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Best JEE Coaching in Jorhat 

Nuclear fusion is the process of the production of energy in stars two light nuclei fusion to create a bigger nucleus and release energy, because the bigger nuclear structure is much more closely bound. In the initial process, protons mix into a deuteron as well as the positron. The release that is 0.42 MeV of energy. In the reaction [13.29(b)(b)), two deuterons combine to create the light isotope helium. In the reaction (13.29c) two deuterrons join to form the triton and proton. To allow fusion to occur the two nuclei need to be close enough that an the attractive short-range nuclear force capable of impacting the two. Since the two are positively charged and therefore, they are both subject to coulomb attraction. Therefore, they must be able to overcome the coulomb barrier. The size of the barrier is determined by the charges and radii of two nuclei in the interaction. It is evident that, for instance, the height of the barrier of two proton nuclei is approximately 400 keV. This it is greater for nuclei that have more charges. Best JEE Coaching in Jorhat.

We can determine that the temperatures at which protons within the proton gas would (averagely) possess enough energy to cross the coulomb-barrier: (3/2)k T = K 400 keV, which yields T 3x 110 K. If fusion can be achieved through raising your system’s temperature such that particles are able to generate sufficient kinetic energy to defeat the coulomb repulsive behavior that is known as thermonuclear fusion.¬†Thermonuclear fusion is the main source of energy produced inside the stars’ interior.¬†The solar interior is at 1.5×107 K which is considerably lower than the expected temperature for the fusion of particles with average energy.¬†It is evident that fusion within the sun involves protons with energies that are over the average.¬†The fusion reaction that occurs in solar fusion is multi-step procedure that involves the hydrogen being transformed to Helium.¬†The fuel used by solar system is hydrogen that lies in its center.¬†Best JEE Coaching in Jorhat.¬†This proton-proton (p, (p,) cycle through which this occurs is represented in the following set of chemical reactions. 1 2 1 1 The H H is a + 1 e + + n 0.42 MeV (i) E + e – + + 1.02 MeV (ii) 2 1 3 1 2 H H G + 5.49 MeV (iii) + – + 3 3 1 1 2 1 He The H-H ratio is 12.86 MeV

To allow 4 reactions to take place the three previous reactions have to occur at least twice, and in this case two light helium nuclei combine to form a normal nucleus of helium.¬†If we look at the combination 2(i) + 2(ii) + 2(iii) +(iv), the result is 4 4 He 2 6. 26.7MeV 1 2 e + + or 1 4 1 2 (4 4 ) ( He 2 ) 2 6 26.7MeV e 26.7MeV e n g + + + (13.31) This means that four hydrogen atoms meld to form a 4 2He Atom that releases 26.7 MeV in energy.¬†Helium isn’t just one element which is synthesized inside the star.¬†When the hydrogen within the core is depleted and turns into helium, the center begins to get cooler.¬†The star is beginning to collapse under the pressure of its self-gravity (c) NCERT is not to be republished in 457 Nuclei which raises the core’s temperature.¬†If the temperature rises to 108 K Then fusion occurs again with the helium nuclei to carbon.¬†This type of process may produce larger and greater mass elements.¬†Best JEE Coaching in Jorhat.¬†However, elements with a mass greater than those that are near the top of the curve of energy binding shown in Fig.¬†13.1 can’t be produced.¬†The sun’s age is approximately 5×109 years, and it is believed that there enough solar hydrogen to continue burning for the next five billion years.¬†

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Following the burning of hydrogen will cease as the sun begins to cool down and will begin to collapse under the pressure of gravity which will increase the temperature of its core.¬†The outer layer of the sun will expand, transforming it into the red giant.¬†NUCLEAR HOLOCAUST Within the course of a single fission in uranium, around 0.9×235 MeV (200 MeV) of energy is released.¬†When each nucleus weighing around 50 kg of 235U goes through fission, the amount of energy released is approximately 4x1015J.¬†This is equivalent to around 220,000 tons of TNT enough to cause an explosion of super-sized proportions.¬†Uncontrolled release of massive nuclear energy is known as an explosion of the atomic type.¬†On August 6, 1945, an atomic weapon was deployed in battle in the very first instance.¬†It was the day that the US dropped an Atom blast on Hiroshima, Japan.¬†The blast was the equivalent of 20000 tonnes of TNT.¬†Best JEE Coaching in Jorhat.¬†The radioactive particles instantly destroyed 10 square kilometers of the city, which was home to 3,43,000 residents.

From this, 66,000 died and 69,000 were wounded; more than 67 percent of the city’s structures were destroyed.¬†The conditions for high temperature the fusion reaction can be made through the explosion of a fission explosive.¬†Super-explosions of 10-megatons explosive power from The nuclear substance TNT have been tested since the year 1954.¬†The bombs that are a result of the fusion of hydrogen isotopes tritium, deuterium and deuterium are referred to as hydrogen bombs.¬†It is believed the nuclear arsenal that is sufficient to eliminate all life on the planet many times over is ready to be activated by the push of one button.¬†A nuclear catastrophe will not just destroy the existence that is present, but also cause the earth to be unfit for living in all time.¬†Based on mathematical calculations, scenarios suggest a long winter for nuclear reactors because the radioactive waste will be clouds in the atmosphere of Earth and absorb solar radiation.¬†Best JEE Coaching in Jorhat.

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The dimensions of the nucleus are significantly smaller than the dimensions for an atom. Studies on the scattering of particles showed it was the case that the radius of the nucleus is less than the radius of an atom , by a factor of around 100 . This means that the size of a nucleus is around 10-12 times the size of an atom. That is the atom is nearly empty. If an atom were to be expanded enough to fill a schoolroom the nucleus will be the size of a pinhead. However, the nucleus is home to the majority (more than 99.9 percent) in the mass and weight of an atom. Does the nucleus possess an atom-like structure, like the atom? If yes What are the constituents in the nucleus? How do they hold them together? What is the mechanism that holds them together? In this section, we will explore the answers to such questions. We will look at various aspects of nuclei, such as their size, mass , and stability, as well as with nuclear phenomena like nuclear fission, radioactivity and nuclear fusion. 

