Radioactivity: stability of the nucleus; Law of radioactive decay; Mean life and half-life;

Alpha decay; Beta decay- energy released, spectrum and Pauli's prediction of neutrino;

Gamma ray emission, energy-momentum conservation: electron-positron pair creation

By gamma photons in the vicinity of a nucleus.

(8 Lectures)

Fission and fusion- mass deficit, relativity and generation of energy; Fission- nature of

Fragments and emission of neutrons. Fusion and thermonuclear reactions driving stellar

Energy (brief qualitative discussions).

Radioactivity

Radioactivity is a process by means of which a nucleus emits various particles such as alpha, beta and gamma without any external perturbation. It is a spontaneous process inside the nucleus and unaffected by changes like

temperature, pressure etc. to which the atom may be subjected.

Discovery of Radioactivity

Way back in 1896, the French Physicists Henry Becquerel observed that a photographic plate wrapped in black paper was affected by some uranium salt

kept outside it. In fact he was studying fluorescence from uranium. He was also amazed to notice that it does not matter whether the photographic plate is kept at dark or sunlight. It was he along with his students Marie and Pierre

Curie realized that the source of the radiation which affects the photographic plate in all situations is the nucleus and the mechanism was named as radioactivity. All three received Nobel Prize in physics in 1903 for this

pioneering work. Eight years later Mme Curie again got the Noble Prize in chemistry for isolating the radioactive element radium.

Law of radioactive decay

When a nucleus disintegrates by emitting charged particles (α, β) or a γ- ray photon, or by capturing electron from the atomic shell, the process is called radioactive decay. This decay is spontaneous. The rate of change of nuclei in a

radioactive sample is given by law of radioactive decay. Let us take a radioactive sample containing N0 nuclei at time t =0, i.e., at the beginning. We want to calculate the number N of these nuclei left after time t. The number of

nuclei of a radioactive sample disintegrating per second is called the activity of that sample.

𝑑𝑁

𝑑𝑡= Rate of decrease of nuclei with time

Activity at time t. Experimentally, it has been observed that the activity at any instant of time t is directly proportional to the number N of parent type nuclei present at that

time.

−𝑑𝑁

𝑑𝑡𝛼 𝑁

−𝑑𝑁

𝑑𝑡= 𝜆𝑁

or where λ>0, is the proportionality constant known as decay constant.

The negative sign indicates that N decreases as t increases. By integrating Eqn. m4.1 we get,

or

Thus we see that the law of radioactive decay is exponential in nature. Number

of radioactive nuclei versus time is plotted in Fig. m4.1.

FIGURE : Number of radioactive nuclei with time showing the

exponential nature of radioactive decay. On the basis of the radioactive decay law the following physical quantities are

defined:

Half Life (T1/2):

It is convenient to define a time interval during which half of a given sample of radioactive substance decays. This interval is called the half-life or half-value

period of that substance, denoted by T1/2,

𝑒−𝜆𝑇1 2⁄ = 2

or

Mean Life (𝝉):

Individual radioactive atoms may have life spans between zero and infinity. Therefore, the average or mean life 𝜏 is defined as,

𝜏 = Total life time of all nuclei in a given sample/ total number of nuclei in that

sample Mathematically, this can be represented as

Where dN1 atoms have a lifetime t1, dN2 atoms a lifetime t2 and so on.

In the notation of integral calculus

From Eqn. (6.2) we can write,

𝑑𝑁 = −𝜆𝑁0 𝑒−𝜆𝑡𝑑𝑡

𝜏 = 𝜆 ∫ 𝑡𝑒−𝜆𝑡𝑑𝑡∞

0

= 𝜆 [−𝑡

𝜆𝑒−𝜆𝑡 +

1

𝜆∫ 𝑒−𝜆𝑡𝑑𝑡

∞

0

]

=1

𝜆

𝑚𝑒𝑎𝑛 𝑙𝑖𝑓𝑒, 𝜏 =1

𝜆

The mean or average life of a radioactive element is thus not the same as its half-life. The mean life is the reciprocal of the decay constant.

