40)/4 BC = (3 x 10) BC = 30 m Hence, the height of the light house is 30 m Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 6 : A man wants to determine the height of a light house.

He A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall.Find the length of ladder.

Solution : First let us draw a figure for the information given in the question. Here AB represents height of the wall, BC represents the distance of the wall from the foot of the ladder. In the right triangle ABC the side which is opposite to the angle 20 degree is known as opposite side (AB),the side which is opposite to 90 degree is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now we need to find the length of the ladder (AC). Cos θ = Adjacent side/Hypotenuse side Cos θ = BC/AC Cos 20° = 3/AC 0.9396 = 3/AC AC = 3/0.9396 AC = 3.192 Hence, the The length of the ladder is 3.192 m. trigono Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 7 : A kite flying at a height of 65 m is attached to a string inclined at 31° to the horizontal.

What is the length of string ? Solution : First let us draw a figure for the information given in the question. Here AB represents height of the kite. In the right triangle ABC the side which is opposite to angle 31 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and the remaining side is called adjacent side (BC).

Now we need to find the length of the string AC. Sin θ = Opposite side/Hypotenuse side Sin θ = AB/AC Sin 31° = AB/AC 0.5150 = 65/AC AC = 65/0.5150 AC = 126.2 m Hence, the length of the string is 126.2 m Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 8 : The length of a string between a kite and a point on the ground is 90 m.

If the string is making an angle θ with the level ground such that tan θ = 15/8, how high will the kite be? Solution : First let us draw a figure for the information given in the question. Here AB represents height of the balloon from the ground. In the right triangle ABC the side which is opposite to angle θ is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now we need to find the length of the side AB. Tan θ = 15/8 --------> Cot θ = 8/15 Csc θ = √(1+ cot²θ) Csc θ = √(1 + 64/225) Csc θ = √(225 + 64)/225 Csc θ = √289/225 Csc θ = 17/15 -------> Sin θ = 15/17 But, sin θ = Opposite side/Hypotenuse side = AB/AC AB/AC = 15/17 AB/90 = 15/17 AB = (15 x 90)/17 AB = 79.41 Hence, the height of the tower is 79.41 m Hence Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 9 : An aeroplane is observed to be approaching the airpoint.

It is at a distance of 12 km from the point of observation and makes an angle of elevation of 50 degree. Find the height above the ground. Solution : First let us draw a figure for the information given in the question. Here AB represents height of the airplane from the ground.In the right triangle ABC the side which is opposite to angle 50 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now we need to find the length of the side AB. From the figure given above, AB stands for the height of the aeroplane above the ground. Sin θ = Opposite side/Hypotenuse side Sin 50° = AB/AC 0.7660 = h/12 0.7660 x 12 = h h = 9.192 km Hence, the height of the aeroplane above the ground is 9.192 km Let us look at the next problem on "Trigonometry word problems worksheet with answers" Problem 10 : A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 degree angle .

Find the height of the balloon from the ground. (Imagine that there is no slack in the cable.) Solution : First let us draw a figure for the information given in the question. Here AB represents height of the balloon from the ground. In the right triangle ABC the side which is opposite to angle 60 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse (AC) and the remaining side is called as adjacent side (BC).

Now we need to find the length of the side AB. From the figure given above, AB stands for the height of the balloon above the ground.

Sin θ = Opposite side/Hypotenuse side Sin θ = AB/AC Sin 60° = AB/200 √3/2 = AB/200 AB = (√3/2) x 200 AB = 100√3 Approximate value of √3 is 1.732 AB = 100 (1.732) AB = 173.2 m Hence, the height of the balloon from the ground is 173.2 m After having gone through the stuff given on "Trigonometry word problems worksheet with answers", we hope that the students would have understood "Trigonometry word problems worksheet with answers".

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best radiometric dating problems worksheet with answers pdf

best radiometric dating problems worksheet with answers pdf - Radiometric dating ~ Learning Geology


best radiometric dating problems worksheet with answers pdf

Contents • • • • • • • • • Types of radiometric dating • • • • • • • • • • • Fundamentals of radiometric dating All ordinary is made up of combinations of , each with its own , indicating the number of in the . Additionally, elements may exist in different , with each isotope of an element differing only in the number of in the nucleus. A particular isotope of a particular element is called a . Some nuclides are inherently unstable. That is, at some point in time, an atom of such a nuclide will be transformed into a different nuclide by the process known as .

