Nuclear Physics and Isotopes

Content of Lecture1. The Structure of Atom2. The nuclear binding energy and The Nuclear Force3. Types of Radioactive Emission4. ALPHA, BETA and GAMMA decay5. The Natural Radioactivity6. Activity and Half-lives7. Carbon dating

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Introduction- The 19th century was the century of the “victory of chemistry”, the 20 th was the century of the

“victory of physics”, the 21st would it be that of the “victory of biology”? At what price? Again poor impact on Earth and now human life and all the live materials through poor genetics?

- Ideally, opinions should be based on understanding (not on prejudice or emotion). That is why we should try to understand the principles of the concerned science before introducing denunciations and accusations against it.

- Nuclear physics? For what doing? For energy, for peace applications in agriculture, medicine and so on, not for another nuclear bombing.

- Through studying the basic concepts of nuclear physics we would learn not only some of its basics and applications, but also admit that with the nuclear era the humanity has started a long way, with the right tools in hand this time, to better grasp the universe, the natural resources and the Nature with its complicated great many phenomena and processes that were, up-to-that time, obscure. We have the theoretical and experimental foundations to develop (through new technologies based on nuclear physics) new inventions that increasingly participate in improving life quality. Sure, the negative impact of nuclear physics is there outdoors, but its risk could be seen as no more than other risks if its equipment is properly treated and closely maintained.

- In this short account, we cannot cover everything about nuclear physics, or go much behind the preliminary statements, so we will be severely selective in dealing with each of the items dealt with here. Some of the given text is presented just for making link between parts and some information is given for general reading for those who could be interested in the subject. It is recommended, however, to get a look to all the contents, then concentrate on what you feel more complicated and needs further attention and more studying effort.

“By convention there is color, by convention sweetness, by convention bitterness, but in reality there are atoms and space”. -Democritus (400 BC)

Today, we know that there is something more fundamental than Earth, water, air, and fire... But is the atom fundamental?

The Structure of Atom- The structure of an atom consists of a positively charged nucleus surrounded by negatively

charged electrons that orbit the nucleus at extremely high speeds. The high speed of the electrons creates a negative cloud. For this reason, the latest theory of atomic structure is referred to as the Electron Cloud Model.

- Most of an atom consists of empty space.- The nucleus, located at the center, contains protons and neutrons. Protons are positively charged

particle, and neutrons have no charge. Therefore, the nucleus is positively charged.- Electrons are negatively charged. Negatively charged electrons are attracted to the positively

charged nucleus. The high-speed spin of the electrons keep the electrons from being drawn into the nucleus.

- Electrons are found in different shells or energy levels. Each energy level is named by letters starting with K. Each energy level can hold a certain number of electrons.- K shell can hold 2 electrons- L shell can hold 8 electrons- M shell can hold 18 electrons- N shell can hold 32 electrons

- Atoms have extremely small dense positively charged centers.- Stability, or instability, of a particular nucleus is controlled by the balance between the protons’

repulsive forces, on one hand, and the attraction among protons and neutrons, on the other hand.- Atom structure change can take place either by natural decay or by induction.- The Sun is a nuclear fusion reactor.

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- We have a spherical model for the nucleus. The particles inside are the nucleons present in a given number (mass number “A”).

- The size of a nucleus is possible to be determined by bombarding it with some other particle and measuring its cross-section by the probability of collision. The radius of the nucleus “ R ”, in meter can be find by the following formula, where A is the mass number :

R (in m) = 1.2 x 10-15 A1/3

- There is a small difference between the nucleus real mass, and the mass number in terms of atomic units (a. m. u.) (why ?).

1 a. m. u. = 1.6605402 x 10-27kg- All nuclei have approximately the same density (calculate that of iron and knowing that bulk

iron density is 7000 kg m-3, show how much the nucleus is more dense).- Masses of proton, neutron and electron

Proton = 1.007276 a. m. u. = 1.672623 x 10-27 kgNeutron = 1.008665 a. m. u. = 1.674929 x 10-27 kgElectron = 0.000548580 a. m. u. = 9.10939 x 10-31 kg

- The atomic number “Z” is the number of protons, whereas “N’ is the number of neutrons, their sum is the atomic mass “A”,

A = Z + N- A single nucleus species having specific value of both “Z” and “N” is called a nuclide- Nuclides having the same “Z” but different “N” are called isotopes of that element.- Due to the small mass difference of isotopes of the same element, they have slight difference in

their physical properties (melting point, boiling point, diffusion rate…). This is the basis of isotope enrichment in nature and industry.

