NUCLEAR ENERGY -

NUCLEAR REACTION

A reaction in which suitable high energy particles bombarding a stable nucleus change it into another nucleus is called a nuclear reaction. Popularly, there are two nuclear reactions named as nuclear fission and nuclear fusion reactions.

It is represented by

$\fn_cm \large X+x\rightarrow Y+y\pm Q$

Here Q is the energy equivalent to a mass defect in the reaction. $\fn_cm \large \left [ E=\Delta m c^2 \right ]$

If Q is +ve, it means that energy is released in the reaction. (Exothermic reaction)

If Q is -ve, it means that extra energy must be supplied to carry out the reaction. (Endothermic reaction)

$\fn_cm \large ex:\;\;\;_{7}^{}\textrm{N}^{14}+_{2}^{}\textrm{He}^4\rightarrow [_{9}^{}\textrm{F}^{18}]\rightarrow _{8}^{}\textrm{O}^{17}+_{1}^{}\textrm{H}^1-1.19Mev$

Here energy 1.19 MeV is -ve. So extra energy is required. Therefore bombarding α- particles must have an energy of 1.19 MeV to carry out this reaction. The above reaction name is (α,p)

In all types of nuclear reactions, the following conservation laws are obeyed.

Conservation of linear momentum
Conservation of total energy
Conservation of charge.
Conservation of a number of nucleons.

Ex.1. $\fn_cm \large _{3}^{}\textrm{Li}^7+ _{1}^{}\textrm{H}^1\rightarrow _{4}^{}\textrm{Be}^8\rightarrow _{2}^{}\textrm{He}^4+_{4}^{}\textrm{He}^4\;\;(p,\alpha)$

2. $\fn_cm \large _{6}^{}\textrm{C}^{12}+ _{1}^{}\textrm{H}^1\rightarrow _{7}^{}\textrm{N}^{13}\rightarrow _{7}^{}\textrm{N}^{13}+\gamma\;\;(p,\gamma)$

3. $\fn_cm \large _{8}^{}\textrm{O}^{16}+ _{1}^{}\textrm{H}^2\rightarrow _{7}^{}\textrm{N}^{14}+_{4}^{}\textrm{He}^4\;\;(d,\alpha)$

4. $\fn_cm \large _{1}^{}\textrm{H}^{2}+ \gamma\rightarrow _{1}^{}\textrm{H}^{1}+_{0}^{}\textrm{n}^1\;\;(photo\;nuclear\;reaction)$

We know that between Z=30 and A= 170 binding energy per nucleon is nearly constant (8.0 MeV). For lighter nuclei, a<30 and for heavier nuclei A> 170, the binding energy per nucleon is less than 8.0 MeV.

Now the greater the binding energy, the less is the total mass of a bound system, such as a nucleus.

i.e we can say that, if nuclei with less total B.E transform to nuclei with greater B.E, there will be a net huge energy release. According to this concept, there are two most important phenomena(nuclear reaction) that occur.

Nuclear fission $\fn_cm \large \rightarrow$ When a heavy nucleus decays into two or more intermediate-mass fragments.

Nuclear fusion $\fn_cm \large \rightarrow$ When light nuclei fuse into a heavier nucleus.

Energy from exothermic chemical reactions such as coal or petroleum involved is in the range of electron volt. On the other hand, in a nuclear reaction, the energy release is of the order of MeV. For example fission of 1 kg of Uranium generates $\fn_cm \large 10^{14}J$ of energy whereas 1 kg of coal generates the energy of $\fn_cm \large 10^{7}J$ .

NUCLEAR FISSION

It is the phenomenon of splitting of a heavy nucleus (usually A>230) into two or more lighter nuclei. In this process, certain mass disappears and release of huge amount of energy due to this mass defect calculated by $\fn_cm \large E=\Delta m c^2$

Thus when a uranium nucleus is bombarded by a slow-moving neutron (called thermal neutron), the uranium nucleus splits into two medium-mass nuclei with the release of a huge amount of energy.