Activity or strength

Differentiating Eqn. (6.2) w.r.t. time we get,

at

Hence from (6.7) we get

or,

Where,

At is called the activity or the strength of the sample and is proportional to the

rate of disintegration. The activity or the strength At of a radioactive sample at any instant of time t is thus defined as, the number of disintegrations occurring in the sample in unit time t, that is,

The activity per unit mass of a sample is called its specific activity. is the initial activity of the sample. The half-life may be defined in terms of activity as well. It is the time in which the activity drops to one-half of the initial activity.

Units of Activity

The customary unit of activity of any radioactive material is curie (Ci), which

is defined as the activity of any radioactive substance that disintegrates at the rate of 3.7 x 1010 disintegrations per second. A thousandth part of a curie is called a millicurie (mCi). A still smalled unit is the microcurie ( μCi). So, by

definition

1 Ci = 1 curie = 3.7 x 1010 disint./sec

1 mCi = 10-3 Ci; 1 μCi = 10-6 Ci Another unit of activity is the rutherford, which is defined as a disintegration rate of 106disintegrations per second.

Example m4.1: Calculate the activity of (i) One gram of radium , whose half- life is 1622 years and

(ii) 3 x 10-9 kg of active gold, whose half- life is 48 minutes.

Solution: (i) Here,

= 2.66 x 1021 atoms.

Activity =

(ii) In the same way number of Au, N = 9.04 x 1015 atoms λ = 2.406 x 10-4 sec-1

and Activity = λN = 58.9 Ci

Let us now discuss how to determine the radioactive constants:

In radioactivity, it is important to determine the values of disintegration

constant, half life, average life etc. Considering the activity of any radioactive nuclide we can write,

where, x = ln At and C = ln A0

Therefore, disintegration constant λ can be determined from the slope of the

straight-line graph of ln At versus t, provided the activity of the nuclide under study is known. The disintegration constant of the radio nuclide having extremely high λ values can be determined from Geiger-Nuttal law, which

connects the range of α particles with the disintegration constant by a simple relationship

where, A, B are constants and have different values for different radioactive series. Therefore, if the value of the range R is known (discussed in chapter 2), λ can be calculated from the above equation. Disintegration constant of the

radioactive nuclide having short λ values can be determined from the condition of secular equilibrium also. The condition for such equilibrium is N1λ1 = N2λ2 . Hence, if N1, λ1 and N2 are known, λ2 can be evaluated. The half-life of a

radioactive nuclide can be found by plotting the activity of a sample with time. The time corresponding to half of the activity is the half-life of that particular radioactive nuclide.

Alpha ,Beta& Gamma Decay

Many nuclei are radioactive; that is, they decompose by emitting particles and in doing so, become a different nucleus. In our studies up to this point, atoms of one element were unable to change into different elements. That is because in all other types of changes we have talked about only the electrons were changing. In these changes, the nucleus, which contains the protons which dictate which element an atom is, is changing. All nuclei with 84 or more protons are radioactive and elements with less than 84 protons have both stable and unstable isotopes. All of these elements can go through nuclear

changes and turn into different elements.

In natural radioactive decay, three common emissions occur. When these emissions were originally observed, scientists were unable to identify them as some already known particles and so named them

• alpha particles (α), • beta particles, (β), and • gamma rays (γ)

herealpha particles were identified as helium-4 nuclei, beta particles were identified as electrons, and gamma rays as a form of electromagnetic radiation like x-rays except much higher in energy and even more dangerous to living systems.

Alpha 1. α-particles have zero spin and positive intrinsic parity.

2. α- particles are positively charged and charged to mass ratio has been found to be 4.826810^7

coulomb /Kg, which shows that these are doubly charged helium atoms

3. The α- particles have great ionizing power and less penetrating power.

4. The α- particles are deflected by magnetic and electric fields. Rosenblum in 1930, showed with

the help of magnetic spectrograph that in many cases several groups of mono energetic α-particles

are emitted by a single nuclide. This led to the conclusion that the emission of α-particles leave the

daughter nuclei an excited state. This fact was later on confirmed by observing ϓ rays along with

α-decays.