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 ATOMIC MASSES AND COMPOSITION OF NUCLEUS The weight of an atom is extremely tiny, when compared to a kilogram. For instance the mass of carbon atoms 12C, for instance, is 1.992647 10-26 kg. Kilogram is not a efficient way to measure these small amounts. So, aNuclei unit is used to describe the mass of atoms. This is the nuclear mass unit (u) that is which is defined as 1/12th of mass of carbon (12C) atom.The mass atomics of different elements, that are expressed in the the atomic mass units (u) are near in relation to the masses of the hydrogen atom. However, there are several remarkable exceptions to this principle. For instance the mass of an atomic the chlorine atom has been measured to be 35.46 u. The precise measurements of the mass of an atom are done using the mass spectrometer. The measurements of the atomic masses reveal the existence of several kinds of atoms from the same element that have similar chemical properties however, they differ in terms of mass. These atomic species of the same element that differ in mass are referred to as isotopes. Best JEE Coaching in Jorhat.

(In Greek, isotope means the same spot, i.e. they share the same location on the periodic table.) It was found that virtually every element is made up of an assortment of isotopes. The amount of isotopes is different between elements. Chlorine for instance has two isotopes with masses of 34.98 U and 36.98 u that are almost an integral multiple of the mass of the hydrogen atom. The relative abundances of these isotopes are 75.4 and 24.6 per cent, respectivelyDiscovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. However, the masses of the nuclei from tritium, deuterium, and hydrogen are approximately 1:2:3. So, the nuclei of tritium and deuterium must have as well as the proton, a small amount of neutral matter. There is a lot of neutral material inside the nuclei of the isotopes measured in mass units of proton, is equivalent to one and two in each. This suggests that the nuclei of atoms are filled with as well as proton particles, neutral matter in many different units of a fundamental unit. 

This theory was confirmed during 1932, by James Chadwick who observed emission of neutral radiation after beryllium nuclei were bombarded by alpha particles.¬†(a-particles constitute helium nuclei to be covered in a future section).¬†It was discovered that this radiation neutral could remove protons from light nuclei , such like those of helium nitrogen and carbon.¬†The only known neutral radiation at the at the time were the photons (electromagnetic radiation).¬†Utilizing the concepts of conservation of energy and momentum proved that if the radiation was composed of photons the energies of photons would need to be significantly higher than the energy available from radiation of the beryllium nuclei by particles called a-particles.¬†The key for this mystery, that Chadwick successfully resolved, was to suppose that neutral radiation consisted of a brand new kind of neutrons, which are neutral particles.¬†Best JEE Coaching in Jorhat.¬†Through the conservation in energy as well as momentum Chadwick was able to calculate the mass of this new particle that is’very nearly similar to the mass of a proton’.¬†A neutron’s mass is recognized to a great level of precision.¬†

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It is m nu equals 1.00866 the u value is 1.6749×10-27 kilograms (13.3)¬†Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron.¬†The neutron free of charge, unlike the proton that is free it is unstable.¬†It degrades into a proton, an electron , and an neutrino (another fundamental particle) as well as an average life of around 1000 years.¬†However, it is solid inside the nucleus.¬†The structure of a nucleus can be described with the following words and symbols: Z – atomic number = the number of proton [13.4(a) N neutron number = the quantity of neutrons [13.4(b) Mass number Z + N = the total number of neutrons and protons [13.4(c)The other employs the term “nucleon” to refer to either a neutron or proton.¬†Therefore, the number of nucleons within an atom is the mass number, A.¬†Nuclides, or nuclear species, are represented by the notation X A Z , where”X” is the symbol for chemical used to describe the species.¬†Best JEE Coaching in Jorhat.¬†For instance the nucleus of gold is represented by the number the number 197 779 Au.¬†It has 197 nucleons including 79 of them being protons, and the rest are neutrons.¬†¬†

Nuclei composition of isotopes in an element is now easily understood.¬†The isotopes of an element have the same amount of protons but differ in the amount of neutrons.¬†Deuterium 2 1 H which is an isotope made of hydrogen, includes one proton as well as one neutron.¬†Its tritium isotope 3 1 H includes one proton as well as two neutrons.¬†The element gold contains 32 isotopes that range from A = 173 to A = A =.¬†As we have mentioned before, the elements’ chemical properties depend from their electron structure.¬†Since the isotopes’ atoms possess the same electronic structure, they share the same chemical properties and are in the same spot within the periodic table.¬†All nuclides having the identical mass numbers A are known as isobars.¬†For instance the nuclides 3 1 H as well as 3 2He, are considered isobars.¬†Nuclides that have the identical neutron numbers N but with different atomic numbers Z, like that 198 80 Hg is 1907 79 Au , are known as isotones.¬†13.3 SIZE of the nucleus As we’ve seen in Chapter 12 Rutherford was the first to proposed and proved that there is an nucleus atomicum.

Based on Rutherford’s suggestions, Geiger and Marsden performed their famous experiment on the scattering of particles of a-particles by the thin foils of gold.¬†Their research showed that the distance of closest contact with a gold nucleus in a-particles that has kinetic energy 5.5 MeV is about 4.0 10-14 meters. The scattering caused by a particle due to the gold sheet could be explained by Rutherford by concluding that coulomb repulsive force is the only one the cause of scattering.¬†Because this positive charge limited to the nucleus itself, the dimensions of the nuclear nucleus must to be smaller than 4.0 10-14 meters.¬†Best JEE Coaching in Jorhat.¬†If we employ particles with greater energies that are greater than 5.5 MeV, the distance from the nearest contact with the gold nucleus is less and eventually the scattering will be affected by forces of nuclear short-range which diverge from Rutherford’s calculations.¬†Rutherford’s calculations are based upon pure coulomb attraction among the positively charged aparticle as well as the nucleus of gold.¬†From the distances that the deviations are set the nucleus, the size of the nuclear can be calculated.¬†

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When conducting scattering tests that use fast electrons instead of particles, are projectiles that pound targets that are composed of various elements, the dimensions of the nuclei from various elements have been precisely determined.¬†It was discovered that a nucleus having mass A is radial in size. of R equals R0 A/3 (13.5) where R0 is 1.2 10-15m. This indicates that the size of the nucleus is proportional to the R3 is proportional to the A.¬†The density of the nucleus is steady, regardless of the A in all the nuclei.¬†Different nuclei are similar to drops of liquid that has a the same density.¬†Nuclear matter has a density 2.3 1017kg m-3.¬†This is quite a lot in comparison to the ordinary matter, for instance water, which has a density of approximately 103 kg m-3.¬†This is natural, since we’ve seen before that the majority of an atoms are empty.¬†The normal matter that is composed of atoms is filled with a huge quantity of empty space.Mass is energy. Einstein proved through the theory of special relativity, that it was important to view mass as a different kind of energy.¬†Best JEE Coaching in Jorhat.