- Example: two isotopes of chlorine(76%) and (24%)

- Remember that the unified atomic mass unit a. m. u. (or u), is based on the assumption that 12 C has a mass of exactly 12 u .

- The sum of the mass of protons, neutrons, and electrons of any given atom is slightly greater than the atomic mass, this is known as the mass defect. The attractive nuclear force causes this small decrease in mass since this mass corresponds to the nuclear binding energy (EB):

EB = [(Z MB + N mp) - ] C2

Z number of protons.MB mass of 1

1H atom(To include the mass of electrons, we used the mass of the neutral 1

1Hinstead of the mass of proton)

N number of neutrons.mp mass of neutrons.

C2 = 931.5 MeV/a. m. u.- For example, to calculate the binding energy of deuterium, we use the following piece of

information:The nucleon of deuterium 1

2H = a proton + a neuron (together, they are called a deuteron), so the binding energy is:

EB=(1.007825 u + 1.008665 u - 2.014102 u) (931.5M e V/u)= 0.002388 u * (931.5M e V/u)= 2.224 MeV

This is the energy needed to keep the proton and the neutron apart in deuteron.

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- To compare the binding energy needed by different nucleus, we divide by (A), in this case:

EB/A= 2.224/2= 1.112 M e V per nucleon.- It has been theoretically shown (then experimentally found) that proton and neutron are not

fundamental particles, but are made-up of simpler particles called the “quarks ” and the “ leptons ”. This discovery was based on the anomalous magnetic moments.

- The force that binds the protons and the neutrons together in the nucleus (despite the electric repulsion of protons) is an example of what is known as the “strong interaction”; in this particular case it is known as the “nuclear force”.

The Nuclear Force- The properties of a nucleus include mass, charge, size, shape, spin, and magnetic moment.- The “nuclear force” is not yet completely understood, so there is no simple equation that

describes it (since no complete theory combining the fundamental forces has yet been developed), but current theoretical work in particle physics is attempting to devise the grand unification theories which will do so. However, we have simple models (e.g. the liquid-drop model and the shell model) for illustrating how the nuclear force is behaving.

- The motions of the particles within a nucleus will be the factor that defines its properties. Unfortunately, these motions cannot be calculated precisely, so physicists have invented these simplified models, from which properties can be predicted.

- One model assumes that the nucleus is equivalent to a charged liquid drop. The mass of a given isotope is mainly the sum of the free masses of its nucleons. Corrections are calculated for the nuclear binding (the surface tension of the drop), and the electrostatic repulsion of the protons in the nucleus, taking note also of the excess of neutrons over protons, and whether or not the nucleon number is odd or even. The resulting formula can be modified to fit closely all the measured masses of the known isotopes.

- Some nucleus properties, however, can only be explained by the shell model, which assumes that every nucleon moves in an average field created by all the other nucleons. Its permitted orbits, or energy levels, can be calculated by quantum mechanics. Corrections are made for the spin-orbit force, which depends on whether a nucleon's spin is parallel (or opposite) to its orbital momentum. The shell model permits the calculation of the energies of many excited levels of nuclei and the way in which the spins and the momentum of the individual nucleons combine to give the net spin of the nucleus.

- Energy levels of some nuclei demonstrate patterns of regularity. So, in the collective model of the nucleus, these properties are to be determined by the motion of the nucleus as a whole.

- Also, the unified model of the nucleus combines the effects of the collective and shell models to provide comprehensive view that is in more agreement with the experimental observations.

- Despite the lack of a generalized theory, we know some major characteristics of nuclear force:

1- It does not depend on charge2- It has a short range (about 10-13 cm, i.e. one fermi) that is to say, it has little effect at a

distance more than a few fermis from a nucleus, but at distances smaller than about 0.4fermi it becomes strongly repulsive.

3- It is much more stronger than the electrical force. This <0.4fermi region is called the hard core of the strong nuclear force. It is responsible for the density of nuclear matter, whereas the weak nuclear force is involved in nuclear decay processes and in interactions involving the basic particle called the neutrino

4- A particular nucleus cannot interact simultaneously with all the other nucleons, but with those few in its immediate vicinity, (this is called saturation)

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5- It favors binding of protons or neutrons with opposite spin and of pairs, each pair having opposite spin [e.g. alpha particle, which is a 4H nucleus (2p+2n) bound together with total spin zero].