This transformation is accomplished by the emission of particles such as (known as ) or . While the moment in time at which a particular nucleus decays is random, a collection of atoms of a radioactive nuclide decays at a rate described by a parameter known as the , usually given in units of years when discussing dating techniques.

After one half-life has elapsed, one half of the atoms of the substance in question will have decayed. Many radioactive substances decay from one nuclide into a final, stable (or "daughter") through a series of steps known as a . In this case, usually the half-life reported is the dominant (longest) for the entire chain, rather than just one step in the chain.

Nuclides useful for radiometric dating have half-lives ranging from a few thousand to a few billion years. In most cases, the half-life of a nuclide depends solely on its nuclear properties; it is not affected by , chemical environment, and , or any other external factors.

The half-life of any nuclide is also believed to be constant through time. Although decay can be accelerated by radioactive bombardment, such bombardment tends to leave evidence of its occurrence. Therefore, in any material containing a radioactive nuclide, the proportion of the original nuclide to its decay product(s) changes in a predictable way as the original nuclide decays.

This predictability allows the relative abundances of related nuclides to be used as a that measures the time from the incorporation of the original nuclide(s) into a material to the present. The processes that form specific materials are often conveniently selective as to what elements they incorporate during their formation. In the ideal case, the material will incorporate a parent nuclide and reject the daughter nuclide. In this case, the only daughter nuclides to be found through examination of a sample must have been created since the sample was formed.

When a material incorporates both the parent and daughter nuclides at the time of formation, it may be necessary to assume that the initial proportions of a radioactive substance and its daughter are known. The daughter product should not be a small-molecule gas that can leak out of the material, and it must itself have a long enough half-life that it will be present in significant amounts. In addition, the initial element and the decay product should not be produced or depleted in significant amounts by other reactions.

The procedures used to isolate and analyze the reaction products must be straightforward and reliable. If a material that selectively rejects the daughter nuclide is heated, any daughter nuclides that have been accumulated over time will be lost through , setting the isotopic "clock" to zero. The temperature at which this happens is known as the "blocking temperature" and is specific to a particular material. In contrast to the most simple radiometric dating techniques, , which can be used for many isotopic decay sequences (e.g.

decay sequence), does not require knowledge of the initial proportions. Also the argon-argon dating technique can be used for the potassium-argon sequence to assure that no initial 40Ar was present. Limitation of techniques Although radiometric dating is accurate in principle, the accuracy is very dependent on the care with which the procedure is performed. The possible confounding effects of initial contamination of parent and daughter isotopes have to be considered, as do the effects of any loss or gain of such isotopes since the sample was created.

Accuracy is enhanced if measurements are taken on different samples taken from the same rock body but at different locations. This permits some compensation for variations. The precision of a method of dating depends in part on the half-life of the radioactive isotope involved.

For instance, carbon-14 has a half-life of less than 6000 years. After an organism has been dead for 60,000 years, so little carbon-14 is left in it that accurate dating becomes impossible. On the other hand, the concentration of carbon-14 falls off so steeply that the age of relatively young remains can be determined precisely to within a few decades.

The isotope used in has a longer half-life, but other factors make it more accurate than radiocarbon dating. Modern dating techniques Radiometric dating can be performed on samples as small as a billionth of a gram using a . The mass spectrometer was invented in the and began to be used in radiometric dating in the .

The mass spectrometer operates by generating a beam of from the sample under test. The ions then travel through a magnetic field, which diverts them into different sampling sensors, known as "", depending on their mass and level of ionization. On impact in the cups, the ions set up a very weak current that can be measured to determine the rate of impacts and the relative concentrations of different atoms in the beams.

The -lead radiometric dating scheme is one of the oldest available, as well as one of the most highly respected. It has been refined to the point that the error in dates of rocks about three billion years old is no more than two million years. Uranium-lead dating is best performed on the "" (ZrSiO 4), though it can be used on other materials.