Prior to the development of this concept of special relativity,, it was thought that energy and mass kept separate during reactions.¬†But, Einstein showed that mass is an alternative form of energy that is able to convert energy from mass into other types of energy, like the kinetic energy, and vice versa.¬†Einstein provided the well-known mass-energy-energy-equivalence relationship E = Mc2 (13.6) The energy equivalent to mass m is calculated by the equation above and c is the speed of light in a vacuum. It is about 3×108. m s-1Nuclei , which only affects the nucleons that are close to it. This is also known as the saturation property of nuclear force.¬†(iii) A very heavy nucleus, for example A of 240 is bound with less energy per nucleon as compared to the nucleus that has A equal to 120.¬†So, if a nuclear the nucleus A = 240 is broken in two A = 120 nuclei, the nucleons become much more closely bound.¬†This means that energy will be released as a result.¬†It is a significant issue for the production of energy through fission. This will be discussed further in section 13.7.1.¬†(iv) Think about two light nuclei (A 10) merging to create an atomic mass that is heavier.¬†

It is apparent that the binding power per unit for the heavier nuclei that have been fused is higher that the energy of binding per nucleon for the nuclei with lighter mass.¬†Best JEE Coaching in Jorhat.¬†This means that the resulting system is more closely bound than the first system.¬†In addition, energy will be released during a process of Fusion.¬†This is the source of energy of the sun, which will be explained in section 13.7.3.¬†13.5 NUCLEAR FORCE The force that governs the movement of electrons within atomic nuclei is known as the Coulomb force.¬†In Section 13.4 we’ve discovered that for nuclei with average mass the energy required to bind a nucleon is about 8 MeV which is a lot more than the energy of binding in the atoms.¬†So, in order to tie two nuclei, there needs to be an attraction force of a completely different type.¬†It has to be sufficient to overcome the attraction between (positively positively charged) protons, and to bring neutrons and protons into the small nucleus.¬†

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We’ve already observed that the steadfastness of the binding energy per nucleon could be seen by its short-range.¬†Numerous aspects that make up the force of nuclear binding can be described below.¬†They are derived from a myriad of experiments that were conducted from 1930 to 1950.¬†(i) It is more powerful that the Coulomb force that is acting between charges, or the gravitational force between masses.¬†The force of nuclear binding has been able to prevail in comparison to the Coulomb repulsive force acting between protons within the nucleus.¬†This happens since the force of nuclear is more powerful than the Coulomb force.¬†It is smaller than even Coulomb force.¬†(ii) the nuclear energy between nucleons decreases quickly to zero when their distance is greater than one femtometer.¬†This causes an oversaturation of forces in a medium or huge nucleus. This is the main reason for the constant nuclear energy binding per.¬†Best JEE Coaching in Jorhat.¬†A rough graph of the energy potential between two nucleons as a result of distance is displayed in Potential energy is minimal at a distance that is approximately 0.8 Fm.¬†

This implies this force will be appealing for distances that are greater than 0.8 Fm, and is repulsive when they(iii) the nuclear forces between neutron-neutron proton-neutron and proton is about the similar.¬†The nuclear force doesn’t have to be dependent on electric charges.¬†In contrast to Coulomb’s laws or Newton’s law of gravitation , there is no mathematical formula of nuclear force.¬†13.6 Radioactivity A. H. Becquerel discovered radioactivity in 1896 completely through accidental.¬†In his research into the phosphorescence and fluorescence of compounds exposed to the visible spectrum, Becquerel discovered an intriguing phenomenon.¬†After shining a few pieces of uranium-potassium sulfurate using visible light He wrapped the pieces in black paper and separated the box from a photographic plate using an element of silver.¬†When, after a few hours of exposure and exposure, the photographic plate developing, it began to show darkening caused by something released by the compound.¬†

It could be absorbed by both the black paper and the silver.¬†The subsequent experiments showed radioactivity to be an atomic phenomenon where unstable nuclei undergoes decay.¬†This is known as radioactive decomposition.¬†Three kinds of radioactive decay can be observed in nature: (i) A-decay, where a Helium nucleus 4 2He emits; (ii) b-decay in which positrons or electrons (particles that have similar mass to electrons, but having an energy charge that is opposite to electrons) are released; (iii) G-decay where high-energy (hundreds in keV or greater) photons are released.¬†Best JEE Coaching in Jorhat.¬†Each of these types of decay will be discussed in the following subsections.¬†13.6.1 The law of decay in radioactive substances in any radioactive substance that undergoes a, b , or decay in g, it is observed that the amount of nuclei that undergo decay in a unit of duration is proportional to number of nuclei present in the sample.¬†When N represents the nuclei present in the sample, and they undergo decay within the time t, then is N t or N/t = lN (13.10) where L is”radioactive decay”, or disintegration constant.¬†

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The variation in the number of nuclei present in the sample is dN = -N in the period t.¬†So the change rate in N (in the range t – 0.) (d – d N). T = l * N is the number of nuclei which change, which it is never negative.¬†The variation in N, which can be expressed in either of two ways.¬†In this instance, it’s negative since, out of the original N nuclei, the nuclei have been destroyed leaving (N-N) Nuclei.¬†(c) These NCERT are separated at smaller than 0.8 fm.NUCLEAR Energy The curve for binding energy per nuclear nucleon, Ebn shown in figure.¬†13.1 It has a lengthy flat middle region that runs between A equal 30 to A = 170.¬†In this region, the bound energy of a nucleon remains almost always the same (8.0 MeV).¬†For the nuclei with lighter mass, A < 30, and for the heavier nuclei region, A > 170, the binding energy per nucleon is lower than 8.0 MeV, as we have previously noted.¬†The higher it is the energy of binding, the smaller is the mass of a system bound, like the nucleus.¬†Best JEE Coaching in Jorhat.¬†Therefore, if nuclei that have less binding energy convert to nuclei with more binding energy that will result in an release of energy.¬†

