The Earth and its atmosphere are constantly exposed to radioactive bombardment by streams of elementary particles from interstellar space. Penetrating into the upper atmosphere, the particles split the atoms there, releasing protons and neutrons, as well as larger atomic structures. Nitrogen atoms in the air absorb neutrons and release protons. These atoms have, as before, a mass of 14, but have less positive charge; now their charge is six. Thus, the original nitrogen atom is converted into a radioactive isotope of carbon:

where n, N, C and p stand for neutron, nitrogen, carbon and proton, respectively.

The formation of radioactive carbon nuclides from atmospheric nitrogen under the influence of cosmic rays occurs at an average rate of approx. 2.4 at./s for every square centimeter of the earth's surface. Changes in solar activity may cause some fluctuations in this value.

Because carbon-14 is radioactive, it is unstable and gradually turns into the nitrogen-14 atoms from which it was formed; in the process of such a transformation, it releases an electron - a negative particle, which makes it possible to record this process itself.

The formation of radiocarbon atoms under the influence of cosmic rays usually occurs in the upper layers of the atmosphere at altitudes from 8 to 18 km. Like regular carbon, radiocarbon oxidizes in the air to form radioactive dioxide (carbon dioxide). Under the influence of wind, the atmosphere is constantly mixed, and ultimately radioactive carbon dioxide, formed under the influence of cosmic rays, is evenly distributed in atmospheric carbon dioxide. However, the relative content of radiocarbon 14 C in the atmosphere remains extremely low - approx. 1.2ґ10 –12 g per gram of ordinary carbon 12 C.

Radiocarbon in living organisms.

All plant and animal tissues contain carbon. Plants get it from the atmosphere, and since animals eat plants, carbon dioxide also enters their bodies indirectly. Thus, cosmic rays are the source of radioactivity for all living organisms.

Death deprives living matter of the ability to absorb radiocarbon. In dead organic tissues, internal changes occur, including the decay of radiocarbon atoms. During this process, over 5730 years, half of the original number of 14 C nuclides turns into 14 N atoms. This time interval is called the half-life of 14 C. After another half-life, the content of 14 C nuclides is only 1/4 of their original number, after the next period half-life – 1/8, etc. As a result, the content of the 14 C isotope in the sample can be compared with the radioactive decay curve and thus establish the period of time that has elapsed since the death of the organism (its exclusion from the carbon cycle). However, for such a determination of the absolute age of a sample, it is necessary to assume that the initial content of 14 C in organisms over the past 50,000 years (radiocarbon dating resource) has not undergone changes. In fact, the formation of 14 C under the influence of cosmic rays and its absorption by organisms changed somewhat. As a result, measuring the 14 C isotope content of a sample provides only an approximate date. To account for the effects of changes in initial 14 C content, dendrochronological data on 14 C content in tree rings can be used.

The radiocarbon dating method was proposed by W. Libby (1950). By 1960, radiocarbon dating had gained widespread acceptance, radiocarbon laboratories had been established throughout the world, and Libby had been awarded the Nobel Prize in Chemistry.

Method.

The sample intended for radiocarbon dating should be collected using absolutely clean instruments and stored dry in a sterile plastic bag. Accurate information about the location and conditions of selection is necessary.

An ideal sample of wood, charcoal or fabric should weigh approximately 30 g. For shells, a weight of 50 g is desirable, and for bones - 500 g (the latest techniques, however, make it possible to determine age from much smaller samples). Each sample must be thoroughly cleaned of older and younger carbon-containing contaminants, for example, from the roots of later-growing plants or from fragments of ancient carbonate rocks. Pre-cleaning of the sample is followed by chemical processing in the laboratory. An acidic or alkaline solution is used to remove foreign carbon-containing minerals and soluble organic matter that may have penetrated the sample. After this, the organic samples are burned and the shells are dissolved in acid. Both of these procedures result in the release of carbon dioxide gas. It contains all the carbon in the purified sample and is sometimes converted into another substance suitable for radiocarbon dating.