Zircon incorporates uranium atoms into its crystalline structure as substitutes for , but strongly rejects lead. It has a very high blocking temperature, and is very chemically inert. One of its great advantages is that any sample provides two clocks, one based on uranium-235's decay to lead-207 with a half-life of about 700 million years, and one based on uranium-238's decay to lead-206 with a half-life of about 4.5 billion years, providing a built-in crosscheck that allows accurate determination of the age of the sample even if some of the lead has been lost.

Two other radiometric techniques are used for long-term dating. - dating involves or decay of potassium-40 to argon-40. Potassium-40 has a half-life of 1.3 billion years, and so this method is applicable to the oldest rocks. Radioactive potassium-40 is common in , , and , though the blocking temperature is fairly low in these materials, about 125 °C.

Rubidium-strontium dating is based on the beta decay of -87 to -87, with a half-life of 50 billion years. This scheme is used to date old igneous and , and has also been used to date lunar samples. Blocking temperatures are so high that they are not a concern.

Rubidium-strontium dating is not as precise as the uranium-lead method, with errors of 30 to 50 million years for a 3-billion-year-old sample. Short-range dating techniques There are a number of other dating techniques that have short ranges and are so used for historical or archaeological studies.

One of the best-known is the . Carbon-14 is a radioactive isotope of carbon-12, with a half-life of 5,730 years (very short compared with the above). In other radiometric dating methods, the heavy parent isotopes were synthesized in the explosions of massive stars that scattered materials through the Galaxy, to be formed into planets and other stars.

The parent isotopes have been decaying since that time, and so any parent isotope with a short half-life should be extinct by now. Carbon-14 is an exception. It is continuously created through collisions of neutrons generated by with nitrogen in the upper atmosphere. The carbon-14 ends up as a trace component in atmospheric (CO 2). An organism acquires carbon from carbon dioxide during its lifetime.

Plants acquire it through and , and animals acquire it from consumption of plants and other animals. When the organism dies, the carbon-14 begins to decay, and the proportion of carbon-14 left when the remains of the organism are examined provides an indication of the date of its death.

Carbon-14 radiometric dating has a range of about 58,000 to 62,000 years. The rate of creation of carbon-14 appears to be roughly constant, as cross-checks of carbon-14 dating with other dating methods show it gives consistent results. However, local eruptions of or other events that give off large amounts of carbon dioxide can reduce local concentrations of carbon-14 and give inaccurate dates. The releases of carbon dioxide into the as a consequence of industrialization have also depressed the proportion of carbon-14 by a few percent; conversely, the amount of carbon-14 was increased by above-ground tests that were conducted into the early .

Also, an increase in the or the earth's above the current value would depress the amount of carbon-14 created in the atmosphere. Another relatively short-range dating technique is based on the decay of uranium-238 into thorium-230, a process with a half-life of 80,000 years It is accompanied by a sister process, in which uranium-235 decays into protactinium-231, which has a half-life of 34,300 years. While is water-soluble, and are not, and so they are selectively precipitated into ocean-floor , from which their ratios are measured.

The scheme has a range of several hundred thousand years. use tree-ring dating () to determine the age of old pieces of wood. Trees grow rings on a yearly basis, with the spacing of rings being wider in good growth years than in bad growth years. These spacings can be used to help pin down the age of old wood samples, and also give some hints to climate change.

The technique is only useful to about 4,000 years in the past, however, because it requires overlapping tree ring series. Although determining geologic time by measuring the rate of deposition of sediments is not reliable over the large scale, it is still useful for certain scenarios, such as the deposition of layers of sediment on the bottom of a stable lake.

The approach is now known as "varve analysis" (the term "" means a layer or layers of sediment). Another technique used by archaeologists is to inspect the depth of penetration of water vapor into chipped (volcanic glass) artifacts.

The water vapor creates a "hydration rind" in the obsidian, and so this approach is known as "hydration dating" or "obsidian dating", and is useful for determining dates as far back as 200,000 years. Natural sources of radiation in the environment knock loose electrons in, say, a piece of pottery, and these electron accumulate in defects in the material's crystal lattice structure.

When the sample is heated, at a certain temperature it will glow from the emission of electrons released from the defects, and this glow can be used to estimate the age of the sample to a threshold of a few hundred thousand years. This is termed .

Finally, involves inspection of a polished slice of a material to determine the density of "track" markings left in it by the of uranium-238 impurities.