Alpha Decay

The nuclear disintegration process that emits alpha particles is called alpha decay. An example of a nucleus that undergoes alpha decay is uranium-238.

The alpha decay of UU-238 is

U92238→He24+Th90234(17.3.1)

In this nuclear change, the uranium atom (U92238)transmuted into an atom of

thorium (Th90234) and, in the process, gave off an alpha particle. Look at the symbol for the alpha particle: He24. Where does an alpha particle get this symbol? The bottom number in a nuclear symbol is the number of protons. That means that the alpha particle has two protons in it which were lost by the uranium atom. The two protons also have a charge of +2. The top number, 4, is the mass number or the total of the protons and neutrons in the particle. Because it has 2 protons, and a total of 4 protons and neutrons, alpha particles must also have two neutrons. Alpha particles always have this same composition: two protons and two neutrons.

Another alpha particle producer is thorium-230.

Th90230→He24+Ra88226(17.3.2)

These types of equations are called nuclear equations and are similar to the chemical equivalent discussed through the previous chapters.

Beta Decay

Another common decay process is beta particle emission, or beta decay. A beta particle is simply a high energy electron that is emitted from the nucleus. It may occur to you that we have a logically difficult situation here. Nuclei do not contain electrons and yet during beta decay, an electron is emitted from a nucleus. At the same time that the electron is being ejected from the nucleus, a neutron is becoming a proton. It is tempting to picture this as a neutron breaking into two pieces with the pieces being a proton and an electron. That would be convenient for simplicity, but unfortunately that is not what happens; more about this at the end of this section. For convenience sake, though, we will treat beta decay as a neutron splitting into a proton and an electron. The proton stays in the nucleus, increasing the atomic number of the atom by one. The electron is ejected from the nucleus and is the particle of radiation called beta.

To insert an electron into a nuclear equation and have the numbers add up properly, an atomic number and a mass number had to be assigned to an electron. The mass number assigned to an electron is zero (0) which is reasonable since the mass number is the number of protons plus neutrons and an electron contains no protons and no neutrons. The atomic number assigned to an electron is negative one (-1), because that allows a nuclear equation containing an electron to balance atomic numbers. Therefore, the nuclear symbol representing an electron (beta particle) is

e−10 or β−10

Thorium-234 is a nucleus that undergoes beta decay. Here is the nuclear equation for this beta decay.

Th90234→e−10+Pa91234(17.3.3)

Gamma Radiation

Frequently, gamma ray production accompanies nuclear reactions of all types. In the alpha decay of U-238, two gamma rays of different energies are emitted in addition to the alpha particle.

U92238→He24+Th90234+2γ00

Virtually all of the nuclear reactions in this chapter also emit gamma rays, but for simplicity the gamma rays are generally not shown. Nuclear reactions produce a great deal more energy than chemical reactions. Chemical reactions release the difference between the chemical bond energy of the reactants and products, and the energies released have an order of magnitude of 1×103kJ/mol. Nuclear reactions release some of the binding energy and may convert tiny amounts of matter into energy. The energy released in a nuclear reaction has an order of magnitude of 1×1018kJ/mol. That means that nuclear changes involve almost a million times more energy per atom than chemical changes!

The Ionizing and Penetration Power of Radiation

With all the radiation from natural and man-made sources, we should quite reasonably be concerned about how all the radiation might affect our health. The damage to living systems is done by radioactive emissions when the particles or rays strike tissue, cells, or molecules and alter them. These interactions can alter molecular structure and function; cells no longer carry out their proper function and molecules, such as DNA, no longer carry the appropriate information. Large amounts of radiation are very dangerous, even deadly. In most cases, radiation will damage a single (or very small number) of cells by breaking the cell wall or otherwise preventing a cell from reproducing.