The traditional method requires much less bulky equipment. First, a counter was used that determined the composition of the gas and was similar in principle to a Geiger counter. The counter was filled with carbon dioxide or other gas (methane or acetylene) obtained from the sample. Any radioactive decay occurring inside the device produces a weak electrical impulse. The energy of environmental background radiation usually varies widely, in contrast to radiation caused by the decay of 14 C, the energy of which is usually close to the lower limit of the background spectrum. The very undesirable ratio of background values ​​to 14 C data can be improved by isolating the counter from external radiation. For this purpose, the counter is covered with screens made of iron or high-purity lead several centimeters thick. In addition, the walls of the counter itself are shielded by Geiger counters located close to one another, which, by delaying all cosmic radiation, deactivate the counter itself containing the sample for about 0.0001 seconds. The screening method reduces the background signal to a few decays per minute (a 3 g wood sample dating back to the 18th century gives ~40 decays of 14 C per minute), which makes it possible to date fairly ancient samples.

Since about 1965, the liquid scintillation method has become widespread in dating. It converts the carbonaceous gas produced from the sample into a liquid that can be stored and examined in a small glass container. A special substance is added to the liquid - a scintillator - which is charged with the energy of electrons released during the decay of 14 C radionuclides. The scintillator almost immediately emits the accumulated energy in the form of flashes of light waves. Light can be captured using a photomultiplier tube. A scintillation counter contains two such tubes. A false signal can be identified and eliminated since it is sent by only one handset. Modern scintillation counters have very low, almost zero, background radiation, allowing highly accurate dating of samples up to 50,000 years old.

The scintillation method requires careful sample preparation because the carbon must be converted to benzene. The process begins with a reaction between carbon dioxide and molten lithium to form lithium carbide. Water is added little by little to the carbide and it dissolves, releasing acetylene. This gas, containing all the carbon in the sample, is converted under the influence of a catalyst into a transparent liquid - benzene. The following chain of chemical formulas shows how carbon is transferred from one compound to another in this process:

All age determinations obtained from laboratory measurements of 14 C are called radiocarbon dates. They are given in the number of years before the present day (BP), and the round modern date (1950 or 2000) is taken as the starting point. Radiocarbon dates are always given with an indication of possible statistical error (for example, 1760 ± 40 BP).

Application.

Typically, several methods are used to determine the age of an event, especially if it is a relatively recent event. The age of a large, well-preserved sample can be determined to within ten years, but repeated analysis of the sample requires several days. Usually the result is obtained with an accuracy of 1% of the determined age.

The importance of radiocarbon dating increases especially in the absence of any historical data. In Europe, Africa and Asia, the earliest traces of primitive man extend beyond the time limits of radiocarbon dating, i.e. turn out to be older than 50,000 years. However, the initial stages of the organization of society and the first permanent settlements, as well as the emergence of ancient cities and states, fall within the scope of radiocarbon dating.

Radiocarbon dating has been particularly successful in developing a timeline for many ancient cultures. Thanks to this, it is now possible to compare the course of development of cultures and societies and establish which groups of people were the first to master certain tools, create a new type of settlement, or pave a new trade route.

Determination of age by radiocarbon has become universal. After formation in the upper layers of the atmosphere, 14 C radionuclides penetrate into different environments. Air currents and turbulence in the lower atmosphere ensure the global distribution of radiocarbon. Passing in air currents over the ocean, 14 C first enters the surface layer of water, and then penetrates into the deep layers. Over the continents, rain and snow bring 14 C to the earth's surface, where it gradually accumulates in rivers and lakes, as well as in glaciers, where it can persist for thousands of years. Studying radiocarbon concentrations in these environments adds to our knowledge of the water cycle in the world's oceans and the climate of past eras, including the last ice age. Radiocarbon dating of the remains of trees felled by the advancing glacier showed that the most recent cold period on Earth ended approximately 11,000 years ago.