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This is the result when a nucleus that is heavy decays into more or less intermediate mass pieces (fission) and when nuclei with light energy combine into a bigger nucleus (fusion.) Exothermic chemical reactions are the basis of the traditional energy sources, such as petroleum or coal. In this case, the energy levels involved are within the range of 13.4 b-decay of the 13.4 b-decay of the 28-60 Ni nucleus, followed by emission of two g-rays due to deexcitation of the daughter nucleus 2860 (c) (c) Ni . Not to be published again. Physics 452 electronvolts. However when you have a nuclear reaction the energy released is on the order of MeV. For the same amount of material, nuclear sources release an energy output that is a million times greater than chemical sources. The fission of one kilogram of uranium for instance produces 10.14J of power. Compare this to burning 1 kg of coke which produces 107 J. 13.7.1 Fission The possibilities for new possibilities are revealed as we explore the limits of natural radioactive decays and investigate nuclear reactions by bombarding nuclei other nuclear particles, such as neutron, proton, a-particle as well as other particles. 

One of the most significant neutron-induced reaction is the fission.¬†A good example of fission is when a uranium-isotope of 235 92U that is bombarded by neutron fragments into two nuclear fragments with intermediate mass. 1 235 236 144 1 0 92 % 92, 56, 36 N U U BaKr 3 n0 + + (13.26) The same reaction could produce additional types with intermediate mass pieces. For instance, 1 236 236 133 99 1. 236 133 99 1 n U U Sb 4 n0 + + – + (13.27) As an example 1 235 140 1 0 54 38 n Nb Xe 2 n0 + – + (13.28) The fragments are radioactive nuclear nuclei.¬†Best JEE Coaching in Jorhat.¬†They release b particles sequentially in order to create stable products.¬†Energy released (the Quantum value ) during the fission process of nuclei such as uranium is around 200 MeV per fissioning nuclear nucleus.¬†It is calculated by following Let’s consider the nucleus that has A = 240, which breaks into two pieces with A equals 120.¬†It is then Ebn in the case of A of 240 is around 7.6 MeV, Ebn for the two A = 120 fragment nuclei is around 8.5 MeV.¬†The gain in binding energy of nucleons is approximately 0.9 MeV.¬†Thus, the increase of binding energy equals 240×0.9 which is 216 MeV.¬†

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The energy of disintegration in fission reactions first manifests as the kinetic energy of neutrons and fragments. Then it transfers to the surrounding matter , appearing as heat. The energy source in nuclear reactors that produce electricity can be found in nuclear fission. The huge amount of energy released in an atom bomb results from an uncontrolled nuclear fission. We will discuss in the following section about how the nuclear reactor works. 13.7.2 Nuclear reactor Note one important aspect in the fission reactions described in the equations. (13.26) to (13.28). There is the release of an extra neutron (s) during the process of fission. In the average 2 1/2 neutrons are released during the fission process of the nuclear uranium. This is a tiny fraction because during certain fission events, two neutrons are produced by nuclei fission. It has the elements of fuel properly fabricated. The fuel can be an enriched form of by uranium (i.e. the one that is more abundant in 235 922 U than the naturally occurring uranium). Top JEE Coaching in Jorhat. The core is equipped with an adsorber to reduce the neutrons. 

The core is enclosed by a reflector in order to limit the leakage.¬†It is important to note that the energy (heat) produced by fission is removed continuously with a suitable coolant.¬†A containment vessel stops the release of radioactive fission products.¬†The entire unit is shielded to prevent radiation that is harmful from escaping.¬†The reactor is shut down using rods (made from, for instance Cadmium, for instance) which absorb a high amount of neutrons.¬†The coolant transfers heat into an operating fluid that produces stream.¬†Steam drives turbines and produces electricity.¬†Similar to any power plant nuclear reactors produce a lot of waste products.¬†However, nuclear wastes require special attention when it comes to treatment because they’re radioactive and dangerous.¬†Comprehensive safety measures to ensure the safety of reactors and dealing with and reprocessing waste fuel, are essential.¬†Top¬†JEE Coaching in Jorhat.¬†This is an important aspect that is part of the Indian Atomic Energy programme.¬†

A plan of action is being developed to investigate the possibility of turning the radioactive material into less active and less long-lasting material. 13.7.3 Nuclear fusion is the process of the production of energy in stars two light nuclei fusion to create a bigger nucleus and release energy, because the bigger nuclear structure is much more closely bound. In the initial process, protons mix into a deuteron as well as the positron. The release that is 0.42 MeV of energy. In the reaction [13.29(b)(b)), two deuterons combine to create the light isotope helium. In the reaction (13.29c) two deuterrons join to form the triton and proton. To allow fusion to occur the two nuclei need to be close enough that an the attractive short-range nuclear force capable of impacting the two. Since the two are positively charged and therefore, they are both subject to coulomb attraction. Therefore, they must be able to overcome the coulomb barrier. The size of the barrier is determined by the charges and radii of two nuclei in the interaction. 

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It is evident that, for instance, the height of the barrier of two proton nuclei is approximately 400 keV. This it is greater for nuclei that have more charges.¬†We can determine that the temperatures at which protons within the proton gas would (averagely) possess enough energy to cross the coulomb-barrier: (3/2)k T = K 400 keV, which yields T 3x 110 K. If fusion can be achieved through raising your system’s temperature such that particles are able to generate sufficient kinetic energy to defeat the coulomb repulsive behavior that is known as thermonuclear fusion.¬†Thermonuclear fusion is the main source of energy produced inside the stars’ interior.¬†The solar interior is at 1.5×107 K which is considerably lower than the expected temperature for the fusion of particles with average energy.¬†Top¬†JEE Coaching in Jorhat.¬†It is evident that fusion within the sun involves protons with energies that are over the average.¬†The fusion reaction that occurs in solar fusion is multi-step procedure that involves the hydrogen being transformed to Helium.¬†