The ability of radiation to damage molecules is analyzed in terms of what is called ionizing power. When a radiation particle interacts with atoms, the interaction can cause the atom to lose electrons and thus become ionized. The greater the likelihood that damage will occur by an interaction is the ionizing

power of the radiation.

Much of the threat from radiation is involved with the ease or difficulty of protecting oneself from the particles. How think of wall do you need to hide behind to be safe? The ability of each type of radiation to pass through matter is expressed in terms of penetration power. The more material the radiation can pass through, the greater the penetration power and the more dangerous they are. In general, the greater mass present the greater the ionizing power and the lower the penetration power.

Comparing only the three common types of ionizing radiation, alpha particles have the greatest mass.

• Alpha particles have approximately four times the mass of a proton or neutron and approximately 8,000 times the mass of a beta particle.

Because of the large mass of the alpha particle, it has the highest ionizing power and the greatest ability to damage tissue. That same large size of alpha particles, however, makes them less able to penetrate matter. They collide with molecules very quickly when striking matter, add two electrons, and become a harmless helium atom. Alpha particles have the least penetration power and can be stopped by a thick sheet of paper or even a layer of clothes. They are also stopped by the outer layer of dead skin on people. This may seem to remove the threat from alpha particles but only from external sources. In a situation like a nuclear explosion or some sort of nuclear accident where radioactive emitters are spread around in the environment, the emitters can be inhaled or taken in with food or water and once the alpha emitter is inside you, you have no protection at all.

Beta particles are much smaller than alpha particles and therefore, have much less ionizing power (less ability to damage tissue), but their small size gives them much greater penetration power. Most resources say that beta particles can be stopped by a one-quarter inch thick sheet of aluminum. Once again, however, the greatest danger occurs when the beta emitting source gets inside

of you.

Gamma rays are not particles but a high energy form of electromagnetic radiation (like x-rays except more powerful). Gamma rays are energy that has no mass or charge. Gamma rays have tremendous penetration power and require several inches of dense material (like lead) to shield them. Gamma rays may pass all the way through a human body without striking anything. They are considered to have the least ionizing power and the greatest penetration power.

The safest amount of radiation to the human body is zero. It isn't possible to be exposed to no ionizing radiation so the next best goal is to be exposed to as little as possible. The two best ways to minimize exposure is to limit time of exposure and to increase distance from the source.

Pauli's prediction of neutrino: The neutrino — the mystery and the discovery

In the 1920s, physicists were confused: the phenomenon of decay (in which an electron is emitted

from the atomic nucleus) seemed to violate conservation laws. The energy spectrum of the electrons,

or -rays, is continuous: if energy is conserved, another, variable, amount of energy must somehow

leave the system. In 1933, Pauli had devised an explanation in terms of another, undetected, particle being emitted by the nucleus; Fermi called it 'the neutrino'.

The known properties of the neutrino are summarized below.

Properties of the Neutrino

Spin: 1/2 Mass: < 1/500 electron mass, if any Charge: 0 Magnetic moment: < 10–9 Bohr magneton

Cross-section for reaction: – + p+ → + + n0 at 3 MeV = 10–43 cm2

Neutrino + not identical with antineutrino -

Pair Production

Pair production is the phenomena Here a photon is incident on the atomic nucleus, electron-positron pair is produced.It is the process to get mass from energy.

Pair Production Process

Pair Production is a method where energy can be converted into mass. In this process a pair of

elementary particle and its antiparticle is formed by high energy photon incident on a heavy nucleus.

It explains the concepts that in what way our internal world is converted into physical world that we

see.

This process can be viewed in two ways :

1. First is as a particle and an anti particle 2. Second as a particle and a hole.

The basic process of pair production includes the interaction of a packet or wave of energy with a

nucleus which forms an electron positron pair. The process of pair production occurs naturally a

photon of energy greater than 1.02 million electron volt passes nearby the electric field of a heavy

atom which has large number of protons with atomic number of about 80 to 90. If the initial energy

photon has its energy greater than 1.02 eV than energy is divided into the kinetic energy of motion of

the two particles.

