Physics 12th Book 1,2

Physics atomic electrons is thus (–Ze), and since the atom is neutral, the charge of the nucleus is (+Ze). The number of protons in the nucleus of the atom is, therefore, exactly Z, the atomic number. Discovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. But the masses of the nuclei of hydrogen, deuterium and tritium are in the ratio of 1:2:3. Therefore, the nuclei of deuterium and tritium must contain, in addition to a proton, some neutral matter. The amount of neutral matter present in the nuclei of these isotopes, expressed in units of mass of a proton, is approximately equal to one and two, respectively. This fact indicates that the nuclei of atoms contain, in addition to protons, neutral matter in multiples of a basic unit. This hypothesis was verified in 1932 by James Chadwick who observed emission of neutral radiation when beryllium nuclei were bombarded with alpha-particles (α-particles are helium nuclei, to be discussed in a later section). It was found that this neutral radiation could knock out protons from light nuclei such as those of helium, carbon and nitrogen. The only neutral radiation known at that time was photons (electromagnetic radiation). Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons, the energy of photons would have to be much higher than is available from the bombardment of beryllium nuclei with α-particles. The clue to this puzzle, which Chadwick satisfactorily solved, was to assume that the neutral radiation consists of a new type of neutral particles called neutrons. From conservation of energy and momentum, he was able to determine the mass of new particle ‘as very nearly the same as mass of proton’. The mass of a neutron is now known to a high degree of accuracy. It is m n = 1.00866 u = 1.6749×10–27 kg (13.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron. A free neutron, unlike a free proton, is unstable. It decays into a proton, an electron and a antineutrino (another elementary particle), and has a mean life of about 1000s. It is, however, stable inside the nucleus. The composition of a nucleus can now be described using the following terms and symbols: Z - atomic number = number of protons [13.4(a)] N - neutron number = number of neutrons [13.4(b)] A - mass number = Z + N = total number of protons and neutrons [13.4(c)] One also uses the term nucleon for a proton or a neutron. Thus the number of nucleons in an atom is its mass number A. Nuclear species or nuclides are shown by the notation A X where X is Z the chemical symbol of the species. For example, the nucleus of gold is 440 denoted by 197 Au . It contains 197 nucleons, of which 79 are protons 79 and the rest118 are neutrons. 2020-21

Nuclei The composition of isotopes of an element can now be readily explained. The nuclei of isotopes of a given element contain the same number of protons, but differ from each other in their number of neutrons. Deuterium, 2 H , which is an isotope of hydrogen, contains one proton 1 and one neutron. Its other isotope tritium, 3 H , contains one proton and 1 two neutrons. The element gold has 32 isotopes, ranging from A =173 to A = 204. We have already mentioned that chemical properties of elements depend on their electronic structure. As the atoms of isotopes have identical electronic structure they have identical chemical behaviour and are placed in the same location in the periodic table. All nuclides with same mass number A are called isobars. For example, the nuclides 3 H and 3 He are isobars. Nuclides with same 1 2 neutron number N but different atomic number Z, for example 198 Hg 80 and 197 Au , are called isotones. 79 13.3 SIZE OF THE NUCLEUS As we have seen in Chapter 12, Rutherford was the pioneer who postulated and established the existence of the atomic nucleus. At Rutherford’s suggestion, Geiger and Marsden performed their classic experiment: on the scattering of α-particles from thin gold foils. Their experiments revealed that the distance of closest approach to a gold nucleus of an α-particle of kinetic energy 5.5 MeV is about 4.0 × 10–14 m. The scattering of α-particle by the gold sheet could be understood by Rutherford by assuming that the coulomb repulsive force was solely responsible for scattering. Since the positive charge is confined to the nucleus, the actual size of the nucleus has to be less than 4.0 × 10–14 m. If we use α-particles of higher energies than 5.5 MeV, the distance of closest approach to the gold nucleus will be smaller and at some point the scattering will begin to be affected by the short range nuclear forces, and differ from Rutherford’s calculations. Rutherford’s calculations are based on pure coulomb repulsion between the positive charges of the α- particle and the gold nucleus. From the distance at which deviations set in, nuclear sizes can be inferred. By performing scattering experiments in which fast electrons, instead of α-particles, are projectiles that bombard targets made up of various elements, the sizes of nuclei of various elements have been accurately measured. It has been found that a nucleus of mass number A has a radius R = R 0 A1/3 (13.5) where R0 = 1.2 × 10–15 m (=1.2 fm; 1 fm = 10–15 m). This means the volume of the nucleus, which is proportional to R 3 is proportional to A. Thus the density of nucleus is a constant, independent of A, for all nuclei. Different nuclei are like a drop of liquid of constant density. The density of nuclear matter is approximately 2.3 × 1017 kg m–3. This density is very large compared to ordinary matter, say water, which is 103 kg m–3. This is understandable, as we have already seen that most of the atom is empty. 441 Ordinary matter consisting of atoms has a large amount of empty space. 2020-21