Nuclear Stability and Radioactivity

In the late 1800s, the German physicist Wilhelm Röntgen discovered a strange new ray produced when an electron beam struck a piece of glass. Since these were rays of an unknown nature, they were named "X-rays".

Two months after this discovery, the French physicist Henri Becquerel was performing an experiment where he wrapped different elements with black-coated photographic plates to measure whether these elements could emit rays. If an element did emit a ray, it would penetrate the black coating and expose the photographic plate. Much to his surprise, Becquerel found that a few elements, including uranium, emitted energetic rays without any energy input.

The significance of Becquerel's experiments is that some natural process was responsible for certain elements releasing energetic X-rays. This suggested that some elements were inherently unstable because these elements would spontaneously release different forms of energy. This release of energetic particles (like X-rays) from the decay of unstable atoms is called radioactivity. - Among about 2500 known nuclides, fewer than 300 are stable, others are unstable and decay (by

emitting particles and electromagnetic radiation, a process called radioactivity), during highly different time scales.

- There is a narrow range (called “valley”) of stability on the diagram relating the Z with the N.- Nuclei containing certain numbers of protons or neutrons are usually stable. The values of

these numbers, are 2, 8, 20, 28, 50, 82, and 126, they're called the “ magic numbers ” .

Types of Radioactive EmissionAlpha decay- Alpha decay transforms an unstable big nuclide into another nuclide.- Alpha particle is a 4H nucleus (2p+2n) bound together with total spin zero.- Using the conservation of mass and energy principle, we understand that alpha decay is possible

whenever the mass of the original neutral atom is greater than the sum of the masses of the final neutral atom and helium-4 atom. In alpha decay, the particle tunnels through a potential energy barrier.

- Example

Can 88226R = 226.025403 a. m. u. be transformed into

86222R = 222.017571 a. m. u. by alpha decay ?

Solution:

Alpha emission is possible if the 88226R mass is greater than the sum of the 88

222R mass and

the 24He mass.

We know that 24He = 4.002603 a. m. u., so the difference in mass between the

original nucleus and the decay products is:

226.025403 - (222.017571 + 4.002603) = + 0.005229 a. m. u.Since this value is positive, alpha decay is energetically possible. The energy equivalent to this mass is

E = 0.005229 a. m. u. x 931.5 M e V/a. m. u.= 4.871 MeV

Thus, we expect the decay products to emerge with total kinetic energy of 4.871 M e V. Out of this energy, the alpha particles will acquire most of the released energy since:

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[222/(222+4)] x 4.871 = 4.78 MeV[222/(222+4)] x 4.871 = 4.784788 MeV[222.07571/(222.017571+4.002503)]x4.871=4.784739MeV

This means that alpha particles are always emitted with definite kinetic energies, determined be conservation of momentum and energy. However, because of their charge and mass, -particles can travel only several centimeters in air, or a few tenth or hundredth of a millimeter through solids before they are brought to rest by collisions.

- It is straightforward that when a nuclide emits particle, its Z and N decrease by 2 and its A decreases by 4, moving it closer to stable territory on the N-Z chart.

- The speed of particle can be determined from the curvature of its path in a transverse magnetic field. It is about 1.52 x 107 m/s. This speed is only 5% of the speed of light, so we can use the Newtonian (non-relativistic) kinetic energy formula to calculate the kinetic energy of alpha-particle:

K = ½ mV2

K = ½ 6.64 x 10-27 x (1.52 x 107 m/s)K = 7.7 x 10-13 Joules

= 4.8 MeV

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READING ONLY PART

- In the following, we give a brief account on the difference between the non-relativistic (Newtonian) kinetic energy and the relativistic kinetic energy. This first is applicable only to particles moving with velocities much smaller than the light velocity, whereas the second is applicable to particles moving with very high speed which is slightly smaller than or equal to light velocity.

- Newtonian Kinetic EnergyThe Newton second law says that:

F = m . am massa acceleration (constant)F forceAnd for a particle that changes its velocity from (v1) to (v2) while making a displacement (d), we also know that:

v22 = v1

2 + 2 a di.e. a = ½ [(v2

2 - v12)/d]

so, F = ½ m [(v22 - v1

2)/d]F d = ½ m v2

2 - ½ m v12 = K

K is the change in kinetic energy, and work done- Here, F d is the work done on the moving particle, it is equal to the change in the particle’s

kinetic energy (and the term ½ m v2 is known as the kinetic energy) between the final state and the initial state.