Pair production is represented by following equation:

E = 2(mo c2) + KE(p) + KE(p')

This equation obeys conservation of energy.

Here,

E is the initial energy of incident photon.

moc2 is rest energy of elementary particle which is equal to the rest energy of its anti particle.

KE (p) is the kinetic energy of elementary particle

KE (p)’ is the kinetic energy of anti particle.

Electron Positron Pair Production

Electron Positron Pair Production includes the formation of electron and its anti particle i.e.

positron. This is formed by the interaction of high energy photon with a heavy nucleus. Photon splits

into electron and positron. Both electron and its anti particle have rest mass energy equal to 0.511

million electron volts.

As mass and energy are similar or we can say equivalent mass energy is the amount of energy that

it needs to convert into mass. The equivalent energy for an electron is 0.511 million electron volts

(i.e. for mass of electron which is 9.11 [Math Processing Error]× 10-24 gms).

For electron positron pair production energy conservation equation can be written as:

E = 2(mo c2) + KE (-e) + KE (+e)

Here,

E is the initial energy of incident photon.

moc2 is rest energy of the electron which is equal to the rest energy of its anti particle i.e. positron. Its

value is 0.511 million electron volts.

KE (-e) is the kinetic energy of electron

KE (+e) is the kinetic energy of anti particle of electron.

Basic Fusion Process, Stellar Evolution and Fusion Reaction Rate

Nuclear fusion, process by which nuclear reactions between lightnelements form

heavier elements (up to iron). In cases where the interacting nuclei belong to elements

with low atomic numbers (e.g., hydrogen [atomic number 1] or its

isotopes deuterium and tritium), substantial amounts of energy are released.

From binding energy curve, it is clear that the B.E./A value of light nuclei on the steep portion of the curve is less than that for nuclei of intermediate mass numbers. So, if two lighter nuclei combine or fuse together to produce a

relatively heavier nucleus, there would be a greater binding energy and a consequent decrease in nuclear mass. This would thus result in a positive Q-

value and release of energy. Such type of nuclear reaction is known as nuclear fusion, a process opposite to that of fission.

Typical examples of fusion reaction are

with energy associated as shown.

The reactions appear to be very simple, still an useful fusion reactor could not

be built, why? The reason behind this is that to drive a fusion a reaction, nuclei needs high kinetic energy to overcome the Coulomb barrier. This kinetic energy can be provided by raising the temperature of the reactant nuclei. Let

me clarify this point. Fusion reaction is possible only when two interacting particles can come near each other within a distance of the order of 10-14 m. This means that they will have to overcome the mutual repulsive Coulomb

barrier acting between them due to positive charges and come within a distance of the above order. Therefore, their relative velocity should be very large, i.e., the temperature of the medium must be very high. Balancing the

kinetic energy and mutual electrostatic energy, one can write

or

where R ~ 10-14 m.

Therefore, the necessary condition for the occurrence of fusion process is that the medium must be very hot. This is why it is called thermonuclear reaction. These thermonuclear reactions have a great impact in energy generation in

stars and also in evolution of stars. Considering sun as a medium size star, it radiates energy at the rate of about 1026 Js-1. This can be considered as typical hot star whose interiors have a temperature of about 20 x 106 K. The age of

the sun is at least 5 x 109 year, so that the total loss of energy during this period is exceedingly high. Therefore, the natural question is how did the sun maintain this energy output for so long - what is the source of all stellar

energy? Helmholtz and Kelvin suggested that a slow gravitational contraction of the stars might be the source; Jeans proposed annihilation of high-energy

protons and electrons in the stars as the source. In 1938, Hans Bethe suggested that the stellar energy is produced by thermonuclear reactions in which protons are combined and transformed into helium nuclei. This is known

as the proton-proton cycle, and is applicable for relatively low stellar temperatures. The cycle is:

by addition, we have

For the main sequence - stars having higher temperatures, Bethe suggested an alternative to the proton-proton cycle - the carbon nitrogen cycle.