Physics Example 13.1 Given the mass of iron nucleus as 55.85u and A=56, find the nuclear density? Solution EXAMPLE 13.1 mFe = 55.85, u = 9.27 × 10–26 kg Nuclear density = mass = 9.27 ×10−26 ×1 volume (4π /3)(1.2 × 10−15 )3 56 = 2.29 × 1017 kg m–3 The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus. 13.4 MASS-ENERGY AND NUCLEAR BINDING ENERGY 13.4.1 Mass – Energy Einstein showed from his theory of special relativity that it is necessary to treat mass as another form of energy. Before the advent of this theory of special relativity it was presumed that mass and energy were conserved separately in a reaction. However, Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa. Einstein gave the famous mass-energy equivalence relation E = mc 2 (13.6) Here the energy equivalent of mass m is related by the above equation and c is the velocity of light in vacuum and is approximately equal to 3×108 m s–1. EXAMPLE 13.2 Example 13.2 Calculate the energy equivalent of 1 g of substance. Solution Energy, E = 10–3 × ( 3 × 108)2 J E = 10–3 × 9 × 1016 = 9 × 1013 J Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy. 442 Experimental verification of the Einstein’s mass-energy relation has been achieved in the study of nuclear reactions amongst nucleons, nuclei, electrons and other more recently discovered particles. In a reaction the conservation law of energy states that the initial energy and the final energy are equal provided the energy associated with mass is also included. This concept is important in understanding nuclear masses and the interaction of nuclei with one another. They form the subject matter of the next few sections. 13.4.2 Nuclear binding energy In Section 13.2 we have seen that the nucleus is made up of neutrons and protons. Therefore it may be expected that the mass of the nucleus is equal to the total mass of its individual protons and neutrons. However, 2020-21

Nuclei the nuclear mass M is found to be always less than this. For example, let us consider 16 O ; a nucleus which has 8 neutrons and 8 protons. We 8 have Mass of 8 neutrons = 8 × 1.00866 u Mass of 8 protons = 8 × 1.00727 u Mass of 8 electrons = 8 × 0.00055 u Therefore the expected mass of 16 O nucleus 8 = 8 × 2.01593 u = 16.12744 u. The atomic mass of 16 O found from mass spectroscopy experiments 8 is seen to be 15.99493 u. Substracting the mass of 8 electrons (8 × 0.00055 u) from this, we get the experimental mass of 16 O nucleus to be 15.99053 u. 8 Thus, we find that the mass of the 16 O nucleus is less than the total 8 mass of its constituents by 0.13691u. The difference in mass of a nucleus and its constituents, ∆M, is called the mass defect, and is given by ∆M = [Zm p + (A − Z )mn ] − M (13.7) What is the meaning of the mass defect? It is here that Einstein’s equivalence of mass and energy plays a role. Since the mass of the oxygen nucleus is less that the sum of the masses of its constituents (8 protons and 8 neutrons, in the unbound state), the equivalent energy of the oxygen nucleus is less than that of the sum of the equivalent energies of its constituents. If one wants to break the oxygen nucleus into 8 protons and 8 neutrons, this extra energy ∆M c2, has to supplied. This energy required Eb is related to the mass defect by Eb = ∆ M c2 (13.8) Example 13.3 Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of 16 O in MeV/c2. 8 Solution 1u = 1.6605 × 10–27 kg To convert it into energy units, we multiply it by c2 and find that energy equivalent = 1.6605 × 10–27 × (2.9979 × 108)2 kg m2/s2 = 1.4924 × 10–10 J = 1.4924 ×10−10 eV 1.602 ×10−19 = 0.9315 × 109 eV = 931.5 MeV EXAMPLE 13.3 or, 1u = 931.5 MeV/c2 For 16 O , ∆M = 0.13691 u = 0.13691×931.5 MeV/c2 8 = 127.5 MeV/c 2 The energy needed to separate 16 O into its constituents is thus 8 127.5 MeV/c2. If a certain number of neutrons and protons are brought together to 443 form a nucleus of a certain charge and mass, an energy Eb will be released 2020-21

Physics in the process. The energy Eb is called the binding energy of the nucleus. If we separate a nucleus into its nucleons, we would have to supply a total energy equal to Eb, to those particles. Although we cannot tear apart a nucleus in this way, the nuclear binding energy is still a convenient measure of how well a nucleus is held together. A more useful measure of the binding between the constituents of the nucleus is the binding energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a nucleus to the number of the nucleons, A, in that nucleus: Ebn = Eb / A (13.9) We can think of binding energy per nucleon as the average energy per nucleon needed to separate a nucleus into its individual nucleons. Figure 13.1 is a plot of the binding energy per nucleon Ebn versus the mass number A for a large number of nuclei. We notice the following main features of the plot: (i) the binding energy per nucleon, Ebn, is practically constant, i.e. practically independent of the atomic number for nuclei of middle mass number ( 30 < A < 170). The curve has a maximum of about 8.75 MeV for A = 56 FIGURE 13.1 The binding energy per nucleon and has a value of 7.6 MeV as a function of mass number. for A = 238. (ii) Ebn is lower for both light nuclei (A<30) and heavy nuclei (A>170). We can draw some conclusions from these two observations: (i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon. (ii) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged. Consider a particular nucleon inside a sufficiently large nucleus. It will be under the influence of only some of its neighbours, which come within the range of the nuclear force. If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon it will have no influence on the binding energy of the nucleon under consideration. If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Let the binding energy of the nucleus be pk, where k is a constant having the dimensions of energy. If we increase A by adding nucleons they will not change the binding energy of a nucleon inside. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is 444 approximately equal to pk. The property that a given nucleon 2020-21

