Structure of Atoms and Nuclei - PDF

15. Structure of Atoms and Nuclei Can you recall? was applied between two electrodes inside an evacuated tube. The cathode was seen to  What is Dalton’s atomic model? emi...

15. Structure of Atoms and Nuclei Can you recall? was applied between two electrodes inside an evacuated tube. The cathode was seen to  What is Dalton’s atomic model? emit rays which produced a glow when they  What are atoms made of? struck the glass behind the anode. By studying  What is wave particle duality? the properties of these rays, he concluded that  What are matter waves? the rays are made up of negatively charged 15.1 Introduction: particles which he called electrons. This Greek philosophers Leucippus (-370 BC) demonstrated that atoms are not indestructible. and Democritus (460 – 370 BC) were the first They contain electrons which are emitted by scientists to propose, in the 5th century BC, that the cathode. matter is made of indivisible parts called atoms. Thomson proposed his model of an atom Dalton (1766-1844) gave his atomic theory in 1903. According to this model an atom is in early nineteenth century. According to his a sphere having a uniform positive charge in theory (i) matter is made up of indestructible which electrons are embedded. This model is particles, (ii) atoms of a given element are referred to as Plum-pudding model. The total identical and (iii) atoms can combine with positive charge is equal to the total negative other atoms to form new substances. That charge of electrons in the atom, rendering atoms were indestructible was shown to be it electrically neutral. As the whole solid wrong by the experiments of J. J. Thomson sphere is uniformly positively charged, the (1856-1940) who discovered electrons in 1887. positive charge cannot come out and only the He then proceeded to give his atomic model negatively charged electrons which are small, which had some deficiencies and was later can be emitted. The model also explained improved upon by Ernest Rutherford (1871- the formation of ions and ionic compounds. 1937) and Niels Bohr (1885-1962). We will However, further experiments on structure discuss these different models in this Chapter. of atoms which are described below, showed You have already studied about atoms and the distribution of charges to be very different nuclei in XIth Std. in chemistry. This chapter than what was proposed in Thomson’s model. will enable you to consolidate your concepts 15.3 Geiger-Marsden Experiment: in this subject. In order to understand the structure of We will learn that, an atom contains atoms, Rutherford suggested an experiment for scattering of alpha particles by atoms. Alpha a tiny nucleus whose size (radius) is about particles are helium nuclei and are positively 100000 times smaller than the size of an atom. charged (having charge of two protons). The The nucleus contains all the positive charge experiment was performed by his colleagues of the atom and also 99.9% of its mass. In Geiger (1882-1945) and Marsden (1889- this Chapter we will also study properties of 1970) between 1908 and 1913. A sketch of the the nucleus, the forces that keep it intact, its radioactive decays and about the energy that experimental set up is shown in Fig.15.1. can be obtained from it. Alpha particles from a source were collimated, 15.2 Thomson’s Atomic Model: i.e., focused into a narrow beam, and were Thomson performed several experiments made to fall on a gold foil. The scattered with glass vacuum tube wherein a voltage particles produced scintillations on the 324  this particle was thus found to be about 10-15 times that of an atom. He called this particle T the nucleus of an atom. He proposed that the entire positive charge and most (99.9%) of the mass of an atom is concentrated in the central nucleus and the Fig.15.1: Geiger-Marsden experiment. electrons revolve around it in circular orbits, surrounding screen. The scintillations could be similar to the revolution of the planets around observed through a microscope which could be the Sun in the Solar system. The revolution of moved to cover different angles with respect the electrons was necessary as without it, the to the incident beam. It was found that most electrons would fall into the positively charged alpha particles passed straight through the foil nucleus and the atom would collapse. The while a few were deflected (scattered) through space between the orbits of the electrons (which various scattering angles. A typical scattering decide the size of the atom) and the nucleus is angle is shown by T in the figure. Only about mostly empty. Thus, most alpha particles pass 0.14% of the incident alpha particles were through this empty space undeflected and a scattered through angles larger than 0.1o. Even very few which are in direct line with the tiny out of these, most were deflected through very nucleus or are extremely close to it, get repelled small angles. About one alpha particle in 8000 and get deflected through large angles. This was deflected through angle larger than 90o model also explains why no positively charged and a fewer still were deflected through angles particles are emitted by atoms while negatively as large as 180o. charged electrons are. This is because of the 15.4 Rutherford’s Atomic Model: large mass of the nucleus which does not get Results of Geiger-Marsden’s experiment affected when force is applied on the atom. could not be explained by Thomson’s model. In 15.4.1. Difficulties with Rutherford’s Model: that model, the positive charge was uniformly Though this model in its basic form is still spread over the large sphere constituting the accepted, it faced certain difficulties. We atom. The volume density of the positive know from Maxwell’s equations that an charge would thus be very small and all of the accelerated charge emits electromagnetic incident alpha particles would get deflected radiation. An electron in Rutherford’s model only through very small angles. Rutherford moves uniformly along a circular orbit around argued that the alpha particles which were the nucleus. Even though the magnitude deflected back must have encountered a of its velocity is constant, its direction massive particle with large positive charge so changes continuously and so the motion is an that it was repelled back. From the fact that accelerated motion. Thus, the electron should extremely small number of alpha particles emit electromagnetic radiation continuously. turned back while most others passed through Also, as it emits radiation, its energy would almost undeflected, he concluded that the decrease and consequently, the radius of its positively charged particle in the atom must be orbit would decrease continuously. It would then spiral into the nucleus, causing the atom very small in size and must contain most of the to collapse and lose its atomic properties. As mass of the atom. From the experimental data, the electron loses energy, its velocity changes the size of this particle was found to be about continuously and the frequency of the radiation 10 fm (one femtometre = 10-15m) which is about emitted would also change continuously as 10-5 times the size of the atom. The volume of 325 it moves towards the nucleus. None of these series, Balmer series, Paschen series, Brackett things are observed. Firstly, most atoms are series, Pfund series, etc. In each