$\fn_cm \large _{92}^{}\textrm{U}^{235}+ _{0}^{}\textrm{n}^1\rightarrow \left [ _{92}^{}\textrm{U}^{236} \right ]\rightarrow _{56}^{}\textrm{Ba}^{141}+_{36}^{}\textrm{Kr}^{92}+3 _{0}^{}\textrm{n}^1+ Q$

Mass defect in this reaction comes out to be 0.2135 a.m.u.

i.e energy released per fission of $\fn_cm \large _{92}^{}\textrm{U}^{235}$ is $\fn_cm \large E=0.2135\times931 MeV=200.4 MeV$

The same reaction can produce other pairs of intermediate-mass fragments as

$\fn_cm \large _{92}^{}\textrm{U}^{235}+ _{0}^{}\textrm{n}^1\rightarrow \left [ _{92}^{}\textrm{U}^{236} \right ]\rightarrow _{51}^{}\textrm{Sb}^{133}+_{41}^{}\textrm{Nb}^{99}+4 _{0}^{}\textrm{n}^1+ Q$

another example

$\fn_cm \large _{92}^{}\textrm{U}^{235}+ _{0}^{}\textrm{n}^1\rightarrow _{54}^{}\textrm{Xe}^{140}+_{38}^{}\textrm{Sr}^{94}+2 _{0}^{}\textrm{n}^1+ Q$

NOTE

The fission process occurs very quickly. ( $\fn_cm \large < 10^{-6}s$ )

The released energy in the fission first appears as the K.E of the fission products and released neutrons. Eventually, it is transferred to the surrounding matter in the form of heat.

The fragment products are radioactive nuclei, they emit β particles in succession to achieve stable end products.

The great advantage of nuclear fission is that it can be controlled and so it is used in a nuclear power station to produce electrical energy.

The enormous energy released in an atom bomb comes from uncontrolled nuclear fission.

CHAIN REACTION

Nuclear fission which once started continues till all the atoms of the fissionable material is disintegrated is called a chain reaction. The chain reaction was first suggested by Enrico Fermi.

Here three secondary neutrons produced fission of three more $\fn_cm \large _{92}^{}\textrm{U}^{235}$ nuclei and produce nine neutrons, one could expect the initiation of chain reaction in the process.

If we can have control over this reaction, the energy released can be utilised for peaceful purposes. This led to the construction of a nuclear reactor.

However, if the reaction cannot be controlled, the energy released will bring about a disastrous effect. This led to the discovery of an atom bomb.

In chain reaction

$\fn_cm \large Neutron\;multiplication\;factor/Reproduction\;factor\;(k)=\frac{Rate\;of\;production\;of\;neutrons}{Rate\;of\;loss\;of\;neutrons}$

Generally the value of k>1

NUCLEAR REACTOR

A nuclear reactor is a device in which a controlled fission chain reaction takes place.

In order to maintain a steady non-explosive chain reaction, several complications arise as

1. The first problem is neutrons emitted during fission is very fast, also fast neutrons liberated of causing would escape instead of causing another fission reaction. Slow neutrons are much more likely to cause fission than are fast neutrons.

Chadwick’s experiments showed that in an elastic collision with Hydrogen the neutron almost comes to rest and the proton carries away the energy. Therefore in the reactor, light nuclei called moderators are provided along with the fissionable nuclei for slowing down fast neutrons.

The moderators are commonly used as water, heavy water ( $\fn_cm \large D_2 O$ ) and graphite.

The APSARA reactor at the BARC, Mumbai, used water as a moderator. The other Indian reactors use heavy water as moderators.

2. The second problem is regarding the fuel used in a nuclear reactor. Natural uranium is mostly $\fn_cm \large _{92}^{}\textrm{U}^{238}$ (about 99.3%) with a small amount of $\fn_cm \large _{92}^{}\textrm{U}^{235}$ (about 0.7%). It is the $\fn_cm \large _{92}^{}\textrm{U}^{235}$ isotope that plays the important role in the chain reaction. To increase the probability of fission of $\fn_cm \large _{92}^{}\textrm{U}^{235}$ nuclei, natural uranium is enriched to increase the percentage of $\fn_cm \large _{92}^{}\textrm{U}^{235}$ .

$\fn_cm \large _{92}^{}\textrm{U}^{238}$ will fission, but only with fast neutrons. The probability of absorbing a fast neutron and producing a fission of $\fn_cm \large _{92}^{}\textrm{U}^{238}$ is too low to produce a self- sustaning chain reaction.

The supply of $\fn_cm \large _{92}^{}\textrm{U}^{235}$ is very much limited, therefore, reactor using $\fn_cm \large _{92}^{}\textrm{U}^{235}$ as fuel cannot meet our energy needs for a long time.