- However, theory of relativity has shown that the Newtonian laws should be rectified in order to apply to high velocity particles. The derivation of the relativistic form of the kinetic energy formula (the general form, with respect to which the Newtonian formula is a special case applicable only for low velocity particles) is out of the scope of this curriculum, so we will give only the final formula:

K = {[1/(1 - (v2/c2))0.5] - 1} m c2

END OF READING ONLY

Beta decay- There are three types of beta decay:

1. - decay2. + decay3. electron capture

- decay- - is en electron. How can the nucleus (which has no electrons inside) emit an electron?

The answer of this dilemma is very simple (it was not such simple when that phenomenon was discovered, however); a neutron is transformed into a proton and - is emitted, nevertheless there must also be an emission of another particle known as “anti-neutrino”.

- the speed of beta particles range up to 0.9995 of the light speed (i.e. its motion is highly relativistic, and they are emitted with a continuous spectrum of energy). (in fact, this high speed would not be possible if there were only - and the nucleus, since the principle of conservation of energy and angular momentum would be violated, and even in this case the emitted particle should have a definite speed - but this is not the case. So, the observation of

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- emission requires, in order not to violate these principles, that a third particle should be also emitted and it must be neutral and of the type spin-1/2 particle, it has been called the “anti-neutrino” - symbol - and has been experimentally observed in 1953; it is the “anti-particle” of the “neutrino” - symbol ). Both the “neutrino” and “anti-neutrino” has zero charge and zero (or very small) mass. The spin angular momentum of a “neutrino” has a component that is opposite to its linear momentum, whereas the “anti-neutrino” has a component of which is parallel to its linear momentum. Until recently, it was believed tat all the neutrinos have zero rest mass, however, there is speculations and some experimental evidence that they may have small non-zero masses.

- in fact, we now know that there are three types of “neutrinos” and each has its corresponding “anti-neutrino”; one is associated with beta decay (e) and the other two are associated with the decay of two other unstable particles known as the “muon” and the “tau”. (the discussion of particle physics is out of the scope of the curriculum, however it is recommended that the student should read about it elsewhere).

- the basic process of - decay isn p + - + e

- - decay usually occurs with nuclides for which the N:Z ratio is too large for stability.- in - decay, “N” decreases by one and “Z” increases by one, so, “A” does not change.- here also, we can use the conservation of mass-energy principle to show that - decay can

occur whenever the neutral atomic mass of the original atom is larger than that of the final products.

- ExampleThe nucleus 27

60Co is an odd-odd unstable nucleus. It is used in many applications of radiation. Show - using the following masses - that this nuclide is unstable relative to - decay.

2760Co = 59.933820 a. m. u.

2860Ni = 59.30788 a. m. u.

Solution:- in - decay, “Z” increases by one (from 27 to 28 in this case) whereas “A” remains 60, so the

final nuclide is 2860Ni which is a mass less than that of 27

60Co by:59.933820 - 59.930788 = 0.003032 a. m. u.so, - decay can occur.

+ decay- nuclides that have a N/Z ratio which is too small for stablity emit a “positron” (the “positron”

is the “anti-particle” of the electron; it identical to the electron but has a positive charge).- + decay process is written as follows:

p n + + + ewhere e is the neutrino of the electron.

- + decay can occur whenever the neutral atomic mass of the original atom is at least two electron masses larger than that of te final atom.

electron capture- the third type of beta decay is know as “electron capture”. There are a few nuclides for which

+ emission is not energetically possible but in which an orbital electron (usually in the K shell) can combine with a proton in the nucleus to form a neutron and a neutrino. The neutron remains in the nucleus and the neutrino is emitted according to the following reaction:

p + - n + e- electron capture can occur whenever the neutral atomic mass of the original atom is larger

than the final atom.- in both + and electron capture, “N” increases by one and “Z” decreases by one, so

shifting the N/Z ratio towards a more stable value.- Example

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The nucleus 2757Co is an odd-even unstable nucleus. It is often used as a source of radiation in a

nuclear process called the “Mössbaur effect”. Show - using the following masses - that this nuclide is stable relative to + decay but can decay by electron capture.