series, the very stable and secondly, they do not constantly separation between successive lines decreases emit electromagnetic radiation and definitely as we go towards shorter wavelength and they not of varying frequency. The atoms have to reach a limiting value. be given energy, e.g., by heating, for them to Schematic diagrams for the first three be able to emit radiation and even then, they series are shown in Fig.15.3. The limiting value emit electromagnetic radiations of particular of the wavelength for each series is shown by frequencies as will be seen in the next section. dotted lines in the figure. Rutherford’s model failed on all these counts. O 15.5 Atomic Spectra: We know that when a metallic object O is heated, it emits radiation of different wavelengths. When this radiation is passed through a prism, we get a continuous O spectrum. However, the case is different when we heat hydrogen gas inside a glass tube to high temperatures. The emitted radiation has Fig.15.3: Lyman, Balmer and Paschen series in hydrogen spectrum. only a few selected wavelengths and when passed through a prism we get what is called The observed wavelengths of the emission a line spectrum as shown for the visible range lines are found to obey the relation. 1 1 1  in Fig.15.2. It shows that hydrogen emits  R * 2  2 + --- (15.1) radiations of wavelengths 410, 434, 486 and / n m  Here O is the wavelength of a line, R is a 656 nm and does not emit any radiation with constant and n and m are integers. n = 1, wavelengths in between these wavelengths. 2, 3,…. respectively, for Lyman, Balmer, The lines seen in the spectrum are called Paschen…. series, while m takes all integral emission lines. values greater than n for that series. The wavelength decreases with increase in m. The difference in wavelengths of successive lines in each series (fixed value UV IR of n) can be calculated from Eq. (15.1) and visible shown to decrease with increase in m. Thus, the successive lines in a given series come closer and closer and ultimately reach the Fig.15.2: Hydrogen spectrum. 2 Hydrogen atom also emits radiation values of /  n in the limit m o f, for R at some other values of wavelengths in the different values of n. Atoms of other elements ultraviolet (UV), the infrared (IR) and at also emit line spectra. The wavelengths of the longer wavelengths. The spectral lines can lines emitted by each element are unique, so be divided into groups known as series with much so that we can identify the element from names of the scientists who studied them. The the wavelengths of the spectral lines that it series, starting from shorter wavelengths and emits. Rutherford’s model could not explain going to larger wavelengths are called Lyman the atomic spectra. 326 15.6 Bohr’s Atomic Model: The positive integer n is called the principal Niels Bohr modified Rutherford’s model quantum number of the electron. The by applying ideas of quantum physics which centripetal force necessary for the circular were being developed at that time. He realized motion of the electron is provided by the that Rutherford’s model is essentially correct electrostatic force of attraction between the and all that it needs is stability of the orbits. electron and the nucleus. Assuming the atomic Also, the electrons in these stable orbits should number (number of electrons) of the atom to be not emit electromagnetic waves as required by Z, the total positive charge on the nucleus is Ze classical (Maxwell’s) electromagnetic theory. and we can write, He made three postulates which defined his me v 2n Ze 2  --- (15.3) atomic model. These are given below. rn 4M 0 rn2 1. The electrons revolve around the nucleus Here, H 0 is the permeability of vacuum and e in circular orbits. is the electron charge. Eliminating vn from the This is the same assumption as in Eq.(15.2) and Eq.(15.3), we get, Rutherford’s model and the centripetal force me n 2 h 2 Ze 2 necessary for the circular motion is provided = 4S 2 me 2 rn3 4M 0 rn2 by the electrostatic force of attraction between n 2 h 2M 0 the electron and the nucleus.  ? rn  --- (15.4)  me Ze 2 2. The radius of the orbit of an electron can Similarly, eliminating rn from Eq.(15.2) and only take certain fixed values such that the Eq.(15.3), we get, angular momentum of the electron in these Ze 2 vn  --- (15.5) orbits is an integral multiple of h/2S, h being 2M 0h n the Planck’s constant. Equation (15.4) shows that the radius of Such orbits are called stable orbits or the orbit is proportional to n 2 , i.e., the stable states of the electrons and electrons square of the principal quantum number. in these orbits do not emit radiation as is The radius increases with increase in n. The demanded by classical physics. Thus, different orbits have different and definite values of hydrogen atom has only one electron, i.e., Z angular momentum and therefore, different is 1. Substituting the values of the constants values of energies. h, H0, m and e in Eq.(15.4), we get, for n = 1, 3. An electron can make a transition from r1 = 0.053 nm. This is called the Bohr radius and is denoted by a0 = h M 0. 2 one of its orbit to another orbit having lower energy. In doing so, it emits a photon of  me e 2 energy equal to the difference in its energies This is the radius of the smallest orbit of the in the two orbits. electron in hydrogen atom. From Eq. (15.4), 15.6.1. Radii of the Orbits: we can write, Using first two postulates we can study rn a 0 n 2 --- (15.6) the entire dynamics of the circular motion of Example 15.1: Calculate the radius of the the electron, including its energy. Let the mass 3rd orbit of the electron in hydrogen atom. of the electron be me, its velocity in the nth stable orbit be vn and the radius of its orbit be Solution: The radius of nth orbit is given by rn. The angular momentum is then mevnrn and rn a 0 n 2. Thus, the radius of the third orbit according to the second postulate above, we (n = 3) is can write r3  a 0 32  9a 0  9  0.053 nm h = 0.477 nm. mevn rn  n --- (15.2) 2 327  Example 15.2: In a Rutherford scattering The negative value of the energy of the experiment, assume that an incident alpha electron indicates that the electron is bound particle (radius 1.80 fm) is moving directly inside the atom and it has to be given energy toward a target gold nucleus (radius 6.23 so as to make the total energy zero, i.e., to fm). If the alpha particle stops right at the make the electron free from the nucleus. The surface of the gold nucleus, how much energy increases (becomes less negative) with energy did it have to start with? increase in n. Substituting the values of the Solution: Initially when the alpha particle constants m,e,h and H 0 in the above equation, is far away from the gold nucleus, its total we get energy is equal to its kinetic energy. As Z2 En  13.6  2 eV --- (15.8) it comes closer to the nucleus, more and n more of its kinetic energy gets converted to The first orbit ( n 1) which has minimum potential energy. By the time it reaches the energy, is called the ground state of the atom. surface of the nucleus, its kinetic energy is Orbits with higher values of n and therefore, completely converted into potential energy higher values of energy are called the excited and it stops moving. Thus, the initial kinetic states of the atom. If the electron is in the nth energy K, of the alpha particle is equal to the orbit, it is said to be in the nth energy state. potential energy when it is at the surface of For hydrogen atom (Z = 1) the energy of the the nucleus, i.e., when the distance between electron in its ground state is -13.6 eV and the the gold nucleus and the alpha particle is equal to the sum of the radii of the gold energies of the excited states increase as given nucleus and alpha particle. by Eq.