The uranium content of the sample has to be known, but that can be determined by placing a plastic film over the polished slice of the material, and bombarding it with . This causes induced fission of U-235, as opposed to the spontaneous fission of U-238. The fission tracks produced by this process are recorded in the plastic film. The uranium content of the material can then be calculated from the number of tracks and the neutron .

This scheme has application over a wide range geologic dates. For dates up to a few million years , (glass fragments from volcanic eruptions), and meteorites are best used.

Older materials can be dated using , , , and which have a variable amount of uranium content. Because the fission tracks are healed by temperatures over about 200 °C the technique has limitations as well as benefits. The technique has potential applications for detailing the thermal history of a deposit.

Dating with shortlived extinct radionuclides At the beginning of the solar system there were several relatively shortlived radionuclides like 26Al, 60Fe, 53Mn, and 129I present within the solar nebula. These radionuclides—possibly produced by the explosion of a supernova—are extinct today but their decay products can be detected in very old material like .

Measuring the decay products of an extinct radionuclides with a and using isochronplots it is possible to determine relative ages between different events in the early history of the solar system.

Dating methods based on extinct radionuclides can also be calibrated with the U-Pb method to give absolute ages. Notes The decay rate is not always constant for , as occurs in nuclides such as 7Be, 85Sr, and 89Zr. For this type of decay, the decay rate may be affected by local electron density. These isotopes are not used, however, for radiometric dating. . See also • • • • • • • • External links •


best radiometric dating problems worksheet with answers pdf

The discovery of radioactivity and the radiogenic decay of isotopes in the early part of the 20th century opened the way for dating rocks by an absolute, rather than relative, method. Up to this time estimates of the age of the Earth had been based on assumptions about rates of evolution, rates of deposition, the thermal behaviour of the Earth and the Sun or interpretation of religious scriptures.

Radiometric dating uses the decay of isotopes of elements present in minerals as a measure of the age of the rock: to do this, the rate of decay must be known, the proportion of different isotopes present when the mineral formed has to be assumed, and the proportions of different isotopes present today must be measured.

This dating method is principally used for determining the age of formation of igneous rocks, including volcanic units that occur within sedimentary strata. It is also possible to use it on authigenic minerals, such as glauconite, in some sedimentary rocks. Radiometric dating of minerals in metamorphic rocks usually indicates the age of the metamorphism. A number of elements have isotopes (forms of the element that have different atomic masses) that are unstable and change by radioactive decay to the isotope of a different element.

Each radioactive decay series takes a characteristic length of time known as the radioactive half-life, which is the time taken for half of the original (parent) isotope to decay to the new (daughter) isotope. The decay series of most interest to geologists are those with half-lives of tens, hundreds or thousands of millions of years. If the proportions of parent and daughter isotopes of these decay series can be measured, periods of geological time in millions to thousands of millions of years can be calculated.

To calculate the age of a rock it is necessary to know the half-life of the radioactive decay series, the amount of the parent and daughter isotopes present in the rock when it formed, and the present proportions of these isotopes. It must also be assumed that all the daughter isotope measured in the rock today formed as a result of decay of the parent.

This may not always be the case because addition or loss of isotopes can occur during weathering, diagenesis and metamorphism and this will lead to errors in the calculation of the age. It is therefore important to try to ensure that decay has taken place in a 'closed system', with no loss or addition of isotopes, by using only unweathered and unaltered material in analyses. The radiometric decay series commonly used in radiometric dating of rocks are detailed in the following sections.

The choice of method of determination of the age of the rock is governed by its age and the abundance of the appropriate elements in minerals. The samples of rock collected for radiometric dating are generally quite large (several kilograms) to eliminate inhomogeneities in the rock.

The samples are crushed to sand and granule size, thoroughly mixed to homogenise the material and a smaller subsample selected. In cases where particular minerals are to be dated, these are separated from the other minerals by using heavy liquids (liquids with densities similar to that of the minerals) in which some minerals will float and others sink, or magnetic separation using the different magnetic properties of minerals. The mineral concentrate may then be dissolved for isotopic or elemental analysis, except for argon isotope analysis, in which case the mineral grains are heated in a vacuum and the composition of the argon gas driven off is measured directly.