Plants annually absorb carbon dioxide from the atmosphere during the growing season, and the isotopes 12 C, 13 C and 14 C are present in plant cells in approximately the same proportion as they are present in the atmosphere. Atoms 12 C and 13 C are contained in the atmosphere in almost constant proportions, but the amount of the isotope 14 C fluctuates depending on the intensity of its formation. Layers of annual growth, called tree rings, reflect these differences. The continuous sequence of annual rings of a single tree can span 500 years in oak and more than 2,000 years in redwood and bristlecone pine. In the arid mountainous regions of the northwestern United States and in the peat bogs of Ireland and Germany, horizons with trunks of dead trees of different ages were discovered. These findings allow us to combine information about fluctuations in the concentration of 14 C in the atmosphere over almost 10,000 years. The correct determination of the age of samples during laboratory research depends on knowledge of the concentration of 14 C during the life of the organism. For the last 10,000 years, such data has been collected and is usually presented in the form of a calibration curve showing the difference between the level of atmospheric 14 C in 1950 and in the past. The discrepancy between the radiocarbon and calibrated dates does not exceed ±150 years for the interval between 1950 AD. and 500 BC For more ancient times, this discrepancy increases and, with a radiocarbon age of 6000 years, reaches 800 years. see also ARCHEOLOGY

Radioactive decay is a random event in the “life” of an atom, one might say an accident. Let us try, based on this very general consideration, to derive a law according to which the concentration of radioactive atoms should change with time.

Let at some point in time t the concentration of the radioactive isotope was equal to P(t), and after a short time D t became equal P(t+D t). It is clear that during time D t fell apart P(t) – P(t+D t) atoms.

If radioactive decay is a random process, then it is quite logical to assume that the number of decays in time D t will be greater, the greater the concentration of atoms P(t) and the longer the time period D t:

P(t) – P(t+D t) ~ P(t) × D t

P(t) – P(t+D t) = l P(t)D t, (1)

where l is the proportionality coefficient. It is clear that each isotope has its own coefficient: if the isotope decays quickly, then the coefficient l is large, if it decays slowly, then it is small.

Let us rewrite equality (1) in the form:

P(t+D t) – P(t) = –l P(t)D t. (2)

Now let's direct D t to zero and note that P(t+D t) – P(t) - This increment of function n(t) in time D t, we get:

We have obtained a differential equation. It is clear that if at the initial moment the isotope concentration was equal to P 0, then P(0) = = P 0 . Let’s “guess” the solution to equation (3):

P(t) = P 0 e–l t. (4)

Let's check by substituting expression (4) into equation (3):

l.h.: ( P 0 e–l t)¢ = P 0 e–l t(–l);

p.h.: –l P 0 e–l t.

It is obvious that the left side is identically equal to the right, in addition, the initial condition is also satisfied:

P(0) = P 0 e– l × 0 = P 0 e 0 = P 0×1 = P 0 .

So, we have obtained the law of radio wave decay:

P(t) = P 0 e–l t. (25.1)

The quantity l is called radioactive decay constant.

Half life

When studying radioactive decay, instead of the decay constant as a characteristic of the rate of the process, another value is often used - half-life.

Half life T is the time during which half of the original amount of a given radioactive isotope decays. Let's find a connection between T and l.

Let's use the mathematical fact that for any number A equality is true.

Indeed,

ln e a = a ln e = a×1 = a And .

Then we rewrite formula (25.1) in the form

.

Let us introduce the notation

If we substitute the value into formula (25.3) t = T, we get

.

Thus, is the half-life of a given isotope.

It must be said that the half-lives of different isotopes can take on very different values. For example:

92 U 238 (a-decay): T= 4.5×10 9 years;

94 Pu 239 (a-decay): T= 24400 years;

89 Ra 236 (a-decay): T= 1600 years;

91 Ac 233 (b – -decay): T= 27 days;

90 Th 233 (b – -decay): T= 22 min.

There are isotopes with a half-life of ten thousandths of a second (some isotopes of polonium 84 Po).