(15.8). The energy levels of hydrogen atom are shown in Fig.15.4. The energies of ? K   1  2e Ze  , where, Z is the atomic the levels are given in eV. 4M 0 r1 r2 number of gold and r1 and r2 are the radii of the gold nucleus and alpha particle respectively. For gold Z = 79. 1 2 Ze 2  K   4M 0 r1 r2 2 2  79  1.6  10 19  9  10  9 6.23 1.80  10 15  4.533  10 12 J  28.33MeV 15.6.2. Energy of the Electrons: The total energy of an orbiting electron is Fig.15.4: Energy levels and transitions between the sum of its kinetic energy and its electrostatic them for hydrogen atom (energy not to scale). potential energy. Thus, The energy levels come closer and closer En  K. E. P. E ,En being the total energy as n increases and their energy reaches a of an electron in the nth orbit. limiting value of zero as n goes to infinity. The 1  Ze 2 . energy required to take an electron from the En  me v 2n    2  4 M 0 rn  ground state to an excited state is called the Using Eq. (15.3) and (15.4) this gives excitation energy of the electron in that state. For hydrogen atom, the minimum excitation me Z 2 e 4 --- (15.7) En   energy (of n = 2 state) is -3.4-(-13.6) =10.2 eV. 8 M 0 h 2 n 2 328 In order to remove or take out the electron in the ground state from a hydrogen atom, i.e., to The first two excited states have n = 2 and make it free (and have zero energy), we have 3. Their energies are 1 to supply 13.6 eV energy to it. This energy is E2  13.6  2  3.4 eV and 2 called the ionization energy of the hydrogen 1 atom. The ionization energy of an atom is E3  13.6  2  1.51eV. 3 the minimum amount of energy required to Excitation energy of an electron in nth be given to an electron in the ground state orbit is the difference between its energy of that atom to set the electron free. It is the in that orbit and the energy of the electron binding energy of hydrogen atom. If we form in its ground state, i.e. -13.6 eV. Thus, the a hydrogen atom by bringing a proton and an excitation energies of the electrons in the electron from infinity and combine them, 13.6 first two excited states are 10.2 eV and eV energy will be released. 12.09 eV respectively. According to the third postulate of Bohr, Example 15.4: Calculate the wavelengths when an electron makes a transition from mth of the first three lines in Paschen series of to nth orbit (m > n), the excess energy Em En hydrogen atom. is emitted in the form of a photon. The energy Solution: The wavelengths of lines in of the photon which can be written as hv, v Paschen series (n=3) are given by being its frequency, is therefore given by, 1  1 1   1 1   RH  2  2   1.097  107  2  2  m Z 2 e 4  1 1  / n m  3 m  hT  e 2  2  2  which can be written 8 M 0 h  n m  -1 m for m = 4,5,…. in terms of the wavelength as For the first three lines in the series, m = 4, 1 me Z 2 e 4  1 1  5 and 6. Substituting in the above formula  3  2  2 ---(15.9) / 8c M 0 h  n m  we get, Here c is the velocity of light in vacuum. 1  1 1   1.097  107  2  2  We define a constant called the Rydberg’s /1 3 4  constant, RH as  1.097  107  7 / ( 9  16 ) me e 4 RH = = 1.097 × 107 m–1. ---(15.10)  0.0533  107 m 1 8c H 0 h 3 /1  1.876 x 10-6 m In terms of RH, the wavelength is given by  1 1  1  1 1  1  RH Z 2  2  2  ---(15.11)  1.097  10  7 2  2  / n m  /2 3 5  This is called the Rydberg’s formula. 1.097  10  16 / (9  25) 7 Remember that for hydrogen Z is 1. Thus,  0.078  10 7 m 1 Eq.(15.11) correctly describes the observed /2  1.282  10 6 m spectrum of hydrogen as given by Eq.(15.1). 1  1 1   1.097  10  7 2  2  Example 15.3: Determine the energies of /3 3 6  the first two excited states of the electron 1.097  10 7  27 / (9  36) in hydrogen atom. What are the excitation energies of the electrons in these orbits?  0.09142  10 7 m 1 Solution: The energy of the electron in the /3  1.094  10 6 m nth orbit is given by E  13.6  1 eV. n n2 329 15.6.3. Limitations of Bohr’s Model: Even though Bohr’s model seemed to explain hydrogen spectrum, it had a few shortcomings which are listed below. (i) It could not explain the line spectra of atoms other than hydrogen. Even for hydrogen, more accurate study of the observed spectra showed multiple components in some lines which could not be explained on the basis of this model. Fig. 15.5: Standing electron wave for the 4th orbit of an electron in an atom. (ii) The intensities of the emission lines 2 rn  n /n  , n = 1,2,3….., giving seemed to differ from line to line and 2 rn Bohr’s model had no explanation for that. /n  --- (15.12) n (iii) On theoretical side also the model was The de Broglie wavelength is related to the not entirely satisfactory as it arbitrarily linear momentum pn ,of the particle by assumed orbits following a particular h h. /n   condition to be stable. There was no pn mv n theoretical basis for that assumption. Substituting this in Eq. (15.12) gives, A full quantum mechanical study is hn pn . required for the complete understanding of the 2 rn structure of atoms which is beyond the scope Thus, the angular momentum of the electron in of this book. Some reasoning for the third nth orbit, Ln, can be written as h shortcoming (i.e., theoretical basis for the Ln  pn rn  n , which is the second second postulate in Bohr’s atomic model) was 2 postulate of Bohr’s atomic model. Therefore, given by de Broglie which we consider next. considering electrons as waves gives some 15.6.4 De Broglie’s Explanation: theoretical basis for the second postulate made We have seen in Chapter 14 that material by Bohr. particles also have dual nature like that for 15.7. Atomic Nucleus: light and there is a wave associated with every 15.7.1 Constituents of a Nucleus: material particle. De Broglie suggested that The atomic nucleus is made up of instead of considering the orbiting electrons subatomic, meaning smaller than an atom, inside atoms as particles, we should view particles called protons and neutrons. them as standing waves. Similar to the case Together, protons and neutrons are referred of standing waves on strings or in pipes as to as nucleons. Mass of a proton is about studied in Chapter 6, the length of the orbit of 1836 times that of an electron. Mass of a an electron has to be an integral multiple of its neutron is nearly same as that of a proton but wavelength. Thus, the length of the first orbit is slightly higher. The proton is a positively will be equal to one de Broglie wavelength, O1 charged particle. The magnitude of its charge of the electron in that orbit, that of the second is equal to the magnitude of the charge of an orbit will be twice the de Broglie wavelength electron. The neutron, as the name suggests, of the electron in that orbit and so on. This is is electrically neutral. The number of protons shown for the 4th orbit in Fig.