The cycle is

Adding,

The conversion of hydrogen to helium will continue until whole of star’s supply

of protons is used up. In the case of the sun, both the above cycles take place with roughly equal probability, and it is estimated that it will be about 3 x 1010 years before the protons have all been converted to helium. Fusion of four

protons into helium and overcoming the coulomb repulsion is only possible because of the extremely high initial temperature.

It is now believed that the carbon-nitrogen cycle starts at a later stage in the

life of a star. According to the theory of stellar evolution, a star is formed by way of an enormous amount of matter at certain point in space. As the mass contracts, its temperature increases, and when it reaches the value of about

2,00,000 0C, the proton-proton cycle starts operating. When the deuterons, formed as a product in the cycle, are completely exhausted, the star shrinks and the temperature again increases. At a temperature of about million

degrees, protons interact with heavier elements such as beryllium, lithium, and boron, forming a helium nucleus. At this stage the star looks very bright and is called a red giant. When the above type is also exhausted, further contraction

takes place and the temperature rises to about 20 x 106 0C. The carbon-nitrogen cycle operates at this stage and supplies the energy for the major portion of the radiating life of a star.

Fusion reactions in stars

Fusion reactions are the primary energy source of stars and the mechanism for

the nucleosynthesis of the light elements. In the late 1930s Hans Bethe first recognized

that the fusion of hydrogen nuclei to form deuterium is exoergic (i.e., there is a net

release of energy) and, together with subsequent nuclear reactions, leads to the

synthesis of helium. The formation of helium is the main source of energy emitted by

normal stars, such as the Sun, where the burning-core plasma has a temperature of less

than 15,000,000 K. However, because the gas from which a star is formed often

contains some heavier elements, notably carbon (C) and nitrogen (N), it is important to

include nuclear reactions between protons and these nuclei. The reaction chain between

protons that ultimately leads to helium is the proton-proton cycle. When protons also

induce the burning of carbon and nitrogen, the CN cycle must be considered; and,

when oxygen (O) is included, still another alternativescheme, the CNO bi-cycle, must

be accounted for. (See carbon cycle.)

The proton-proton nuclear fusion cycle in a star containing only hydrogen begins with

the reaction

H + H → D + β+ + ν; Q = 1.44 MeV,

where the Q-value assumes annihilation of the positron by an electron. The deuterium

could react with other deuterium nuclei, but, because there is so much hydrogen, the

D/H ratio is held to very low values, typically 10−18. Thus, the next step is

H + D → 3He + γ; Q = 5.49 MeV,

where γ indicates that gamma rays carry off some of the energy yield. The burning of

the helium-3 isotope then gives rise to ordinary helium and hydrogen via the last step

in the chain: 3He + 3He → 4He + 2(H); Q = 12.86 MeV.

At equilibrium, helium-3 burns predominantly by reactions with itself because its

reaction rate with hydrogen is small, while burning with deuterium is negligible due to

the very low deuterium concentration. Once helium-4 builds up, reactions with helium-

3 can lead to the production of still-heavier elements, including beryllium-7, beryllium-

8, lithium-7, and boron-8, if the temperature is greater than about 10,000,000 K.

The stages of stellar evolution are the result of compositional changes over very long

periods. The size of a star, on the other hand, is determined by a balance between the

pressure exerted by the hot plasma and the gravitational force of the star’s mass. The

energy of the burning core is transported toward the surface of the star, where it is

radiated at an effective temperature. The effective temperature of the Sun’s surface is

about 6,000 K, and significant amounts of radiation in the visible and infrared

wavelength ranges are emitted.