Problem 25.2. The radioactive isotope of carbon in an old piece of wood is 0.0416 times the mass of that isotope in living plants. How old is this piece of wood? The half-life of the isotope is 5570 years.

then the mass changes according to the same law as the concentration

m(t) = m 0 . (1)

Let us express from equation (1) the unknown t.

Natalia asks
Answered by Elena Titova, 04/26/2013

Natalia asks: “Please tell me, what about radiocarbon analysis, which dates the finds to an age much older than the biblical age of the earth?”

Greetings, Natalia!

Radiometric methods, including radiocarbon dating, in determining the ages of archaeological and paleontological finds have colossal errors due to many assumptions that cannot be verified. Therefore, such methods are a very dubious tool in the hands of researchers.

Learn more about radiocarbon dating, which only applies to finds that were once living organisms. The method is based on the following. In the atmosphere, radioactive carbon (C-14) is formed from nitrogen atoms under the influence of cosmic radiation. Unlike regular carbon (C-12), C-14 is radioactive, meaning it is unstable and slowly decays to nitrogen. Both forms of carbon are included in carbon dioxide (CO2), which enters living organisms through photosynthesis. The ratio of C-14 and C-12 is approximately the same both in the atmosphere and in the biosphere. After the death of the organism, decaying C-14 is no longer replaced by carbon from the external environment, and its proportion gradually decreases. Knowing the ratio of C-14 and C-12 at the present time, the same ratio in the sample under study, as well as the decay rate (the half-life of radioactive carbon, that is, the time during which the amount of the element is halved - it is 5730 years), we can determine the age finds. It is believed that if, for example, in the sample under study this ratio is half as much as in a modern one, then the sample is about 5,730 years old, if it is four times less, then 11,460 years old, etc. Theoretically, modern methods can measure carbon-14 concentrations in samples no older than 50 thousand years.

However, there is a serious problem here. The fact is that the decrease in the proportion of radioactive carbon in the studied samples can be attributed exclusively to its decay only if the ratio of C-14 and C-12 is the same for both modern conditions and for the ancient era. If the proportion of radioactive carbon at that distant time was lower, then it is impossible to determine what caused the low ratio of C-14 and C-12 in the sample under study - the decay of radioactive carbon or, in addition, the small initial amount of C-14. The researchers therefore make the following arbitrary assumption: the ratio of C-14 to C-12 has always been the same and constant. The low ratio of C-14 to C-12 in the finds is perceived solely as a result of the decay of radioactive carbon. There is reason to believe that the share of C-14 was actually lower in the antediluvian era (in the atmosphere and biosphere) due to the presence of a water shell above the atmosphere and a stronger magnetic field that screened cosmic radiation. It is clear that radiocarbon analysis greatly overestimates the age of the finds in this case: after all, the lower the level of carbon-14 in them, the more time is believed to have passed since the beginning of the decay of the element.

In addition, the method assumes a constant decay rate (we don't actually know this), and also that C-14 did not enter the samples from outside (we also don't know this). There are other factors that influence the balance of both forms of carbon, for example, the total amount of carbon in the atmosphere and biosphere decreased after the Flood, because countless quantities of animals and plants were buried and turned into fossils, oil, coal, and gas.

As you can see, the radiocarbon dating method is an equation with many unknowns, which makes this analysis unsuitable for research. I will give examples of its “accuracy”. The method showed that the seals that had just been killed died 1,300 years ago; The age of the Shroud of Turin, in which the body of Christ was wrapped after the crucifixion, dates back to the 14th century. At the same time, the fact of the presence of C-14 in fossil remains that are believed to be millions of years old clearly excludes this age, since radiocarbon would have decayed long ago over millions of years.

God's blessings!

Read more on the topic "Creation":

120. During the decay of 94 Pu 239 → 92 U 235 + 2 He 4, energy is released, most of which is the kinetic energy of α particles. 0.09 meV is carried away by γ-rays emitted by uranium nuclei. Determine the speed of α-particles, m P u =±239.05122 amu, m U =235.04299 amu, m A,=4.00260 amu.