(15.5) in an atom is called its atomic number and In general, we can write, is designated by Z. The number of electrons 330 in an atom is also equal to Z. Thus, the total proton and neutron, me , mp and mn respectively, positive and total negative charges in an atom in this unit are: are equal in magnitude and the atom as a me = 9.109383 × 10-31 kg whole is electrically neutral. The number of mp = 1.672623 × 10-27 kg neutrons in a nucleus is written as N. The total mn = 1.674927 × 10-27 kg number of nucleons in a nucleus is called the Obviously, kg is not a convenient unit mass number of the atom and is designated by to measure masses of atoms or subatomic A = Z + N. The mass number determines the particles which are extremely small compared mass of a nucleus and of the atom. The atoms of to one kg. Therefore, another unit called the an element X are represented by the symbol for unified atomic mass unit (u) is used for the the element and its atomic and mass numbers purpose. One u is equal to 1/12th of the mass A as Z X. For example, symbols for hydrogen, of a neutral carbon atom having atomic carbon and oxygen atoms are written as 11 H , number 12, in its lowest electronic state. 1 u = 12 16 6 C and 8 O. The chemical properties of an 1.6605402 × 10-27 kg. In this unit, the masses atom are decided by the number of electrons of the above three particles are present in it, i.e., by Z. me = 0.00055 u The number of protons and electrons in mp = 1.007825 u the atoms of a given element are fixed. For mn = 1.008665 u. example, hydrogen atom has one proton and The third unit for measuring masses of one electron, carbon atom has six protons atoms and subatomic particles is in terms of and six electrons. The number of neutrons in the amount of energy that their masses are the atoms of a given element can vary. For equivalent to. According to Einstein’s famous example, hydrogen nucleus can have zero, one mass-energy relation, a particle having or two neutrons. These varieties of hydrogen mass m is equivalent to an amount of energy are referred to as 11 H , 12 H and 13 H and are E = mc2. The unit used to measure masses in respectively called hydrogen, deuterium and terms of their energy equivalent is the eV/c2. tritium. Atoms having the same number of One atomic mass unit is equal to 931.5 MeV/ protons but different number of neutrons are c2. The masses of the three particles in this unit called iosotopes. Thus, deuterium and tritium are are isotopes of hydrogen. They have the same me = 0.511 MeV/c2 chemical properties as those of hydrogen. mp = 938.28 MeV/c2 Similarly, helium nucleus can have one or two mn = 939.57 MeV/c2 neutrons and are referred as 32 He and 42 He. 15.7.2. Sizes of Nuclei: The atoms having the same mass number A, The size of an atom is decided by the sizes are called isobars. Thus, 13 H and 32 He are of the orbits of the electrons in the atom. Larger isobars. Atoms having the same number of the number of electrons in an atom, higher neutrons but different values of atomic number are the orbits occupied by them and larger is Z, are called iosotones. Thus, 13 H and 42 He the size of the atom. Similarly, all nuclei do are isotones. not have the same size. Obviously, the size of Units for measuring masses of atoms and a nucleus depends on the number of nucleons subatomic particles present in it, i.e., on its atomic number A. From Masses of atoms and subatomic particles experimental observations it has been found that are measured in three different units. First unit the radius RX of a nucleus X is related to A as 1 is the usual unit kg. The masses of electron, RX R0 A 3 --- (15.13) 331 where R0 = 1.2 x 10-15 m. distance up to which it is effective. Over short distances of about a few fm, the strength of the The density U inside a nucleus is given by 4 nuclear force is much higher than that of the 3 x A where, we have assumed m to other two forces. Its range is very small and 3 be the average mass of a nucleon (proton and its strength goes to zero when two nucleons neutron) as the difference in their masses is are at a distance larger than a few fm. This is rather small. The density is then given by, in contrast to the ranges of electrostatic and 3mA ' gravitational forces which are infinite. 4 RX 3 The protons in the nucleus repel one Substituting for RX from Eq.(15.13), we get, 3m another due to their similar (positive) charges. '  constant. The nuclear forces between the nucleons 4 R03 Thus, the density of a nucleus does not counter the forces of electrostatic repulsion. depend on the atomic number of the nucleus As nuclear force is much stronger than the and is the same for all nuclei. Substituting the electrostatic force for the distances between values of the constants m, S and R0 the value nucleons in a typical nucleus, it overcomes of the density is obtained as 2.3 x 1017 kg m-3 the repulsive force and keeps the nucleons which is extremely large. Among all known together, making the nucleus stable. The nuclear force is not yet well elements, osmium is known to have the highest understood. What we know about its properties density which is only 2.2 x 104 kg m-3. This can be summarized as follows. is smaller than the nuclear density by thirteen 1. It is the strongest force among subatomic orders of magnitude. particles. Its strength is about 50-60 times Example 15.5: Calculate the radius and larger than that of the electrostatic force. density of 70Ge nucleus, given its mass to be 2. Unlike the electromagnetic and approximately 69.924 u. gravitational forces which act over Solution: The radius of a nucleus X with large distances (their range is infinity), 1 mass number A is given by RX R0 A 3 , the nuclear force has a range of about a where R0 = 1.2 × 10-15 m few fm and the force is negligible when Thus, the radius of 70Ge is two nucleons are separated by larger RGe 1.2 × 10-15 × 701/3 = 4.945 × 10-15 m. distances. 3mGe 3. The nuclear force is independent of the The density is given by ρ = 4π R 3. charge of the nucleons, i.e., the nuclear Ge force between two neutrons with a given 10 27 ? '  3  69.924  1.66  3 separation is the same as that between 4 4.95  10 15 two protons or between a neutron and a = 2.292 × 1017 kg m-3. proton at the same separation. 15.7.3 Nuclear Forces: 15.8. Nuclear Binding Energy: You have learnt about the four fundamental We have seen that in a hydrogen atom, forces that occur in nature. Out of these four, the energy with which the electron in its the force that determines the structure of the ground state is bound to the nucleus (which is nucleus is the strong force, also called the a single proton in this case) is 13.6 eV. This is nuclear force. This acts between protons and the amount of energy which is released when neutrons and is mostly attractive. It is different a proton and an electron are brought from from the electrostatic and gravitational force infinity to form the atom in its ground state. In in terms of its strength and range, i.e., the other words, this is the amount of energy which 332 has to be supplied to the atom to separate the species and therefore, compare their stabilities. electron and the proton, i.e., to make them free. Nuclei with higher values of EB/A are more The protons and the neutrons inside a nucleus stable as compared to nuclei having smaller are also bound to one another. Energy has to values of this quantity. Binding energy per be supplied to the nucleus to make the nucleons nucleon for different values of A (i.e., for nuclei free, i.e., separate them and take them to large of different elements) are plotted in Fig.15.6. distances from one another. This energy is the binding energy of the nucleus. Same amount of energy is released if we bring individual nucleons from infinity to form the nucleus. Where does this released energy come from? It comes from the masses of the nucleons. The mass of a nucleus is smaller than the total mass of its constituent nucleons. Let the mass of a nucleus having atomic number Z and mass Fig.15.6: Binding energy per nucleon as a number A be M. It is smaller than the sum of function of mass number. masses of Z protons and N (= A-Z) neutrons. Deuterium nucleus has the minimum We can write, value of EB/A and is therefore, the least & M  Z mp N mn  M ---(15.14) stable nucleus. The value of EB/A increases ' M is called the mass defect of the nucleus. with increase in atomic number and reaches The binding energy EB , of the nucleus is given a plateau for A between 50 to 80. Thus, the by nuclei of these elements are the most stable EB  & M c 2 = ( Z mp N mn  M )c ---(15.15) 2 among all the species. The peak occurs around On the right hand side of Eq.