The fuel used by solar system is hydrogen that lies in its center.¬†This proton-proton (p, (p,) cycle through which this occurs is represented in the following set of chemical reactions. 1 2 1 1 The H H is a + 1 e + + n 0.42 MeV (i) E + e – + + 1.02 MeV (ii) 2 1 3 1 2 H H G + 5.49 MeV (iii) + – + 3 3 1 1 2 1 He The H-H ratio is 12.86 MeV (iv) (13.30) To allow 4 reactions to take place the three previous reactions have to occur at least twice, and in this case two light helium nuclei combine to form a normal nucleus of helium.¬†If we look at the combination 2(i) + 2(ii) + 2(iii) +(iv), the result is 4 4 He 2 6. 26.7MeV 1 2 e + + or 1 4 1 2 (4 4 ) ( He 2 ) 2 6 26.7MeV e 26.7MeV e n g + + + (13.31) This means that four hydrogen atoms meld to form a 4 2He Atom that releases 26.7 MeV in energy.¬†Helium isn’t just one element which is synthesized inside the star.¬†When the hydrogen within the core is depleted and turns into helium, the center begins to get cooler.¬†The star is beginning to collapse under the pressure of its self-gravity which raises the core’s temperature.¬†If the temperature rises to 108 K Then fusion occurs again with the helium nuclei to carbon.¬†

This type of process may produce larger and greater mass elements.¬†However, elements with a mass greater than those that are near the top of the curve of energy binding shown in Fig.¬†13.1 can’t be produced.¬†The sun’s age is approximately 5×109 years, and it is believed that there enough solar hydrogen to continue burning for the next five billion years.¬†Following the burning of hydrogen will cease as the sun begins to cool down and will begin to collapse under the pressure of gravity which will increase the temperature of its core.¬†The outer layer of the sun will expand, transforming it into the red giant.¬†NUCLEAR HOLOCAUST Within the course of a single fission in uranium, around 0.9×235 MeV (200 MeV) of energy is released.¬†Top¬†JEE Coaching in Jorhat.¬†When each nucleus weighing around 50 kg of 235U goes through fission, the amount of energy released is approximately 4x1015J.¬†This is equivalent to around 220,000 tons of TNT enough to cause an explosion of super-sized proportions.¬†

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Uncontrolled release of massive nuclear energy is known as an explosion of the atomic type.¬†On August 6, 1945, an atomic weapon was deployed in battle in the very first instance.¬†It was the day that the US dropped an Atom blast on Hiroshima, Japan.¬†The blast was the equivalent of 20000 tonnes of TNT.¬†The radioactive particles instantly destroyed 10 square kilometers of the city, which was home to 3,43,000 residents.¬†From this, 66,000 died and 69,000 were wounded; more than 67 percent of the city’s structures were destroyed.¬†The conditions for high temperature the fusion reaction can be made through the explosion of a fission explosive.¬†Super-explosions of 10-megatons explosive power from The nuclear substance TNT have been tested since the year 1954.¬†The bombs that are a result of the fusion of hydrogen isotopes tritium, deuterium and deuterium are referred to as hydrogen bombs.¬†It is believed the nuclear arsenal that is sufficient to eliminate all life on the planet many times over is ready to be activated by the push of one button.¬†Top¬†JEE Coaching in Jorhat.

A nuclear catastrophe will not just destroy the existence that is present, but also cause the earth to be unfit for living in all time. Based on mathematical calculations, scenarios suggest a long winter for nuclear reactors because the radioactive waste will be clouds in the atmosphere of Earth and absorb solar radiation.The dimensions of the nucleus are significantly smaller than the dimensions for an atom. Studies on the scattering of particles showed it was the case that the radius of the nucleus is less than the radius of an atom , by a factor of around 100 . This means that the size of a nucleus is around 10-12 times the size of an atom. That is the atom is nearly empty. If an atom were to be expanded enough to fill a schoolroom the nucleus will be the size of a pinhead. However, the nucleus is home to the majority (more than 99.9 percent) in the mass and weight of an atom. Does the nucleus possess an atom-like structure, like the atom? If yes What are the constituents in the nucleus? How do they hold them together? 

What is the mechanism that holds them together? In this section, we will explore the answers to such questions. We will look at various aspects of nuclei, such as their size, mass , and stability, as well as with nuclear phenomena like nuclear fission, radioactivity and nuclear fusion. 13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS The weight of an atom is extremely tiny, when compared to a kilogram. For instance the mass of carbon atoms 12C, for instance, is 1.992647 10-26 kg. Top JEE Coaching in Jorhat. Kilogram is not a efficient way to measure these small amounts. So, aNuclei unit is used to describe the mass of atoms. This is the nuclear mass unit (u) that is which is defined as 1/12th of mass of carbon (12C) atom.The mass atomics of different elements, that are expressed in the the atomic mass units (u) are near in relation to the masses of the hydrogen atom. However, there are several remarkable exceptions to this principle. For instance the mass of an atomic the chlorine atom has been measured to be 35.46 u. 

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The precise measurements of the mass of an atom are done using the mass spectrometer. The measurements of the atomic masses reveal the existence of several kinds of atoms from the same element that have similar chemical properties however, they differ in terms of mass. These atomic species of the same element that differ in mass are referred to as isotopes. (In Greek, isotope means the same spot, i.e. they share the same location on the periodic table.) It was found that virtually every element is made up of an assortment of isotopes. The amount of isotopes is different between elements. Chlorine for instance has two isotopes with masses of 34.98 U and 36.98 u that are almost an integral multiple of the mass of the hydrogen atom. The relative abundances of these isotopes are 75.4 and 24.6 per cent, respectivelyDiscovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. Top JEE Coaching in Jorhat.

However, the masses of the nuclei from tritium, deuterium, and hydrogen are approximately 1:2:3. So, the nuclei of tritium and deuterium must have as well as the proton, a small amount of neutral matter. There is a lot of neutral material inside the nuclei of the isotopes measured in mass units of proton, is equivalent to one and two in each. This suggests that the nuclei of atoms are filled with as well as proton particles, neutral matter in many different units of a fundamental unit. This theory was confirmed during 1932, by James Chadwick who observed emission of neutral radiation after beryllium nuclei were bombarded by alpha particles. (a-particles constitute helium nuclei to be covered in a future section). It was discovered that this radiation neutral could remove protons from light nuclei , such like those of helium nitrogen and carbon. The only known neutral radiation at the at the time were the photons (electromagnetic radiation). 