121. During the fission process, the uranium nucleus breaks up into two parts, the total mass of which is less than the initial mass of the nucleus by approximately 0.2 the rest mass of one proton. How much energy is released when one uranium nucleus fissions?

123. Determine the number of uranium atoms 92 U 238 decayed during the year, if the initial mass of uranium is 1 kg. Calculate the decay constant of uranium.

124. Calculate the number of radon atoms that decayed during the first day, if the initial mass of radon is 1 g. Calculate the decay constant of uranium.

125. In the human body, 0.36 of the mass is potassium. The radioactive isotope of potassium 19 K 40 makes up 0.012% of the total mass of potassium. What is the activity of potassium if the person weighs 75 kg? Its half-life is 1.42 * 10 8 years.

126. 100 g of a radioactive substance lies on the scales. After how many days will a scale with a sensitivity of 0.01 g show the absence of a radioactive substance? The half-life of the substance is 2 days.

127. Over two days, the radioactivity of the radon preparation decreased by 1.45 times. Determine the half-life.

128. Determine the number of radioactive nuclei in a freshly prepared 53 J 131 preparation, if it is known that after a day its activity became 0.20 Curie. The half-life of iodine is 8 days.

129. The relative proportion of radioactive carbon 6 C 14 in an old piece of wood is 0.0416 of its proportion in living plants. How old is this piece of wood? The half-life of 6 C 14 is 5570 years.

130. It was found that in a radioactive preparation 6.4 * 10 8 nuclear decays occur per minute. Determine the activity of this drug.

131. What fraction of the initial number of 38 Sg 90 nuclei remains after 10 and 100 years, decays in one day, in 15 years? Half-life 28 years

132. There are 26 * 10 6 radium atoms. How many of them will undergo radioactive decay in one day, if the half-life of radium is 1620 years?

133. The capsule contains 0.16 mol of the isotope 94 Pu 238. Its half-life is 2.44*10 4 years. Determine the activity of plutonium.

134 There is a uranium preparation with an activity of 20.7 * 10 6 dispersion/s. Determine the mass of the isotope 92 U 235 in the preparation with a half-life of 7.1 * 10 8 years.

135. How will the activity of the cobalt drug change over 3 years? Half-life 5.2 years.

136. A lead capsule contains 4.5 * 10 18 radium atoms. Determine the activity of radium if its half-life is 1620 years.

137. How long does it take for 80% of the atoms of the radioactive isotope of chromium 24 Cr 51 to decay if its half-life is 27.8 days?

138. The mass of the radioactive isotope sodium 11 Na 25 is 0.248*10 -8 kg. Half-life 62 s. What is the initial activity of the drug and its activity after 10 minutes?

139. How much radioactive substance remains after one or two days, if at first there was 0.1 kg of it? The half-life of the substance is 2 days.

140. The activity of a uranium preparation with a mass number of 238 is 2.5 * 10 4 dispersion/s, the mass of the preparation is 1 g. Find the half-life.

141. What fraction of atoms of a radioactive isotope
90 Th 234, which has a half-life of 24.1 days, decays -
in 1 second, in a day, in a month?

142. What fraction of atoms of the radioactive isotope co-
balta decays in 20 days if its half-life is
yes 72 days?

143 How long does it take for a preparation with a constant activity of 8.3*10 6 decay/s to decay 25*10 8 nuclei?

144. Find the activity of 1 µg of tungsten 74 W 185 whose half-life is 73 days

145. How many nuclear decays per minute occur in a preparation whose activity is 1.04 * 10 8 dispersion/s?

146. What fraction of the initial amount of radioactive substance remains undecayed after 1.5 half-lives?

147. What fraction of the initial amount of a radioactive isotope decays during the lifetime of this isotope?

148. What is the activity of radon formed from 1 g of radium in one hour? The half-life of radium is 1620 years, radon is 3.8 days.