(15.15), we can A = 56 corresponding to iron, which is thus one add and subtract the mass of Z electrons which of the most stable nuclei. The value of EB/A will enable us to use atomic masses in the decreases gradually for values of A greater calculation of binding energy. The Eq.(15.15) than 80, making the nuclei of those elements slightly less stable. Note that the binding thus becomes energy of hydrogen nucleus having a single EB =  Z mp Z me N mn  M Z me  c 2 proton is zero. EB   Z mH N mn  ZA M  c 2 ---(15.16) Example 15.6: Calculate the binding Here, mH is the mass of a hydrogen atom energy of 73 Li , the masses of hydrogen and and zZA M is the atomic mass of the element lithium atoms being 1.007825 u and 7.016 being considered. We will be using atomic u respectively. masses in what follows, unless otherwise Solution: The binding energy is given by specified. E B z = ( 3mH 4 mn  mLi )c 2 An important quantity in this regard is = (3 u1.007825 + 4 u1.00866 - 7.016) u the binding energy per nucleon (=EB/A) of 931.5 = 39.23 MeV a nucleus. This can be considered to be the 15.9 Radioactive Decays: average energy which has to be supplied to Many of the nuclei are stable, i.e., they can a nucleon to remove it from the nucleus and remain unchanged for a very long time. These make it free. This quantity thus, allows us have a particular ratio of the mass number and to compare the relative strengths with which the atomic number. Other nuclei occurring in nucleons are bound in a nucleus for different nature, are not so stable and undergo changes 333 in their structure by emission of some particles. mX , mY and mHe being the masses of the parent They change or decay to other nuclei (with atom, the daughter atom and the helium atom. different A and Z) in the process. The decaying Note that we have used atomic masses to nucleus is called the parent nucleus while calculate the Q factor. the nucleus produced after the decay is called the daughter nucleus. The process is called Do you know? radioactive decay or radioactivity and was discovered by Becquerel (1852-1908) in 1876. Becquerel discovered the radioactive decay Radioactive decays occur because the parent by chance. He was studying the X-rays nuclei are unstable and get converted to more emitted by naturally occurring materials stable daughter nuclei by the emission of some when exposed to Sunlight. He kept a particles. These decays are of three types as photographic plate covered in black paper, described below. separated from the material by a silver foil. Alpha Decay: In this type of decay, the parent When the plates were developed, he found nucleus emits an alpha particle which is the images of the material on them, showing nucleus of helium atom. The parent nucleus that the X-rays could penetrate the black thus loses two protons and two neutrons. The paper and silver foil. Once while studying uranium-potassium phosphate in a similar decay can be expressed as A4 way, the Sun was behind the clouds so no Z X  Z2 Y  A --- (15.17) exposure to Sunlight was possible. In spite X is the parent nucleus and Y is the of this, he went ahead and developed the daughter nucleus. All nuclei with A > 210 plates and found images to have formed. undergo alpha decay. The reason is that these With further experimentation he concluded nuclei have a large number of protons. The that some rays were emitted by uranium electrostatic repulsion between them is very itself for which no exposure to Sunlight was large and the attractive nuclear forces between necessary. He then passed the rays through the nucleons are not able to cope with it. This magnetic field and found that the rays were makes the nucleus unstable and it tries to affected by the magnetic field. He concluded reduce the number of its protons by ejecting that the rays must be charged particles and them in the form of alpha particles. An hence were different from the X-rays. example of this is the alpha decay of bismuth The term radioactivity was coined by which is the parent nucleus with A = 212 and Z Marie Curie who made further studies and = 83. The daughter nucleus has A= 208 and Z later discovered element radium along with her husband. The Nobel Prize for the year = 81, which is thallium. The reaction is 1903 was awarded jointly to Becquerel, 83 Bi → 81 T + α 212 208 Marie Curie and Pierre Curie for their The total mass of the products of an alpha contributions to radioactivity. decay is always less than the mass of the parent atom. The excess mass appears as the Beta Decay: In this type of decay the nucleus kinetic energy of the products. The difference emits an electron produced by converting a in the energy equivalent of the mass of the neutron in the nucleus into a proton. Thus, parent atom and that of the sum of masses of the basic process which takes place inside the the products is called the Q-value, Q, of the parent nucleus is decay and is equal to the kinetic energy of the n o p + e- + antineutrino --- (15.19) products. We can write, Neutrino and antineutrino are particles Q = [mX – mY – mHe]c2, --- (15.18) 334  In beta decay also, the total mass of the Example 15.7: Calculate the energy products of the decay is less than the mass of released in the alpha decay of 238Pu to 234U, the parent atom. The excess mass is converted the masses involved being mPu = 238.04955 into kinetic energy of the products. The Q u, mU = 234.04095 u and mHe = 4.002603 u. value for the decay can be written as Solution: The decay can be written as 238Pu Q = [mX – mY – me]c2 --- (15.23) o 234U + 4He. Its Q value, i.e., the energy Here, we have ignored the mass of the neutrino released is given by as it is negligible compared to the masses of Q = [mPu -mU – mHe]c2 the nuclei. = [238.04955 - 234.04095 - 4.002603]c2 u = 0.005997 u 931.5 MeV = 5.5862 MeV. Example 15.8: Calculate the maximum which have very little mass and no charge. kinetic energy of the beta particle (positron) 22 During beta decay, the number of nucleons emitted in the decay of 11 Na , given the 22 22 i.e., the mass number of the nucleus remains mass of 11 Na = 21.994437 u, 10 Ne = unchanged. The daughter nucleus has one less 21.991385 u and me = 0.00055 u. neutron and one extra proton. Thus, Z increases Solution: The decay can be written as 11 Na  10 Ne 22 22 by one and N decreases by one, A remaining e neutrino. The energy constant. The