In order to overcome this situation, breeder reactors have been developed. In a breeder reactor, natural uranium is used as fuel. When fast neutrons strike the uranium fuel, the relatively abundant but non-fissionable $\fn_cm \large _{92}^{}\textrm{U}^{238}$ nucleus captures a neutron and become $\fn_cm \large _{92}^{}\textrm{U}^{239}$ . This is radioactive and undergoes β decay twice to produce $\fn_cm \large _{94}^{}\textrm{Pu}^{239}$ (plutonium)

$\fn_cm \large _{92}^{}\textrm{U}^{238}+n\rightarrow _{92}^{}\textrm{U}^{239}\rightarrow _{93}^{}\textrm{Np}^{239} + \bar{e} +\bar{v}$

$\fn_cm \large _{93}^{}\textrm{Np}^{239}\rightarrow _{42}^{}\textrm{Pu}^{239}+\bar{e}+ \bar{v}$

Now $\fn_cm \large _{94}^{}\textrm{Pu}^{239}$ is fissionable by slow neutrons, so that it can serve the same purpose as $\fn_cm \large _{92}^{}\textrm{U}^{235}$ fuel.

But for the best chain reaction, the fuel used is enriched uranium (i.e one that has a greater abundance of $\fn_cm \large _{92}^{}\textrm{U}^{235}$ than naturally occurring uranium.)

3. The third problem is the number of neutrons liberated in fission. If the value of the multiplication factor (K) is about 1 then, the operation of the reactor is said to be critical, which is what we wish it to be for steady power operation. If K becomes greater than one, the reaction rate and the reactor will become supercritical and can even explode.

The explosion of the Chernobyl reactor in Ukraine in 1986 is a sad reminder that accidents in a nuclear reactor can be catastrophic.

So multiplication factor (K) should be nearly equal to 1. This is controlled through control rods made out of neutron absorbing material such as cadmium.

The essential parts of a nuclear reactor are

Nuclear Fuel

Moderator

Control Rods

Coolant and

Protective Shield

Nuclear Fuel:- The nuclear fuel is used in the reactor is generally enriched $\fn_cm \large _{92}^{}\textrm{U}^{235}$ . The nuclear fuel is sealed in the long, narrow metal aluminium tubes called fuel rods.

Moderator:– the neutrons released by fission normally move very fast (2 MeV). This high-speed neutron is less chance to be captured by another $\fn_cm \large _{92}^{}\textrm{U}^{235}$ nucleus. If the neutron is slowed, its chance of capture is much better. The moderator is used to slow down the fast fission neutrons (0.0235 eV). The commonly used moderators are graphite and heavy water.

Control Rods:- In order to control the rate at which fission reaction occurs, control rods of neutron-absorbing material (e.g Cadmium) are used. It is movable.

Coolant:- the purpose of a coolant is to remove heat from the reactor core and take it to the place of its utilisation (e.g steam turbine). Generally, gases or molten solids are used as coolants.

Protective Shield:- In a nuclear reactor, many types of harmful radiation are emitted. In order to prevent these radiations the reactor is enclosed in a thick concrete wall.

WORKING PRINCIPLE

A single slow neutron (thermal neutron) can initiate the chain reaction. When a single slow neutron enters $\fn_cm \large _{92}^{}\textrm{U}^{235}$ nucleus, the nucleus undergoes fission. This results in the release of about 200 MeV energy and 3 fast neutrons. The control rods absorb 2 neutrons while the moderator slows down the remaining neutron to cause further fission.

The rate of fission reaction can be controlled by the adjustment of the control rods. When the control rods are completely inserted, all the emitted neutrons are absorbed and the reaction stopped. If the rods are withdrawn, fewer neutrons are absorbed and the rate of reaction increases.

USES

Nuclear reactors are used to generate electric power.

The nuclear reactor produced isotopes that are used in many places (e.g medical science)

It also produced neutrons, which is used further in nuclear physics.

NUCLEAR FUSION

The process of combining two light nuclei to form a heavy nucleus with the release of a huge amount of energy due to mass defect is known as nuclear fusion.

for example

$\fn_cm \large _{1}^{}\textrm{H}^1+_{1}\textrm{H}^1\rightarrow _{1}^{}\textrm{H}^2+_{}^{}\textrm{e}^+ +\nu +0.42MeV$

Two protons combine to form a deuteron and positron

$\fn_cm \large _{1}^{}\textrm{H}^2+ _{1}^{}\textrm{H}^2\rightarrow _{2}^{}\textrm{He}^3+n+3.27 MeV$

Two deuterons combine to form the light isotope of helium.

$\fn_cm \large _{1}^{}\textrm{H}^2+_{1}^{}\textrm{H}^2\rightarrow _{1}^{}\textrm{H}^3+_{1}^{}\textrm{H}^1+4.03 MeV$

Two deuterons combine to form a triton and a proton.