149. A certain radioactive drug has a decay constant of 1.44*10 -3 h -1 . How long does it take for 70% of the original number of atoms 7 to decay?

150. Find the specific activity of the artificially obtained radioactive isotope of strontium 38 Sg 90. Its half-life is 28 years.

151. Can a silicon core turn into a nucleus?
aluminum, thereby ejecting a proton? Why?

152. During the bombardment of aluminum 13 Al 27 α -
phosphorus 15 P 30 is formed by particles. Write down this reaction and
calculate the energy released.

153. When a proton collides with a berylium nucleus,
the nuclear reaction 4 Be 9 + 1 P 1 → 3 Li 6 + α occurred. Find the reaction energy.

154. Find the average binding energy per
per 1 nucleon, in nuclei 3 Li 6, 7 N 14.

155. When fluorine nuclei are bombarded with 9 F 19 protons, oxygen x O 16 is formed. How much energy is released during this reaction and what nuclei are formed?

156. Find the energy released in the following nuclear reaction 4 Ве 9 + 1 Н 2 → 5 В 10 + 0 n 1

157. An isotope of radium with a mass number of 226 turned into an isotope of lead with a mass number of 206. How many α and β decays occurred in this case?

158. The initial and final elements of four radioactive families are given:

92 U 238 → 82 Pb 206

90 Th 232 → 82 Pb 207

92 U 235 → 82 Pb 207

95 Am 241 → 83 Bi 209

How many α and β transformations occurred in each family?

159. Find the binding energy per nucleon in the nucleus of the oxygen atom 8 O 16.

160. Find the energy released during a nuclear reaction:

1 H 2 + 1 H 2 → 1 H 1 + 1 H 3

161. What energy will be released when 1 g of helium 2 He 4 is formed from protons and neutrons?

162. What does the thorium isotope 90 Th 234, whose nuclei undergo three successive α-decays, turn into?

163. Complete the nuclear reactions:

h Li b + 1 P 1 →?+ 2 He 4;

13 A1 27 + o n 1 →?+ 2 Not 4

164. The uranium nucleus 92 U 235, having captured one neutron, once
split into two fragments, releasing two neutrons. One of the fragments turned out to be a xenon nucleus 54 Xe 140. What is the second shard? Write the reaction equation.

165. Calculate the binding energy of the helium nucleus 2 He 3.

166. Find the energy released during a nuclear reaction:

20 Ca 44 + 1 P 1 → 19 K 41 +α

167. Write the missing symbols in the following
common nuclear reactions:

1 Р 1 →α+ 11 Nа 22

13 Al 27 + 0 p 1 →α+...

168. Determine the specific binding energy of tritin,

169. Change in mass during the formation of the 7 N 15 nucleus
equals 0.12396 a.a.m. Determine the mass of an atom

170 Find the binding energy of 1 H 3 and 2 He 4 nuclei. Which of these nuclei is the most stable?

171 When lithium 3 Li 7 is bombarded with protons, helium is produced. Write down this reaction. How much energy is released during this reaction?

172. Find the energy absorbed during the reaction:

7 N 14 + 2 He 4 → 1 P 1 + ?

173. Calculate the binding energy of the helium nucleus 2 He 4.

174. Find the energy released in the following nuclear reaction:

3 Li 7 + 2 He 4 → 5 V 10 + o n 1

175. Complete the nuclear reactions:

1 Р 1 → 11 Nа 22 + 2 He 4, 25 Mn 55 + ?→ 27 Co 58 + 0 n 1

176. Find the energy released during the following
nuclear reaction.

з Li 6 + 1 Н 2 →2α

177. The nuclei of the isotope 90 Th 232 undergo α decay, two β decays and another α decay. What kernels do you get after this?

178 Determine the binding energy of the deuterium nucleus.

179. The nucleus of the isotope 83 Bi 211 was obtained from another nucleus after one α-decay and one β-decay. What kind of core is this?