decay can be written as, released is Z X  Z 1Y A A e  antineutrino --- (15.20) Q = [mNa -mNe -me]c2 An example is = [21.994437-21.991385-0.00055]c2 e  antineutrino. = 0.002502 u 931.5 MeV = 2.3306 MeV 27 Co  28 Ni 60 60 There is another type of beta decay called This is the maximum energy that the beta the beta plus decay in which a proton gets particle (e+) can have, the neutrino having converted to a neutron by emitting a positron zero energy in that case. and a neutrino. A positron is a particle with Gamma Decay: In this type of decay, gamma the same properties as an electron except rays are emitted by the parent nucleus. As you that its charge is positive. It is known as the know, gamma ray is a high energy photon. The antiparticle of electron. This decay can be daughter nucleus is same as the parent nucleus written as, as no other particle is emitted, but it has less p o n + e+ + neutrino --- (15.21) energy as some energy goes out in the form of The mass number remains unchanged during the emitted gamma ray. the decay but Z decreases by one and N We have seen that the electrons in an atom increases by one. The decay can be written as are arranged in different energy levels (orbits) Z X  Z 1Y A A e neutrino --- (15.22) and an electron from a higher orbit can make a An example is transition to the lower orbit emitting a photon 11 Na  10 Ne 22 22 e neutrino. in the process. The situation in a nucleus is An interesting thing about beta plus decay is similar. The nucleons occupy energy levels that the mass of a neutron is higher than the with different energies. A nucleon can make a mass of a proton. Thus the decay described by transition from a higher energy level to a lower Eq. (15.21) cannot take place for a free proton. energy level, emitting a photon in the process. However, it can take place when the proton is The difference between atomic and nuclear inside the nucleus as the extra energy needed energy levels is in their energies and energy to produce a neutron can be obtained from the separations. Energies and the differences in the rest of the nucleus. 335 energies of different levels in an atom are of that if we have one atom of the radioactive the order of a few eVs, while those in the case material, we can never predict how long it of a nucleus are of the order of a few keV to will take to decay. If we have N 0 number of a few MeV. Therefore, whereas the radiations radioactive atoms (parent atoms or nuclei) of emitted by atoms are in the ultraviolet to radio a particular kind say uranium, at time t = 0, all region, the radiations emitted by nuclei are in we can say is that their number will decrease the range of gamma rays. with time as some nuclei (we cannot say which Usually, the nucleons in a nucleus are ones) will decay. Let us assume that at time t, in the lowest possible energy state. They number of parent nuclei which are left is N(t). cannot easily get excited as a large amount How many of these will decay in the interval of energy (in keVs or MeVs) is required for between t and t +dt ? We can guess that the their excitation. A nucleon however may end larger the value of N(t), larger will be the up in an excited state as a result of the parent number of decays dN in time dt. Thus, we can nucleus undergoing alpha or beta decay. Thus, say that dN will be proportional to N(t). Also, gamma decays usually occur after one of these we can guess that the larger the interval dt, decays. For example, 57Co undergoes beta larger will be the number of particles decaying plus decay to form the daughter nucleus 57Fe in that interval. Thus, we can write, which is in an excited state having energy of dN )  N t dt , 136 keV. There are two ways in which it can or, dN  z/ zN t zdt --- (15.24) make a transition to its ground state. One is where, O is a constant of proportionality by emitting a gamma ray of energy 136 keV and the other is by emitting a gamma ray of called the decay constant. The negative sign energy 122 keV and going to an intermediate in Eq.(15.24) indicates that the change in the state first and then emitting a photon of energy number of parent nuclei dN, is negative, i.e., 14 keV to reach the ground state. Both these N(t) is decreasing with time. We can integrate emissions have been observed experimentally. this equation as N t t Which type of decay a nucleus will undergo dN  / dt depends on which of the resulting daughter N t , N0 0 nucleus is more stable. Often, the daughter Here, N0 is the number of parent atoms at time nucleus is also not stable and it undergoes t = 0. Integration gives, further decay. A chain of decays may take N t place until the final daughter nucleus is stable. ln  / t , An example of such a series decay is that of N0 N t  N 0 e  /t 238 U, which undergoes a series of alpha and or, ---(15.25) beta decays, a total of 14 times, to finally This is the decay law of radioactivity. The rate reach a stable daughter nucleus of 206Pb. of decay, i.e., the number of decays per unit Use your brain power dN t time  , also called the activity A(t), dt Why don’t heavy nuclei decay by emitting can be written using Eq.(15.24) and (15.25) as, a single proton or a single neutron? dN A t    / N t  / N 0 e  /t --- (15.26) 15.10. Law of Radioactive Decay: dt Materials which undergo alpha, beta At t = 0, the activity is given by A0  / N 0. or gamma decays are called radioactive Using this, Eq.(15.26) can be written as materials. The nature of radioactivity is such A t  A0 e  /t  --- (15.27) 336 Activity is measured in units of becquerel (Bq) Example 15.9: The half-life of a nuclear in SI units. One becquerel is equal to one decay species NX is 3.2 days. Calculate its per second. Another unit to measure activity is (i) decay constant, (ii) average life and curie (Ci). One curie is 3.7 x 1010 decays per (iii) the activity of its sample of mass 1.5 mg. second. Thus, 1 Ci = 3.7 x 1010 Bq. Solution: The half-life (T1/2) is related to 15.10.1. Half-life of Radioactive Material: the decay constant (O) by The time taken for the number of parent T1/ 2  0.693 / / giving, radioactive nuclei of a particular species to /  0.693 / T1/2 reduce to half its value is called the half-life = 0.693/3.2 T1/2, of the species. This can be obtained from = 0.2166 per day Eq. (15.25) = 0.2166 /(24 u3600) s-1 N0  N 0 e  /T1/ 2 , giving = 2.507 u 10-6 s-1. 2 Average life is related to decay constant by e /T1/ 2  2, %  1 / / = 1/0.2166 per day = 4.617 days ln 2  or T1/ 2   0.693 / / --- (15.28) The activity is given by A = ON(t), where / N(t) is the number of nuclei in the given The interesting thing about half-life is that sample. This is given by even though the number goes down from N 0 N N(t)= 6.02 u 1023 u 1.5 u 10-3/Y = 9.03 u 1020/Y to 0 in time T1/2 , after another time interval Here, Y is the atomic mass of nuclear 2 T1/2, the number of parent nuclei will not go to species X in g per mol. zero. It will go to half of the value at t =T1/2 , ? A = 9.03 u 1020 u 2.5 u 10-6/Y N = 2.257 u 1015 /Y i.e., to 0. Thus, in a time interval equal to 4 = 2.257 u 1015/(Y x 3.7 u 1010 ) Ci half-life, the number of parent nuclei reduces = 6.1 u 104/Y Ci. by a factor of ½. Example 15.10: The activity of a 15.10.2 Average Life of a Radioactive Species: radioactive sample decreased from 350 s-1 We have seen that different nuclei of a to 175 s-1 in one hour. Determine the half- given radioactive species decay at different life of the species. times, i.e., they have different life times. We Solution: The time dependence of activity can calculate the average life time of a nucleus is given by A t  A0 e  /t , where, A(t) of the material using Eq.