Utilizing the concepts of conservation of energy and momentum proved that if the radiation was composed of photons the energies of photons would need to be significantly higher than the energy available from radiation of the beryllium nuclei by particles called a-particles.¬†The key for this mystery, that Chadwick successfully resolved, was to suppose that neutral radiation consisted of a brand new kind of neutrons, which are neutral particles.¬†Through the conservation in energy as well as momentum Chadwick was able to calculate the mass of this new particle that is’very nearly similar to the mass of a proton’.¬†Top¬†JEE Coaching in Jorhat.¬†A neutron’s mass is recognized to a great level of precision.¬†It is m nu equals 1.00866 the u value is 1.6749×10-27 kilograms (13.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron.¬†The neutron free of charge, unlike the proton that is free it is unstable.¬†It degrades into a proton, an electron , and an neutrino (another fundamental particle) as well as an average life of around 1000 years.¬†However, it is solid inside the nucleus.¬†

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The structure of a nucleus can be described with the following words and symbols: Z – atomic number = the number of proton [13.4(a) N neutron number = the quantity of neutrons [13.4(b) Mass number Z + N = the total number of neutrons and protons [13.4(c)The other employs the term “nucleon” to refer to either a neutron or proton.¬†Therefore, the number of nucleons within an atom is the mass number, A.¬†Nuclides, or nuclear species, are represented by the notation X A Z , where”X” is the symbol for chemical used to describe the species.¬†For instance the nucleus of gold is represented by the number the number 197 779 Au.¬†It has 197 nucleons including 79 of them being protons, and the rest are neutrons should not be republished as 441 Nuclei. composition of isotopes in an element is now easily understood.¬†The isotopes of an element have the same amount of protons but differ in the amount of neutrons.¬†Deuterium 2 1 H which is an isotope made of hydrogen, includes one proton as well as one neutron.¬†Its tritium isotope 3 1 H includes one proton as well as two neutrons.¬†Top¬†JEE Coaching in Jorhat.

The element gold contains 32 isotopes that range from A = 173 to A = A =.¬†As we have mentioned before, the elements’ chemical properties depend from their electron structure.¬†Since the isotopes’ atoms possess the same electronic structure, they share the same chemical properties and are in the same spot within the periodic table.¬†All nuclides having the identical mass numbers A are known as isobars.¬†For instance the nuclides 3 1 H as well as 3 2He, are considered isobars.¬†Nuclides that have the identical neutron numbers N but with different atomic numbers Z, like that 198 80 Hg is 1907 79 Au , are known as isotones.¬†13.3 SIZE of the nucleus As we’ve seen in Chapter 12 Rutherford was the first to proposed and proved that there is an nucleus atomicum.¬†Based on Rutherford’s suggestions, Geiger and Marsden performed their famous experiment on the scattering of particles of a-particles by the thin foils of gold.¬†Their research showed that the distance of closest contact with a gold nucleus in a-particles that has kinetic energy 5.5 MeV is about 4.0 10-14 meters.¬†

The scattering caused by a particle due to the gold sheet could be explained by Rutherford by concluding that coulomb repulsive force is the only one the cause of scattering.¬†Because this positive charge limited to the nucleus itself, the dimensions of the nuclear nucleus must to be smaller than 4.0 10-14 meters. If we employ particles with greater energies that are greater than 5.5 MeV, the distance from the nearest contact with the gold nucleus is less and eventually the scattering will be affected by forces of nuclear short-range which diverge from Rutherford’s calculations.¬†Rutherford’s calculations are based upon pure coulomb attraction among the positively charged aparticle as well as the nucleus of gold.¬†From the distances that the deviations are set the nucleus, the size of the nuclear can be calculated.¬†Top¬†JEE Coaching in Jorhat.¬†When conducting scattering tests that use fast electrons instead of particles, are projectiles that pound targets that are composed of various elements, the dimensions of the nuclei from various elements have been precisely determined.¬†

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It was discovered that a nucleus having mass A is radial in size. of R equals R0 A/3 (13.5) where R0 is 1.2 10-15m. This indicates that the size of the nucleus is proportional to the R3 is proportional to the A.¬†The density of the nucleus is steady, regardless of the A in all the nuclei.¬†Different nuclei are similar to drops of liquid that has a the same density.¬†Nuclear matter has a density 2.3 1017kg m-3.¬†This is quite a lot in comparison to the ordinary matter, for instance water, which has a density of approximately 103 kg m-3.¬†This is natural, since we’ve seen before that the majority of an atoms are empty.¬†The normal matter that is composed of atoms is filled with a huge quantity of empty space.Mass is energy. Einstein proved through the theory of special relativity, that it was important to view mass as a different kind of energy.¬†Prior to the development of this concept of special relativity,, it was thought that energy and mass kept separate during reactions.¬†Top¬†JEE Coaching in Jorhat.

But, Einstein showed that mass is an alternative form of energy that is able to convert energy from mass into other types of energy, like the kinetic energy, and vice versa.¬†Einstein provided the well-known mass-energy-energy-equivalence relationship E = Mc2 (13.6) The energy equivalent to mass m is calculated by the equation above and c is the speed of light in a vacuum. It is about 3×108. m s-1Nuclei , which only affects the nucleons that are close to it. This is also known as the saturation property of nuclear force.¬†(iii) A very heavy nucleus, for example A of 240 is bound with less energy per nucleon as compared to the nucleus that has A equal to 120.¬†So, if a nuclear the nucleus A = 240 is broken in two A = 120 nuclei, the nucleons become much more closely bound.¬†This means that energy will be released as a result.¬†It is a significant issue for the production of energy through fission. This will be discussed further in section 13.7.1.¬†(iv) Think about two light nuclei (A 10) merging to create an atomic mass that is heavier.¬†

It is apparent that the binding power per unit for the heavier nuclei that have been fused is higher that the energy of binding per nucleon for the nuclei with lighter mass.¬†This means that the resulting system is more closely bound than the first system.¬†In addition, energy will be released during a process of Fusion.¬†This is the source of energy of the sun, which will be explained in section 13.7.3.¬†13.5 NUCLEAR FORCE The force that governs the movement of electrons within atomic nuclei is known as the Coulomb force.¬†In Section 13.4 we’ve discovered that for nuclei with average mass the energy required to bind a nucleon is about 8 MeV which is a lot more than the energy of binding in the atoms.¬†Top¬†JEE Coaching in Jorhat.¬†So, in order to tie two nuclei, there needs to be an attraction force of a completely different type.¬†It has to be sufficient to overcome the attraction between (positively positively charged) protons, and to bring neutrons and protons into the small nucleus.¬†We’ve already observed that the steadfastness of the binding energy per nucleon could be seen by its short-range.¬†