180. Which isotope is formed from radioactive thorium 90 Th 232 as a result of 4 α-decays and 2 β-decays?

181. In a radioactive drug with a decay constant λ=0.0546 years -1, to=36.36% of the nuclei of their original number decayed. Determine the half-life, average life time. How long did it take for the nuclei to decay?

182. The half-life of a radioactive substance is 86 years. How long will it take for 43.12% of the original number of nuclei to decay? Determine the decay constant λ and the average lifetime of a radioactive nucleus.

183. In one year, 64.46% of the nuclei of their original amount of radioactive drug decayed. Determine the average lifetime and half-life.

184. The average lifetime of a radioactive substance is τ=8266.6 years. Determine the time during which 51.32% of the nuclei from their original number decay, half-life, decay constant.

185. In a radioactive substance with a decay constant λ=0.025 years -1, 52.76% of the nuclei of their original number decayed. How long did the breakup last? What is the average lifetime of nuclei?

186. Determine the activity of a mass of 0.15 μg with a half-life of 3.8 days after two days. Analyze dependency A =f(t)

187. The half-life of bismuth (83 Bi 210) is 5
days. What is the activity of this 0.25 mcg drug after 24 hours? Assume that all atoms of the isotope are radioactive.

188. Isotope 82 Ru 210 has a half-life of 22 years. Determine the activity of this isotope weighing 0.25 μg after 24 hours?

189. Flux of thermal neutrons passing through aluminum
distance d= 79,4 cm, weakened three times. Define
effective cross sections for the reaction of neutron capture by an atom nucleus
ma of aluminum: Density of aluminum ρ=2699 kg/m.

190. The neutron flux is weakened by 50 times after traveling a distance d in plutonium, the density of which is ρ = 19860 kg/m3. Determine d if the effective cross section for capture by a plutonium nucleus is σ = 1025 bars.

191. How many times is the flux of thermal neutrons weakened after traveling a distance d=6 cm in zirconium, if the density of zirconium is ρ = 6510 kg/m 3, and the effective cross section of the capture reaction is σ = 0.18 bars.

192. Determine the activity of 85 Ra 228 with a half-life of 6.7 years after 5 years, if the mass of the drug is m = 0.4 μg and all atoms of the isotope are radioactive.

193. How long did it take for 44.62% of the original number of nuclei to decay, if the half-life is m=17.6 years. Determine the decay constant λ, the average lifetime of a radioactive nucleus.

194. Determine the age of an archaeological find made of wood if the isotope activity of the sample is 80% of the sample from fresh plants. The half-life is 5730 years.

195. Liquid potassium ρ= 800kg !m weakens the neutron flux by half. Determine the effective cross section for the reaction of neutron capture by the nucleus of a potassium atom if the neutron flux passes a distance d = 28.56 cm in liquid potassium.

196. Determine the age of ancient tissue if active
The isotope content of the sample is 72% activity
sample from fresh plants. Half-life T=5730 years.

197. Write down in full form the equation of the nuclear reaction (ρ,α) 22 Na. Determine the energy released as a result of a nuclear reaction.

198. Uranium, whose density is ρ = 18950 kg/m 2, weakens the flux of thermal neutrons by 2 times with a layer thickness d = 1.88 cm. Determine the effective cross section for the reaction of neutron capture by a uranium nucleus

199. Determine the activity of the isotope 89 Ac 225 with a half-life T = 10 days after a time t = 30 days, if the initial mass of the drug is m = 0.05 μg.

200. Determine the age of an archaeological find made of wood if the 6 C 14 activity of the sample is 10% of the activity of the sample from fresh plants. Half-life T=5730 years.

201. Determine the thickness of the mercury layer if the neutron flux, having passed through this flux, is weakened by 50 times, the effective cross section for the reaction of neutron capture by a nucleus σ = 38 barn, density of mercury ρ = 13546 kg/m 3.

202. Isotope 81 Tℓ 207 has a half-life T = 4.8 million. What is the activity of this isotope weighing 0.16 μg after time t = 5 million. Assume that all atoms of the isotope Tℓ 207 radioactive.