(15.25) as described and A0 are the activities at time t and 0 below. respectively. The number of nuclei decaying between 175 = 350 e  / 3600 , time t and t + dt is given by / N 0 e  /t dt. The O z or, 3600 = ln (350/175) = ln 2 = 0.6931 life time of these nuclei is t. Thus, the average O = 0.6931/3600 = 1.925 u 10-4 s-1. lifetime W of a nucleus is The half-life is given by T1/ 2  0.693 / /. N N 1 %  t / N 0 e dt  / t e  /t dt   /t 0.693 N0 0 , ? T1/ 2  4  3.6  103 s 0  1.925  10 Integrating the above we get Example 15.11: In an alpha decay, the %  1/ / --- (15.29) daughter nucleus produced is itself unstable The relation between the average life and half- and undergoes further decay. If the number life can be obtained using Eq.(15.28) as of parent and daughter nuclei at time t T1/ 2  % ln 2  0.693% --- (15.30) are Np and Nd respectively and their decay 337  constants are Op and Od respectively. What of fuel, the nuclear energy released is about a million times that released through chemical condition needs to be satisfied in order for reactions. However, nuclear energy generation Nd to remain constant? is a very complex and expensive process and Solution: The number of parent nuclei it can also be extremely harmful. Let us learn decaying between time 0 and dt, for small more about it. values of dt is given by N p Op dt. This is We have seen in section 15.8 that the the number of daughter nuclei produced mass of a nucleus is smaller than the sum in time dt. The number of daughter nuclei of masses of its constituents. The difference decaying in the same interval is N d Od dt. in these two masses is the binding energy of For the number of daughter nuclei to remain the nucleus. It would be the energy released constant, these two quantities, i.e., the if the nucleus is formed by bringing together number of daughter nuclei produced in time its constituents from infinity. This energy dt and the number decaying in time dt have is large (in MeV), and this process can be a to be equal. Thus, the required condition is good source of energy. In practice, we never given by form nuclei starting from individual nucleons. N p /p dt N d /d dt , However, we can obtain nuclear energy by or, N p /p N d /d two other processes (i) nuclear fission in which 15.11. Nuclear Energy: a heavy nucleus is broken into two nuclei of You are familiar with the naturally smaller masses and (ii) nuclear fusion in which two light nuclei undergo nuclear reaction and occurring, conventional sources of energy. fuse together to form a heavier nucleus. Both These include the fossil fuels, i.e., coal, fission and fusion are nuclear reactions. Let us petroleum, natural gas, and fire wood. The understand how nuclear energy is released in energy generation from these fuels is through the two processes. chemical reactions. It takes millions of years 15.11.1. Nuclear Fission: for these fuels to form. Naturally, the supply We have seen in Fig.15.6 that the binding of these conventional sources is limited and energy per nucleon (EB/A) depends on the mass with indiscriminate use, they are bound to number of the nuclei. This quantity is a measure get over in a couple of hundred years from of the stability of the nucleus. As seen from the now. Therefore, we have to use alternative figure, the middle weight nuclei (mass number sources of energy. The ones already in use are ranging from 50 to 80) have highest binding hydroelectric power, solar energy, wind energy energy per nucleon and are most stable, while and nuclear energy, nuclear energy being the nuclei with higher and lower atomic masses largest source among these. have smaller values of EB/A. The value of Nuclear energy is the energy released EB/A goes on decreasing till A~238 which is when nuclei undergo a nuclear reaction, the mass number of the heaviest naturally i.e., when one nucleus or a pair of nuclei, occurring element which is uranium. Many of due to their interaction, undergo a change the heavy nuclei are unstable and decay into in their structure resulting in new nuclei and two smaller mass nuclei. generating energy in the process. While the Let us consider a case when a heavy energy generated in chemical reactions is of nucleus, say with A ~230, breaks into two the order of few eV per reaction, the amount of nuclei having A between 50 and 150. The EB/A energy released in a nuclear reaction is of the of the product nuclei will be higher than that order of a few MeV. Thus, for the same weight 338 of the parent nucleus. This means that the in a controlled manner to produce energy combined masses of the two product nuclei in the form heat which is then converted to will be smaller than the mass of the parent electricity. In a uranium reactor, 235 92 U is used nucleus. The difference in the mass of the as the fuel. It is bombarded by slow neutrons to parent nucleus and that of the product nuclei produce 236 which undergoes fission. 92 U taken together will be released in the form of energy in the process. This process in which a Example 15.12: Calculate the energy heavy nucleus breaks into two lighter nuclei released in the reaction with the release of energy is called nuclear 236 92 U o 137 97 53 I + 39 Y + 2n. fission and is a source of nuclear energy. The masses of 92 U , 137 236 97 53 I and 39 Y are One of the nuclei used in nuclear energy 236.04557, 136.91787 and 96.91827 generation by fission is 236. This has a half- respectively. 92 U 7 life of 2.3 x 10 years and an activity of 6.5 x Solution: Energy released is given by 10-5 Ci/g. However, it being fissionable, most Q = [mU – mI – mY – 2mn]c2 of its nuclei have already decayed and it is = [236.04557 – 136.91787 – not found in nature. More than 99% of natural 96.91827 -2 x 1.00865] c2 uranium is in the form of 238 92 U and less than = 0.19011 u 931.5 MeV 1% is in the form of 92 U. 238 235 also decays, = 177.0875 MeV 92 U 3 but its half-life is about 10 times higher than Chain Reaction: that of 23692 U and is therefore not very useful Neutrons are produced in the fission for energy generation. The species needed for reaction shown in Eq. (15.31). Some reactions nuclear energy generation, i.e., 236 92 U can be produce 2 neutrons while others produce 3 or obtained from the naturally occurring 235 92 U 4 neutrons. The average number of neutrons by bombarding it with slow neutrons. 