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Numerous aspects that make up the force of nuclear binding can be described below. They are derived from a myriad of experiments that were conducted from 1930 to 1950. (i) It is more powerful that the Coulomb force that is acting between charges, or the gravitational force between masses. The force of nuclear binding has been able to prevail in comparison to the Coulomb repulsive force acting between protons within the nucleus. This happens since the force of nuclear is more powerful than the Coulomb force. It is smaller than even Coulomb force. (ii) the nuclear energy between nucleons decreases quickly to zero when their distance is greater than one femtometer. This causes an oversaturation of forces in a medium or huge nucleus. This is the main reason for the constant nuclear energy binding per. A rough graph of the energy potential between two nucleons as a result of distance is displayed in Fig. 13.2. Potential energy is minimal at a distance that is approximately 0.8 Fm. Top JEE Coaching in Jorhat. This implies this force will be appealing for distances that are greater than 0.8 Fm, and is repulsive when they(iii) the nuclear forces between neutron-neutron proton-neutron and proton is about the similar. 

The nuclear force doesn’t have to be dependent on electric charges.¬†In contrast to Coulomb’s laws or Newton’s law of gravitation , there is no mathematical formula of nuclear force.¬†13.6 Radioactivity A. H. Becquerel discovered radioactivity in 1896 completely through accidental.¬†In his research into the phosphorescence and fluorescence of compounds exposed to the visible spectrum, Becquerel discovered an intriguing phenomenon.¬†After shining a few pieces of uranium-potassium sulfurate using visible light He wrapped the pieces in black paper and separated the box from a photographic plate using an element of silver.¬†When, after a few hours of exposure and exposure, the photographic plate developing, it began to show darkening caused by something released by the compound. It could be absorbed by both the black paper and the silver.¬†The subsequent experiments showed radioactivity to be an atomic phenomenon where unstable nuclei undergoes decay.¬†This is known as radioactive decomposition.¬†

Three kinds of radioactive decay can be observed in nature: (i) A-decay, where a Helium nucleus 4 2He emits; (ii) b-decay in which positrons or electrons (particles that have similar mass to electrons, but having an energy charge that is opposite to electrons) are released; (iii) G-decay where high-energy (hundreds in keV or greater) photons are released.¬†Each of these types of decay will be discussed in the following subsections.¬†Top¬†JEE Coaching in Jorhat.¬†The law of decay in radioactive substances in any radioactive substance that undergoes a, b , or decay in g, it is observed that the amount of nuclei that undergo decay in a unit of duration is proportional to number of nuclei present in the sample.¬†When N represents the nuclei present in the sample, and they undergo decay within the time t, then is N t or N/t = lN (13.10) where L is”radioactive decay”, or disintegration constant.¬†The variation in the number of nuclei present in the sample is dN = -N in the period t.¬†So the change rate in N (in the range t – 0.) (d – d N). T = l * N is the number of nuclei which change, which it is never negative.¬†

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The variation in N, which can be expressed in either of two ways.¬†In this instance, it’s negative since, out of the original N nuclei, the nuclei have been destroyed leaving (N-N) Nuclei.¬†(c) These NCERT are separated at smaller than 0.8 fm.NUCLEAR Energy The curve for binding energy per nuclear nucleon, Ebn shown in figure.¬†13.1 It has a lengthy flat middle region that runs between A equal 30 to A = 170.¬†In this region, the bound energy of a nucleon remains almost always the same (8.0 MeV).¬†For the nuclei with lighter mass, A < 30, and for the heavier nuclei region, A > 170, the binding energy per nucleon is lower than 8.0 MeV, as we have previously noted.¬†The higher it is the energy of binding, the smaller is the mass of a system bound, like the nucleus.¬†Therefore, if nuclei that have less binding energy convert to nuclei with more binding energy that will result in an release of energy.¬†This is the result when a nucleus that is heavy decays into more or less intermediate mass pieces (fission) and when nuclei with light energy combine into a bigger nucleus (fusion.)¬†Top¬†JEE Coaching in Jorhat.¬†Exothermic chemical reactions are the basis of the traditional energy sources, such as petroleum or coal.¬†

In this case, the energy levels involved are within the range of 13.4 b-decay of the 13.4 b-decay of the 28-60 Ni nucleus, followed by emission of two g-rays due to deexcitation of the daughter nucleus 2860 (c) (c) Ni . Not to be published again. Physics 452 electronvolts. However when you have a nuclear reaction the energy released is on the order of MeV. For the same amount of material, nuclear sources release an energy output that is a million times greater than chemical sources. The fission of one kilogram of uranium for instance produces 10.14J of power. Compare this to burning 1 kg of coke which produces 107 J. 13.7.1 Fission The possibilities for new possibilities are revealed as we explore the limits of natural radioactive decays and investigate nuclear reactions by bombarding nuclei other nuclear particles, such as neutron, proton, a-particle as well as other particles. One of the most significant neutron-induced reaction is the fission. A good example of fission is when a uranium-isotope of 235 92U that is bombarded by neutron fragments into two nuclear fragments with intermediate mass. 1 235 236 144 1 0 92 % 92, 56, 36 N U U BaKr 3 n0 + + (13.26) 

The same reaction could produce additional types with intermediate mass pieces. For instance, 1 236 236 133 99 1. 236 133 99 1 n U U Sb 4 n0 + + – + (13.27) As an example 1 235 140 1 0 54 38 n Nb Xe 2 n0 + – + (13.28) The fragments are radioactive nuclear nuclei. They release b particles sequentially in order to create stable products.¬†Energy released (the Quantum value ) during the fission process of nuclei such as uranium is around 200 MeV per fissioning nuclear nucleus.¬†It is calculated by following Let’s consider the nucleus that has A = 240, which breaks into two pieces with A equals 120.¬†Top¬†JEE Coaching in Jorhat.¬†It is then Ebn in the case of A of 240 is around 7.6 MeV, Ebn for the two A = 120 fragment nuclei is around 8.5 MeV.¬†The gain in binding energy of nucleons is approximately 0.9 MeV.¬†Thus, the increase of binding energy equals 240×0.9 which is 216 MeV.¬†The energy of disintegration in fission reactions first manifests as the kinetic energy of neutrons and fragments.¬†Then it transfers to the surrounding matter , appearing as heat.¬†