203. How many nuclei from their initial amount of matter decay in 5 years, if the decay constant λ = 0.1318 years -1. Determine the half-life, the average lifetime of nuclei.

204. Determine the activity of 87 Fr 221 weighing 0.16 μg with a half-life T = 4.8 million after a time t = 5 min. Analyze the dependence of activity on mass (A=f(m)).

205. The half-life of the carbon isotope 6 C 14 T = 5730 years, the activity of wood for the isotope 6 C 14 is 0.01% of the activity of samples from fresh plants. Determine the age of the wood.

206. Neutron flux passing through sulfur (ρ = 2000 kg/m 3.)
distance d=37.67 cm is weakened by 2 times. Define
effective cross section for the reaction of neutron capture by an atom nucleus
ma sulfur.

207. Comparison of the activity of drugs 89 Ac 227 and 82 Рb 210 if the drug masses are m=0.16 µg, after 25 years. The half-lives of the isotopes are the same and equal to 21.8 years.

208. In a radioactive substance, 49.66% of the nuclei of their original number decayed over t=300 days. Determine the decay constant, half-life, and average lifetime of the isotope nucleus.

209. Analyze the dependence of the activity of the radioactive isotope 89 Ac 225 from the mass after t = 30 days, if the half-life is T = 10 days. Take the initial mass of the isotope, respectively, m 1 = 0.05 μg, m 2 = 0.1 μg, m 3 = 0.15 μg.

210. Iridium weakens the flux of thermal neutrons in
2 times. Determine the thickness of the iridium layer if its density
ity ρ=22400 kg/m 3, and the effective reaction cross section for
neutron capture by an iridium nucleus σ=430 barn

MOSCOW, June 3 - RIA Novosti. Elevated levels of radioactive carbon-14 in the growth rings of two Japanese cedar trees may indicate that the Earth was bombarded by cosmic rays in 774-775 AD, physicists say in a paper published in the journal Nature.

Trees and other types of vegetation react very sensitively to the slightest changes in living conditions - an increase or decrease in temperature, solar radiation energy and other factors. All these events are reflected in the shape and thickness of annual rings - layers of wood in the trunk, which are formed during the growing season. It is believed that dark rings correspond to unfavorable environmental conditions, and light rings correspond to favorable ones.

A group of physicists led by Fusa Miyake from Nagoya University (Japan) examined the growth rings of two ancient Japanese cedar trees to determine the exact date of the cosmic ray “raid” on Earth, which supposedly occurred between 750 and 820 AD.

As physicists explain, episodes of prolonged “bombardment” by particles of extraterrestrial origin are usually accompanied by an increase in the proportion of the heavy and radioactive isotope carbon-14 in wood and soft tissues of plants.

Guided by this idea, physicists divided thin cuts of two Japanese cedars that grew in the land of the rising sun during the Middle Ages into separate growth rings.

In one case, they used pieces of wood to calculate annual variations in carbon-14 between 770 and 779 AD, and in the second, they used them to observe changes in the average concentration of a heavy isotope of carbon for every two years between 750 and 820 AD .

In both cases, scientists recorded a sharp increase in the proportion of radiogenic carbon in the rings dating back to 774 and 775 AD. According to them, this concentration peak cannot be explained by seasonal variations in the strength of solar radiation, since carbon-14 in the rings of 774 and 775 was about 20 times more than in the layers of wood formed during increased solar activity.

According to the researchers, this conclusion is in good agreement with the results of Antarctic studies. Thus, in snow samples of 774 and 775, obtained from the Antarctic Fuji Dome station, a similar peak was recorded in the concentration of another “cosmic” element - beryllium-10.

Scientists believe that the source of cosmic rays could have been a powerful supernova that exploded at a relatively close distance - 6.5 thousand light years - from the solar system. Another possible reason for this could be a “super flare” on the Sun with a power several tens of times greater than the typical power of solar flares.