The energy released per unit mass in nuclear fusion is much greater than the energy released per unit mass in nuclear fission, and energy can be calculated by $\fn_cm \large E=\Delta m c^2$ where $\fn_cm \large \Delta m$ is a mass defect.

Energy obtained in the form of K.E of fused nuclei.

How to cause nuclear fusion?

For fusion to take place, the two nuclei must come close enough so that attractive short-range nuclear force is able to affect them. However, since they are both positively charged particles, they experience coulomb repulsion. They, therefore, must have enough energy to overcome this Coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei. It can be shown, for example, that the barrier height for two protons is ~ 400 keV, and is higher for nuclei with higher charges. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough energy to overcome the coulomb barrier:
$\fn_cm \large K.E= \frac{3}{2}kT$

$\fn_cm \large \Rightarrow \frac{3}{2}kT=400 KeV$ , which gives $\fn_cm \large T\approx 3\times 10^9 K$ .
When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb
repulsive behaviour, it is called thermonuclear fusion.

Thermonuclear fusion is the source of energy output in the interior of stars. The interior of the sun has a temperature of $\fn_cm \large 1.5\times 10^7 K$ , which is considerably less than the estimated temperature required for the fusion of particles of average energy. Clearly, fusion in the sun involves protons whose energies are much above the average energy.

But on earth, such high temperatures are not attained in a controlled manner. However, the temperature produced by fission (atom bomb) is close to $\fn_cm \large 10^8$ K. therefore fission can be used to cause the fusion process.

The fusion reaction in the sun is a multi-step process in which the Hydrogen is burning into Helium. Thus, the fuel in the sun is the hydrogen in its core.

$\fn_cm \large _{1}^{}\textrm{H}^1+_{1}\textrm{H}^1\rightarrow _{1}^{}\textrm{H}^2+_{}^{}\textrm{e}^+ +\nu +0.42MeV$

$\fn_cm \large e^+ + e^-\rightarrow \gamma + \gamma +1.02 MeV$

$\fn_cm \large _{1}^{}\textrm{H}^2 +_{1}^{}\textrm{H}^1\rightarrow _{2}^{}\textrm{He}^3 + \gamma +5.49 MeV$

$\fn_cm \large _{2}^{}\textrm{He}^3 + _{2}^{}\textrm{He}^3\rightarrow _{2}^{}\textrm{He}^4+_{1}^{}\textrm{H}^1 +_{1}^{}\textrm{H}^1 + 12.86 MeV$

The net effect is

$\fn_cm \large 4_{1}^{}\textrm{H}^1 + 2 e^-\rightarrow _{2}^{}\textrm{He}^4 + 2\nu + 6\gamma + 26.7 MeV$

Thus, four hydrogen atoms combine to form an $\fn_cm \large _{2}^{}\textrm{He}^4$ atom with a release of 26.7 MeV of energy.

CONTROLLED THERMONUCLEAR FUSION

Nuclear fusion is an uncontrolled process. Till today, there is no available method of controlling the release of fusion energy. Therefore, its only use so far is destructive i.e for making a Hydrogen bomb which is based on the phenomenon of nuclear fusion.

In controlled fusion reactors, the aim is to generate steady power by heating the nuclear fuel to a temperature in the range of $\fn_cm \large 10^8 K$ . At this temperature, the fuel is a mixture of +ve ions and electrons clouds (Plasma). The challenge is to confine this plasma since no container can stand such temperature.

Several countries are developing techniques in this connection. If successful, fusion reactors will hopefully supply almost unlimited power to humanity.

NOTE:

Nuclear fusion is a more attractive source of energy than nuclear fission due to three basic reasons.

For a given mass, the energy produced in nuclear fusion is greater than produced in nuclear fission.

The fuel deuterium used in fusion is available in unlimited quantities in seawater. and

The final products in nuclear fusion are not radioactive and do not present radioactive waste disposal problems.

BIRTH AND DEATH OF STAR

Stars are born in vast, slowly rotating, clouds of cold gas and dust called nebulae. These large clouds are enormous, they have masses somewhere between 100 thousand and two million times the mass of the sun and their diameters range from 50 to 300 light-years across.

These massive clouds of gases (H, He and other organic complexes) starts contracting due to gravitational pull. Due to this pull, molecules will move fastly towards the centre (core). A stage comes that their K.E is so high and along with the K.E, its temperature also increases. ( $\fn_cm \large K.E= \frac{3}{2}kT$ )

In this process, the temperature increases so much that the atom is completely ionized and the bare nuclei with electrons cloud are formed, this stage itself is called plasma.