235 92 U per reaction can be shown to be 2.7. These absorbs a neutron and yields 236 92 U. This neutrons are in turn absorbed by other 235 92 U 235 reaction can be written as 92 U + no 236 92 U. nuclei to produce 236 92 U which undergo fission 236 92 U can undergo fission in several ways and produce further 2.7 neutrons per fission. producing different pairs of daughter nuclei This can have a cascading effect and the and generating different amounts of energy in number of neutrons produced and therefore the the process. Some of its decays are number of 23692 U nuclei produced can increase 236 o 137 97 quickly. This is called a chain reaction. Such 92 U 53 I + 39 Y + 2n 236 o 140 94 a reaction will lead to a fast increase in the 92 U 56 Ba + 36 Kr + 2n 236 o 133 Sb + 99 Nb + 4n --- (15.31) number of fissions and thereby in a rapid 92 U 51 41 Some of the daughter nuclei produced are increase in the amount of energy produced. not stable and they further decay to produce This will lead to an explosion. In a nuclear more stable nuclei. The energy produced in the reactor, methods are employed to stop a chain fission is in the form of kinetic energy of the reaction from occurring and fission and energy products, i.e., in the form of heat which can generation is allowed to occur in a controlled fashion. The energy generated, which is in the be collected and converted to other forms of form of heat, is carried away and converted to energy as needed. electricity by using turbines etc. Uranium Nuclear Reactor: More than 15 countries have nuclear A nuclear reactor is an apparatus or a reactors and use nuclear power. India is one of device in which nuclear fission is carried out them. There are 22 nuclear reactors in India, 339 the largest one being at Kudankulam, Tamil Nuclear fusion is taking place all the Nadu. Maximum nuclear power is generated time in the universe. It mostly takes place at by the USA. the centres of stars where the temperatures are 15.11.2. Nuclear Fusion: high enough for nuclear reactions to take place. We have seen that light nuclei (A < 40) There, light nuclei fuse into heavier nuclei have lower EB/A as compared to heavier ones. generating energy in the process. Nuclear If any two of the lighter nuclei come sufficiently fusion is in fact the source of energy for stars. close, within about one fm of each other, then Most of the elements heavier than boron till they can undergo nuclear reaction and form a iron, that we see around us today have been heavier nucleus. The heavier nucleus will have produced through nuclear fusion inside stars. higher EB/A than the reactants. The mass of the product nucleus will therefore be lower than the Do you know? total mass of the reactants, and energy of the order of MeV will be released in the process. Light elements, i.e., deuterium, helium, This process wherein two nuclei fuse together lithium, beryllium and boron, have not been to form a heavier nucleus accompanied by a created inside stars, but are believed to have release of nuclear energy is called nuclear been created within the first 200 second fusion. in the life of the universe, i.e., within 200 For a nuclear reaction to take place, seconds of the big bang which marked the it is necessary for two nuclei to come to beginning of the universe. The temperature within about 1 fm of each other so that they at that time was very high and some can experience the nuclear forces. It is very nuclear reactions could take place. After difficult for two atoms to come that close to about 200 s, the temperature decreased and each other due to the electrostatic repulsion nuclear reactions were no longer possible. between the electrons of the two atoms. This problem can be solved by stripping the atoms The temperature at the centre of the Sun is of their electrons and producing bare nuclei. about 107 K. The nuclear reactions taking place It is possible to do so by giving the electrons at the centre of the Sun are the fusion of four energies larger than the ionization energies hydrogen nuclei, i.e., protons to form a helium of the atoms by heating a gas of atoms. But nucleus. Of course, because of the electrostatic even after this, the two bare nuclei find it repulsion and the values of densities at the very difficult to go near each other due to the centre of the Sun, it is extremely unlikely repulsive force between their positive charges. that four protons will come sufficiently close For nuclear fusion to occur, we have to heat to one another at a given time so that they the gas to very high temperature thereby can combine to form helium. Instead, the providing the nuclei with very high kinetic fusion proceeds in several steps. The effective energies. These high energies can help them to reaction can be written as overcome the electrostatic repulsion and come 4 p oD + 2e+ + neutrinos + 26.7 MeV. close to one another. As the positive charge These reactions have been going on inside of a nucleus goes on increasing with increase the Sun since past 4.5 billion years and are in its atomic number, the kinetic energies of expected to continue for similar time period in the nuclei, i.e., the temperature of the gas the future. At the centres of other stars where necessary for nuclear fusion to occur goes on temperatures are higher, nuclei heavier than increasing with increase in Z. hydrogen can fuse generating energy. 340  tested such nuclear devices. America remains Do you know? the only country to have actually used two atom bombs which completely destroyed the The fusion inside stars can only take place cities of Hiroshima and Nagasaki in Japan in between nuclei having mass number smaller early August 1945. than that of iron, i.e., 56. The reason for this is that iron has the highest EB/A value Do you know? among all elements as seen from Fig.15.6. If an iron nucleus fuses into another nucleus, the atomic number of the resulting We have seen that the activity of radioactive nucleus will be higher than that of iron and material decreases exponentially with time. hence it will have smaller EB/A. The mass Other examples of exponential decay are of the resultant nucleus will hence be larger Amplitude of a simple pendulum decays than the sum of masses of the reactants exponentially as A =A0 e-bt, where b is and energy will have to be supplied to the damping factor. reactants for the reaction to take place. Cooling of an object in an open The elements heavier than iron which are surrounding is exponential. Temperature present in the universe are produced via T = T 0e-kt where k depends upon the other type of nuclear reaction which take object and the surrounding. place during stellar explosions. Discharging of a capacitor through a pure resistor is exponential. Charge Q on the capacitor at a given instant is Example 15.13: Calculate the energy Q = Q0 e-[t/RC] where RC is called time released in the fusion reaction taking place constant. inside the Sun, 4 p o D + 2e+ + neutrinos, Charging of a capacitor is also neglecting the energy given to the neutrinos. exponential but, it is called exponential Mass of alpha particle being 4.001506 u. growth. Solution: The energy released in the process, ignoring the energy taken by the neutrinos is given by Internet my friend Q   4  mp  m  2  me  c 2 , Q  # 4  1.00728  4.001506  2  0.00055$ c 2 1. https://www.siyavula.com/read/ science/grade-10/the-atom/04-the-  0.026514  931.5  24.70 MeV atom-02 The discussion on nuclear energy will not 2. https://en.wikipedia.org/wiki/Bohr_ be complete without mentioning its harmful model 3. http://hyperphysics.phy-astr.gsu.edu/ effects. If an uncontrolled chain reaction sets hbase/quantum/atomstructcon.html up in a nuclear fuel, an extremely large amount 4. https://en.wikipedia.org/wiki/Atomic_ of energy can be generated in a very short nucleus time. This fact has been used to produce what 5. h t t p s : / / e n. w i k i p e d i a. o r g / w i k i / are called atom bombs or nuclear devices. Radioactive_decay Either fission alone or both fission and fusion are used in these bombs. The first such devices were made towards the end of the second world war by America. By now, several countries including India have successfully made and 341