Measurement of the concentrations of different isotopes is carried out with a mass spectrometer. In these instruments a small amount (micrograms) of the sample is heated in a vacuum to ionise the isotopes and these charged particles are then accelerated along a tube in a vacuum by a potential difference.

Part-way along the tube a magnetic field induced by an electromagnet deflects the charged particles. The amount of deflection will depend upon the atomic mass of the particles so different isotopes are separated by their different masses. Detectors at the end of the tube record the number of charged particles of a particular atomic mass and provide a ratio of the isotopes present in a sample. This is the most widely used system for radiometric dating of sedimentary strata, because it can be used to date the potassium-rich authigenic mineral glauconite and volcanic rocks (lavas and tuffs) that contain potassium in minerals such as some feldspars and micas.

One of the isotopes of potassium, 40 K, decays partly by electron capture (a proton becomes a neutron) to an isotope of the gaseous element argon, 40 Ar, the other product being an isotope of calcium, 40 Ca. The half-life of this decay is 11.93 billion years. Potassium is a very common element in the Earth’s crust and its concentration in rocks is easily measured. However, the proportion of potassium present as 40 K is very small at only 0.012%, and most of this decays to 40 Ca, with only 11% forming 40 Ar.

Argon is an inert rare gas and the isotopes of very small quantities of argon can be measured by a mass spectrometer by driving the gas out of the minerals.

K–Ar dating has therefore been widely used in dating rocks but there is a significant problem with the method, which is that the daughter isotope can escape from the rock by diffusion because it is a gas. The amount of argon measured is therefore commonly less than the total amount produced by the radioactive decay of potassium.

This results in an underestimate of the age of the rock. The problems of argon loss can be overcome by using the argon–argon method. The first step in this technique is the irradiation of the sample by neutron bombardment to form 39 Ar from 39 K occurring in the rock. The ratio of 39 K to 40 K is a known constant so if the amount of 39 Ar produced from 39 K can be measured, this provides an indirect method of calculating the 40 K present in the rock.

Measurement of the 39 Ar produced by bombardment is made by mass spectrometer at the same time as measuring the amount of 40 Ar present. Before an age can be calculated from the proportions of 39 Ar and 40 Ar present it is necessary to find out the proportion of 39 K that has been converted to 39 Ar by the neutron bombardment.

This can be achieved by bombarding a sample of known age (a 'standard') along with the samples to be measured and comparing the results of the isotope analysis. The principle of the Ar–Ar method is therefore the use of 39 Ar as a proxy for 40 K. Although a more difficult and expensive method, Ar–Ar is now preferred to K–Ar.

The effects of alteration can be eliminated by step-heating the sample during determination of the amounts of 39 Ar and 40 Ar present by mass spectrometer. Alteration (and hence 40 Ar loss) occurs at lower temperatures than the original crystallisation so the isotope ratios measured at different temperatures will be different. The sample is heated until there is no change in ratio with increase in temperature (a 'plateau' is reached): this ratio is then used to calculate the age.

If no 'plateau' is achieved and the ratio changes with each temperature step the sample is known to be too altered to provide a reliable date. This is a widely used method for dating igneous rocks because the parent element, rubidium, is common as a trace element in many silicate minerals. The isotope 87 Rb decays by shedding an electron (beta decay) to 87 Sr with a half-life of 48 billion years.

The proportions of two of the isotopes of strontium, 86 Sr and 87 Sr, are measured and the ratio of 86 Sr to 87 Sr will depend on two factors. First, this ratio will depend on the proportions in the original magma: this will be constant for a particular magma body but will vary between different bodies. Second, the amount of 87 Sr present will vary according to the amount produced by the decay of 87 Rb: this depends on the amount of rubidium present in the rock and the age.

The rubidium and strontium concentrations in the rock can be measured by geochemical analytical techniques such as XRF (X-ray fluorescence).

Two unknowns remain: the original 86 Sr/87 Sr ratio and the 87 Sr formed by decay of 87 Rb (which provides the information needed to determine the age).

The principle of solving simultaneous equations can be used to resolve these two unknowns. If the determination of the ratios of 86 Sr/87 Sr and Rb/Sr is carried out for two different minerals (e.g. orthoclase and muscovite), each will start with different proportions of strontium and rubidium because they are chemically different.