In this stage, the young star starts converting hydrogen to helium via a nuclear fusion reaction (Born of stars). Nuclear reactions in stars produce vast amounts of energy in the form of heat and light, which is radiated into space.

When heat is released, expansion of gas happens and an outward thermal pressure acts and stars become stable because this thermal pressure balances gravitational pull. This stage is known as the main-sequence star phase.

Stars spend most of their lives as main-sequence stars, converting hydrogen into helium at their centre or core. A star may remain as a main-sequence star for millions or billions of years. Most of the stars in the universe, about 90% are the main-sequence stars.

Main-sequence stars are not all the same. They have different masses when they are born, depending on how much matter is available in the nebula from which they formed. Main-sequence stars also have different colours, depending on the temperatures of their surfaces. Bluer colour has more temperature than reddish colour (VIBGYOR).

How long a main-sequence stars lives depends on how massive is. More massive stars move onto the next stages of their lives more quickly than lower mass stars, because due to their higher temperature. For example, the sun will spend about 10 billion years as a main-sequence star, but a star 10 times as massive will last for only 20 million years.

When the hydrogen in the core of the star is depleted, the star’s core shrinks and heat up. There is still some hydrogen left in the outer shell of the core. This hydrogen now fuses to give helium and the outer layer of the star expands due to the heat of fusion. The star becomes larger and brighter and its surface temperature cools so it glows red. The star is now a red giant star.

FOR MEDIUM-SIZED STARS (mass less than 8 times of our sun):- The core of stars still contracts due to gravitational pull. Molecules keep speeding up until the K.E is so high and with that, the temperature also rises a lot. at that time helium-helium fuse to form carbon. Here again, the same process happens which happened during hydrogen-hydrogen fusion. And at this time again the size becomes stable. This continues for 1 billion years until all helium depletes. We can say that Helium fuses in the red giant star phase.

Again the core contracts without any thermal pressure. Now the K.E is never so high in such stars that it can cause fusion of carbon nuclei.

After it has become a red giant, the star becomes unstable and expands and contracts over and over. This causes the star’s outer layers to become detached from the central part of the star and they gently puff off into space. When the last of the gas in the star’s outer layers are blown away. It forms an expanding shell around the core of the star called a planetary nebula.

NOTE

A planetary nebula is different to a stellar nebula. A stellar nebula is where stars are born whereas a planetary nebula is what some stars form at the end of their lives.
A planetary nebula was so named because early astronomers thought they look a bit like planets through a small telescope.
A planetary nebula is relatively short-lived and last just a few tens of thousands of years.

Sometimes after puffing off its outer layers, the central star will run out of fuel. When this happens the central star begins to die. Gravity causes the star to collapse inward and the star becomes incredibly dense and compact about the size of the earth. It is very hot ( $\fn_cm \large 100000^0 C$ ) and very bright. The star has then become a white dwarf star. We can say that also white dwarf is the remaining central part of the star (core). White dwarfs no longer produce energy via nuclear reaction and so they radiate their energy into space in the form of light and heat. They slowly cool down over time. Eventually, once all of their energy is gone, they no longer emit any light.

The star is now a dead black dwarf star (black lumps of carbon) and will remain like this forever. White dwarf/ Black dwarf stars are so dense that one teaspoon of material from a white dwarf would weigh up to 100000 Kg.

FOR LARGE-SIZED STAR( mass more than 8 times of our sun):- When the hydrogen at their cores becomes depleted, they swell into red supergiants which are even larger than red giants. A red supergiant can fuse successively heavier and heavier elements for a few million years until its core is filled with iron.

C-C fuse to N (C depletes)

N-N fuse to O ( N depletes)

………………………

……………………….

…….. fuse to Fe (Most stable nuclei)

At this point, the nuclear reaction stop and the star collapse rapidly under its own gravity. The collapsing outer layers of the star hit the small central core with such a force that they rebound and send a ripple outwards through the star blowing the outer layers of the star into space in a huge explosion called a supernova.

NOTE

The elements made inside stars are scattered through space when the outer layers of the stars are blown off either in planetary nebula or supernova. This stardust is recycled and used in forming the next generation of stars and planets.

For a week or so, a supernova can outshine all of the other stars in its galaxy. However, they quickly fade over time. The central star left behind is either made of neutrons and it is called a neutron star or if the initial star was really massive a black hole forms.

The diagram below shows the life of our Sun. The Sun is a common type of star of average size and mass