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The energy source in nuclear reactors that produce electricity can be found in nuclear fission. The huge amount of energy released in an atom bomb results from an uncontrolled nuclear fission. We will discuss in the following section about how the nuclear reactor works. 13.7.2 Nuclear reactor Note one important aspect in the fission reactions described in the equations. (13.26) to (13.28). There is the release of an extra neutron (s) during the process of fission. In the average 2 1/2 neutrons are released during the fission process of the nuclear uranium. This is a tiny fraction because during certain fission events, two neutrons are produced by nuclei fission. It has the elements of fuel properly fabricated. The fuel can be an enriched form of by uranium (i.e. the one that is more abundant in 235 922 U than the naturally occurring uranium).The core is equipped with an adsorber to reduce the neutrons. Top JEE Coaching in Jorhat. The core is enclosed by a reflector in order to limit the leakage. It is important to note that the energy (heat) produced by fission is removed continuously with a suitable coolant. 

A containment vessel stops the release of radioactive fission products.¬†The entire unit is shielded to prevent radiation that is harmful from escaping.¬†The reactor is shut down using rods (made from, for instance Cadmium, for instance) which absorb a high amount of neutrons.¬†The coolant transfers heat into an operating fluid that produces stream.¬†Steam drives turbines and produces electricity.¬†Similar to any power plant nuclear reactors produce a lot of waste products.¬†However, nuclear wastes require special attention when it comes to treatment because they’re radioactive and dangerous.¬†Comprehensive safety measures to ensure the safety of reactors and dealing with and reprocessing waste fuel, are essential.¬†This is an important aspect that is part of the Indian Atomic Energy programme.¬†A plan of action is being developed to investigate the possibility of turning the radioactive material into less active and less long-lasting material.¬†13.7.3 Nuclear fusion is the process of the production of energy in stars two light nuclei fusion to create a bigger nucleus and release energy, because the bigger nuclear structure is much more closely bound.¬†

In the initial process, protons mix into a deuteron as well as the positron. The release that is 0.42 MeV of energy.¬†In the reaction [13.29(b)(b)), two deuterons combine to create the light isotope helium.¬†In the reaction (13.29c) two deuterrons join to form the triton and proton.¬†To allow fusion to occur the two nuclei need to be close enough that an the attractive short-range nuclear force capable of impacting the two.¬†Since the two are positively charged and therefore, they are both subject to coulomb attraction.¬†Top¬†JEE Coaching in Jorhat.¬†Therefore, they must be able to overcome the coulomb barrier.¬†The size of the barrier is determined by the charges and radii of two nuclei in the interaction.¬†It is evident that, for instance, the height of the barrier of two proton nuclei is approximately 400 keV. This it is greater for nuclei that have more charges.¬†We can determine that the temperatures at which protons within the proton gas would (averagely) possess enough energy to cross the coulomb-barrier: (3/2)k T = K 400 keV, which yields T 3x 110 K. If fusion can be achieved through raising your system’s temperature such that particles are able to generate sufficient kinetic energy to defeat the coulomb repulsive behavior that is known as thermonuclear fusion.

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Thermonuclear fusion is the main source of energy produced inside the stars’ interior.¬†The solar interior is at 1.5×107 K which is considerably lower than the expected temperature for the fusion of particles with average energy.¬†It is evident that fusion within the sun involves protons with energies that are over the average.¬†The fusion reaction that occurs in solar fusion is multi-step procedure that involves the hydrogen being transformed to Helium.¬†The fuel used by solar system is hydrogen that lies in its center.¬†This proton-proton (p, (p,) cycle through which this occurs is represented in the following set of chemical reactions. 1 2 1 1 The H H is a + 1 e + + n 0.42 MeV (i) E + e – + + 1.02 MeV (ii) 2 1 3 1 2 H H G + 5.49 MeV (iii) + – + 3 3 1 1 2 1 He The H-H ratio is 12.86 MeV (iv) (13.30) To allow 4 reactions to take place the three previous reactions have to occur at least twice, and in this case two light helium nuclei combine to form a normal nucleus of helium.¬†

If we look at the combination 2(i) + 2(ii) + 2(iii) +(iv), the result is 4 4 He 2 6. 26.7MeV 1 2 e + + or 1 4 1 2 (4 4 ) ( He 2 ) 2 6 26.7MeV e 26.7MeV e n g + + + (13.31) This means that four hydrogen atoms meld to form a 4 2He Atom that releases 26.7 MeV in energy.¬†Helium isn’t just one element which is synthesized inside the star.¬†When the hydrogen within the core is depleted and turns into helium, the center begins to get cooler.¬†The star is beginning to collapse under the pressure of its self-gravity (c) NCERT is not to be republished in 457 Nuclei which raises the core’s temperature. Best¬†JEE Coaching in Jorhat.¬†If the temperature rises to 108 K Then fusion occurs again with the helium nuclei to carbon.¬†This type of process may produce larger and greater mass elements.¬†However, elements with a mass greater than those that are near the top of the curve of energy binding shown in Fig.¬†13.1 can’t be produced.¬†The sun’s age is approximately 5×109 years, and it is believed that there enough solar hydrogen to continue burning for the next five billion years.¬†

Following the burning of hydrogen will cease as the sun begins to cool down and will begin to collapse under the pressure of gravity which will increase the temperature of its core.¬†The outer layer of the sun will expand, transforming it into the red giant.¬†NUCLEAR HOLOCAUST Within the course of a single fission in uranium, around 0.9×235 MeV (200 MeV) of energy is released.¬†When each nucleus weighing around 50 kg of 235U goes through fission, the amount of energy released is approximately 4x1015J.¬†This is equivalent to around 220,000 tons of TNT enough to cause an explosion of super-sized proportions.¬†Uncontrolled release of massive nuclear energy is known as an explosion of the atomic type.¬†On August 6, 1945, an atomic weapon was deployed in battle in the very first instance.¬†It was the day that the US dropped an Atom blast on Hiroshima, Japan.¬†The blast was the equivalent of 20000 tonnes of TNT.¬†Best¬†JEE Coaching in Jorhat.