Fission Reaction Mechanism, Energy in Fission Reaction and Basic Formulation on Fission

Reactor

Let us first to understand the mechanism of nuclear fission process

based on simple liquid drop picture of a nucleus. Bohr and Wheeler explained the fission process considering the liquid-

drop model, where nucleus resembles a drop of liquid in the respect that each constituent particle interacts equally with its nearest neighbors. In a liquid droplet, the total energy of the molecules inside

the droplet is not sufficient enough to overcome the forces that hold the molecules in the form of a droplet. However, if an external force is applied so that it is set into oscillation, the system passes through a

series of stages, of which the most significant ones are shown in Fig. m3.1. This drop is at first spherical (S-1) and then elongated into an ellipsoid (S-2). Although the volume remains constant, the surface

area has increased. But provided the volume energy exceeds the surface energy, the drop will return to its original form. However, if the deforming force is sufficiently large, the drop will acquire a shape

similar to a dumb-bell as in S-3. In this state, the surface energy will generally exceed the volume energy, which provides the cohesive force for the liquid drop. Consequently, the drop will not return to its initial

shape but will rather split into two droplets. These will, at first, be somewhat deformed, as in S-4, but finally they will become spherical (S-5).

FIGURE m3.1 Schematic representation of nuclear fission process based on liquid drop model.

Now let us calculate energy production in nuclear fission reaction and also establish basic equations related to power generation through this reaction.

Let us consider the following fission reaction

The fission fragments decay to their radioactive isotopes through β-decay in the following manner:

and

The total mass before fission is The isotopic mass after fission is

Mass difference = 0.223 amu. Therefore, Q = 0.223 X 931 = 208 MeV.

In MKS unit, Q = 208 X 1.6 X 10-13 = 3.32 X 10-11 joules. Considering U of mass equal to 1 gm, the energy released is (6.02 X

1023/235) X 3.32 X 10-11 = 8.50 X 1010 joules, which is indeed very large. In the unit of power, 1 gm U generates 8.50 X 1010watt -second i.e. 2.36 X 104 kw-hr. Therefore, one gm-atom U produces a power of

5.54 X 106 kw-hr.

Let us see how many fission reactions are required to generate 1 Watt power.

Example m3.3: When a U-235 nucleus undergoes fission, 200 MeV energy is released. How many fissions per sec are needed for

generating power of 1 Watt?

Solution: 1 Watt = 1 joule per sec = 6.242 X 1012 MeV/s

No. of fission needed per sec =

Next important issue related to the realization of generation of power from a reactor is to satisfy the condition of criticality. In a finite size reactor, some neutrons are lost by leaking out. The criticality

condition is then defined as the effective multiplication factor, which is basically the ratio of number of neutrons resulting from fission in each generation to the total number lost by both absorption and leakage in

the preceding generation. Let, for fissile material, which is a combination of U-238 and U-235, at an instant there be N0 fast neutrons available to produce fission. A few of them may produce

fission in U-238 and hence the number of neutrons increases by a factor ν and become N0ν. These neutrons must be slowed down before they produce fission in U-235. The slowing down process is done in

moderator (discussed later), which reduces the energy of the fast neutrons to thermal energy. Considering p to be the fraction slowed

down, the available neutrons are N0νp. Of these again, a fraction f may succeed in producing fission in U-235 before being lost by diffusion or otherwise. Therefore the number of U-235 nuclei undergoing fission

is N0νpf. If at each fission of U-235, fast neutrons are generated to start the cycle again, the total number of neutrons after one cycle is

The ration N/N0 (=νpfe) is also known as reproduction factor, k. The requirement of criticality in a finite system is thus k = 1; in these

circumstances a steady-state fission chain would be possible. If the conditions are such that k < 1, the chain would be convergent and gradually die out. In each generation, more neutrons would be lost in

one way or another than are produced by fission, and so the neutron density, and hence the fission rate would decrease steadily. Such system is known to be subcritical. Finally, if k > 1, the chain is

divergent and the system is known to be supercritical. More neutrons are produced than are lost in each generation, so that both neutron population and fission rate increase continuously.