An alternative method is whole-rock dating, in which samples from different parts of an igneous body are taken, which, if they have crystallised at different times, will contain different amounts of rubidium and strontium present. This is more straightforward than dating individual minerals as it does not require the separation of these minerals. Isotopes of uranium are all unstable and decay to daughter elements that include thorium, radon and lead.

Two decays are important in radiometric dating: 238 U to206 Pb with a half-life of 4.47 billion years and 235 U to 207 Pb with a half-life of 704 million years. The naturally occurring proportions of 238 U and 235 U are constant, with the former the most abundant at 99% and the latter 0.7%. By measuring the proportions of the parent and daughter isotopes in the two decay series it is possible to determine the amount of lead in a mineral produced by radioactive decay and hence calculate the age of the mineral.

Trace amounts of uranium are to be found in minerals such as zircon, monazite, sphene and apatite: these occur as accessory minerals in igneous rocks and as heavy minerals in sediments. Dating of zircon grains using uranium–lead dating provides information about provenance of the sediment.

Dating of zircons has been used to establish the age of the oldest rocks in the world. Other parts of the uranium decay series are used in dating in the Quaternary. These two rare earth elements in this decay series are normally only present in parts per million in rocks. The parent isotope is 147 Sm and this decays by alpha particle emission to 143 Nd with a half-life of 106 billion years.

The slow generation of 143 Nd means that this technique is best suited to older rocks as the effects of analytical errors are less significant. The advantage of using this decay series is that the two elements behave almost identically in geochemical reactions and any alteration of the rock is likely to affect the two isotopes to equal degrees.

This eliminates some of the problems encountered with Rb–Sr caused by the different reactivity and mobility of the two elements in the decay series. This dating technique has been used on sediments to provide information about the age of the rocks that the sediment was derived from: different provenance areas, for example continental cratons of different ages, can be distinguished by analysis of mud and mudstones. Rhenium occurs in low concentrations in most rocks, but its most abundant naturally occurring isotope 187 Re undergoes beta decay to an isotope of osmium 187 Os with a half-life of 42 Ga.

This dating technique has been used mainly on sulphide ore bodies and basalts, but there have also been some successful attempts to date the depositional age of mudrocks with a high organic content.

Osmium isotopes in seawater have also been shown to have varied through time. Radiometric dating is the only technique that can provide absolute ages of rocks through the stratigraphic record, but it is limited in application by the types of rocks which can be dated.

The age of formation of minerals is determined by this method, so if orthoclase feldspar grains in a sandstone are dated radiometrically, the date obtained would be that of the granite the grains were eroded from.

It is therefore not possible to date the formation of rocks made up from detrital grains and this excludes most sandstones, mudrocks and conglomerates. Limestones are formed largely from the remains of organisms with calcium carbonate hard parts, and the minerals aragonite and calcite cannot be dated radiometrically on a geological time scale. Hence almost all sedimentary rocks are excluded from this method of dating and correlation.

An exception to this is the mineral glauconite, an authigenic mineral that forms in shallow marine environments: glauconite contains potassium and may be dated by K–Ar or Ar–Ar methods, but the mineral is readily altered and limited in occurrence. The formation of igneous rocks usually can be dated successfully provided that they have not been severely altered or metamorphosed.

Intrusive bodies, including dykes and sills, and the products of volcanic activity (lavas and tuff) may be dated and these dates used to constrain the ages of the rocks around them by the laws of stratigraphic relationships. Dates from metamorphic rocks may provide the age of metamorphism, although complications can arise if the degree of metamorphism has not been high enough to reset the radiometric 'clock', or if there have been multiple phases of metamorphism.

General stratigraphic relations and isotopic ages are the principal means of correlating intrusive igneous bodies. Geographically separate units of igneous rock can be shown to be part of the same igneous suite or complex by determining the isotopic ages of the rocks at each locality. Radiometric dating can also be very useful for demonstrating correspondence between extrusive igneous bodies.

The main drawbacks of correlation by this method are the limited range of lithologies that can be dated and problems of precision of the results, particularly with older rocks. For example, if two lava beds were formed only a million years apart and there is a margin of error in the dating methods of one million years, correlation of a lava bed of unknown affinity to one or the other cannot be certain.

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Radioactive Dating, Accurate or Not?
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