2757Co = 56.936294 a. m. u.

2657Co = 56.935396 a. m. u.

Solution:- the original nuclide is 27

57Co. In + decay and electron capture, “Z” decreases by one (from 27 to 26 in this case) and “A” remains at 57. Thus, the final nuclide is 26

57Fe. Its mass is less than that of 27

57Co by 0.000898 a. m. u., a value smaller than 0.001097 (two electron masses). So, + decay cannot occur . However, the mass of the original atom is greater than the mass of the final atom, so electron capture can occur.

Gamma decay- the energy of internal motion of a nucleus is quantized. A typical nucleus has a set o allowed

energy levels, including a ground state (a state of lowest energy) and several exited states. Because of the greatest strength of nuclear interactions, excitation energies of nuclei are

typically of the order of 1 M e v, compared with a few e V for atomic energy levels.In ordinary physics and chemical transformations, the nucleus always remains in its ground

state. When a nucleus is placed in an excited state, either by bombardment of high energy particles or by a radioactive transformation, it can decay to the ground state by emission of 1 or more photons, called photons, with typical energies of 10 k e V to 5 M e V. This process is called gamma decay.

For example, alpha particles emitted from 226Ra have two possible kinetic energies, either 4.784 or 4.602 M e V. Including the recoil energy of the resulting 222Ra nucleus, these correspond to a total energy of 4.871 or 4.685 M e V, respectively. When an alpha particle with the smaller energy is emitted, the 222Ra nucleus is left in an excited state. It then decays to its ground state by emitting a gamma ray photon with energy:(4.871 - 4.685) = 0.186 M e Vand a photon with this energy is observed during the decay.

- in decay, the element does not change into another one, the nucleus merely goes from an excited state to a less excited state.

Natural Radioactivity- your body has a certain level of radioactivity since it has a certain content of 14C and 40K which

are natural radioactive isotopes.parent nucleus daughter nucleus ... ?

- the most abundant radioactive nuclides found on Earth is the uranium isotope 238U.- it undergoes a series of 14 decays, including 8 emissions and 6 - emissions, terminating

at a stable isotope (206Pb). The first two decays in the series are written as:238U 224Th + 234Th 234Pa + -

- in the second process, the beta decay leaves the daughter nucleus 234Pa in an excited state, from which it decays to the ground state by emitting a ray photon. An excited state is denoted by an asterisk (*), so we can represent the emission as:

234Pa* 234Pa + - an interesting feature of the 238U decay series is the branching that occurs at 214Bi. This nuclide

decays to 210Pb by emission of an and a - (which can occur in either order).We also note that the series includes unstable isotopes of several elements that also have stable isotopes, including thalium (TL), lead (Pb) and bismuth (Bi). The unstable isotopes of these elements that occur in the 238U series all have too many neutrons to be stable.

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- many other decay series are known. Two of these occur in nature, one starting with the uncommon isotope 235U and ending with 207Pb, the other starting with thorium 232 and ending with 208Pb.

Activity and Half-lives- the decrease of the original number of a radioactive isotope by radioactive decay is a statistical

process and it obeys an equation of the type known as the first order kinetic equation.- however, there is no way to predict when a particular individual nucleus will decay; there

is only a probability in this respect (see definition below).- let N(t) be the (very large) number of radioactive nuclei in a sample at time t, and let d N(t) be the

(negative) change in the number during a short time interval d t. The number of decays during the interval d t is -d N(t). The rate of change of N(t) is the negative quantity d N(t)/d t; thus -d N(t)/d t is called the decay rate or the activity of the specimen. The larger the number of nuclei in the specimen, the more nuclei decay during any time interval. That is to say that the activity is directly proportional to N(t)); in fact it equals a constant () multiplied by N(t):

-d N(t)/d t = N(t)

- the constant is called the decay constant, (it is a constant for any given nuclide, but it has different values for different nuclides). A large value of corresponds to rapid decay; a small value corresponds to slower decay. Solving this equation for shows that is the ratio of the number of decays per time to the number of remaining radioactive nuclei; can be interpreted as the probability (per time) that any individual nucleus will decay.

- we can write the following radioactive decay equation and use it to measure the time interval during which a certain initial content of the radioactive material has resulted in an observed (smaller) content:

N(t) = No e- t

- this equation can be directly used to calculate the number of the remaining nuclei N(t) starting from the initial number No (that was present at time zero). It is obvious that this equation means that activity -d N(t)/d t depends also on time.

- the half-life is the time required for the number of radioactive nuclei to decrease to one-half the original number No.

- to get the relation between the half-life t1/2 and the decay constant , we set N(t)/No = ½ and t = T½ in the last equation, so we obtain:

½ = e - t½

then we take logarithms of both sides and solve for T½:T½ = l n 2 / = 0.693 /

- the mean life time (T mean) (generally called the lifetime of a nucleus or unstable particle) is proportional to the half-life T½:

T mean= 1/ = T½ / l n 2 = T½ / 0.693- the number of radioactive nuclei and the activity approaches zero only as time t approaches

infinity.- a common unit of activity is the Curie (C i) = 3.7 x 1010 decays per second (approximately equal

to the activity of 1 gram radium). The SI unit of activity is the Becquerel (B q) = 1 decay per second, i.e. 1 Ci = 3.7 x 1010 B q

Example (Carbon dating) - Before 1900, the activity per mass of atmospheric carbon due to the presence of 14C averaged

about 0.22 Bq per gram of carbon.a) what fraction of carbon atoms were 14C?b) in analyzing an archeological specimen containing 500 mg of carbon, you observe 174

decays in one hour. What is the age of the specimen? (assuming that its activity per mass of carbon when it died was that average value of air).

- Solution

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a) we’ll use the equation -d N(t)/d t = N(t)and compare it with No. First, we find form the equation:

T½ = 0.693 / = 0.693 / T½

to the transform half life years into seconds, we calculate y = 5730 y * (3.156 * 107 s/y)

= 1.08 * 1011 sand substitution gives

= 0.693 / T½ = 0.693 / 1.808 * 1011 s= 3.83 * 1012 s-1

or alternatively using the half-life in years directly, we have = 0.693 / 5730 = 1.209 * 10-4 y-1

thus, from the equation:-[d N(t)/d t]*(1/) = N(t)

activity / = N(t)

we getN(t) atoms = 0.255 s-1 / 3.83 * 10-12 = 6.66 * 1010

the total number of carbon atoms in one gram (which is 1/12.011 mole) can be calculated using Avogadro number:

1/12.011 * 6.022* 1023 = 5.01 * 1023 atomsthe ratio of 14C atoms to the total carbon atoms is the 14C atoms corresponding to 1 carbon atom, it is:

6.66 * 1010 / 5.01 * 1023 = 1.33 * 10-12

but it is obvious that it is more practical to multiply both sides by 3 * 1012 to get the nearest integer number of 14 C atoms in a number of 3 * 10 12 carbon atoms :

6.66 * 1010 / 5.01 * 1023 * 3 * 1012 = 3 * 1012 * 1.33 * 10-12

= 3.99= 4 atoms of 14C

in 3 * 1012 total carbon atomsNote that instead of multiplying by 3 * 1012 we could also multiply both sides by the number of carbon atoms in one mole (Avogadro number) in order to calculate the number of 14C atoms in one mole of carbon:

6.66 * 1010 / 5.01 * 1023 * 6.022 * 1023=6.022 * 1023 * 1.33 * 10-12

= 800.926 * 1011

14C atoms inin one mole of carbon atoms

b) assuming that the activity per gram if carbon is the specimen when it died was 0.255 Bq/g, this is equal to

0.255 s-1 g-1 x 3600 sec/hour = 918 h-1 g-1

so, the activity of 500 mg of carbon was0.5 g x 918 h-1 g-1 = 499 h-1

but we know that activity of these organic remnants at the time of measurement in only 174 h-1.Since the activity is proportional to the number of radioactive nuclei, the activity ratio 174/499 = 0.379 equals the atom number ratio N(t) / No.Now, we solve the following equation for time:

N(t) = No e - t

and insert values for N(t) / No and :l n N(t) = ln No - tl n N(t) - ln No = - tl n N(t) / ln No = - t

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t = -[l n N(t)/No] / = -l n 0.379 / 1.209 x 10-4 y-1

= 8020 yearsi.e. the specimen died and stopped taking CO2 out of the air about 8000 years ago

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References

Hewitt, Paul. Conceptual Physics. Harper Collins College Publishers (San Francisco, 1993) pp 596.

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