Nuclei influences only nucleons close to it is also referred to as saturation property of the nuclear force. (iii) A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process. It has very important implications for energy production through fission, to be discussed later in Section 13.7.1. (iv) Consider two very light nuclei (A ≤ 10) joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion. This is the energy source of sun, to be discussed later in Section 13.7.3. 13.5 NUCLEAR FORCE The force that determines the motion of atomic electrons is the familiar Coulomb force. In Section 13.4, we have seen that for average mass nuclei the binding energy per nucleon is approximately 8 MeV, which is much larger than the binding energy in atoms. Therefore, to bind a nucleus together there must be a strong attractive force of a totally different kind. It must be strong enough to overcome the repulsion between the (positively charged) protons and to bind both protons and neutrons into the tiny nuclear volume. We have already seen that the constancy of binding energy per nucleon can be understood in terms of its short-range. Many features of the nuclear binding force are summarised below. These are obtained from a variety of experiments carried out during 1930 to 1950. (i) The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. This happens only because the nuclear force is much stronger than the coulomb force. The gravitational force is much weaker than even Coulomb force. FIGURE 13.2 Potential energy (ii) The nuclear force between two nucleons falls rapidly to of a pair of nucleons as a zero as their distance is more than a few femtometres. This function of their separation. leads to saturation of forces in a medium or a large-sized For a separation greater nucleus, which is the reason for the constancy of the binding energy per nucleon. than r0, the force is attractive and for separations less A rough plot of the potential energy between two nucleons than r0, the force is as a function of distance is shown in the Fig. 13.2. The strongly repulsive. potential energy is a minimum at a distance r0 of about 0.8 fm. This means that the force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less 445 than 0.8 fm. 2020-21

Physics (iii) The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge. Unlike Coulomb’s law or the Newton’s law of gravitation there is no simple mathematical form of the nuclear force. 13.6 RADIOACTIVITY A. H. Becquerel discovered radioactivity in 1896 purely by accident. While studying the fluorescence and phosphorescence of compounds irradiated with visible light, Becquerel observed an interesting phenomenon. After illuminating some pieces of uranium-potassium sulphate with visible light, he wrapped them in black paper and separated the package from a photographic plate by a piece of silver. When, after several hours of exposure, the photographic plate was developed, it showed blackening due to something that must have been emitted by the compound and was able to penetrate both black paper and the silver. Experiments performed subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. This is referred to as radioactive decay. Three types of radioactive decay occur in nature : (i) α-decay in which a helium nucleus 4 He is emitted; 2 (ii) β-decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted; (iii) γ-decay in which high energy (hundreds of keV or more) photons are emitted. Each of these decay will be considered in subsequent sub-sections. 13.6.1 Law of radioactive decay In any radioactive sample, which undergoes α, β or γ-decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and ∆N undergo decay in time ∆t then ∆N ∝ N ∆t or, ∆N/∆t = λN, (13.10) where λ is called the radioactive decay constant or disintegration constant. The change in the number of nuclei in the sample* is dN = – ∆N in time ∆t. Thus the rate of change of N is (in the limit ∆t → 0) dN = – λN dt 446 * ∆N is the number of nuclei that decay, and hence is always positive. dN is the change in N, which may have either sign. Here it is negative, because out of original N nuclei, ∆N have decayed, leaving (N–∆N ) nuclei. 2020-21

Nuclei or, dN = –λdt N Now, integrating both sides of the above equation,we get, (13.11) or, ln N − ln N0 = −λ (t – t0) (13.12) Here N0 is the number of radioactive nuclei in the sample at some arbitrary time t0 and N is the number of radioactive nuclei at any subsequent time t. Setting t0 = 0 and rearranging Eq. (13.12) gives us ln N = −λt (13.13) N0 which gives (13.14) N(t) = N0 e −λ t Note, for example, the light bulbs follow no such exponential decay law. If we test 1000 bulbs for their life (time span before they burn out or fuse), we expect that they will ‘decay’ (that is, burn out) at more or less the same time. The decay of radionuclides follows quite a different law, the law of radioactive decay represented by Eq. (13.14). The total decay rate R of a sample is the number of nuclei disintegrating per unit time. Suppose in a time interval dt, the decay count measured is ∆N. Then dN = – ∆N. The positive quantity R is then defined as R = – dN dt Differentiating Eq. (13.14), we get R = λN0 e −λt or, R = R0 e –λt (13.15) This is equivalant to the law of radioactivity decay, FIGURE 13.3 Exponential decay of a radioactive species. After a lapse of since you can integrate Eq. (13.15) to get back Eq. T1/2 , population of the given species (13.14). Clearly, R0 = λN0 is the decay rate at t = 0. The drops by a factor of 2. decay rate R at a certain time t and the number of undecayed nuclei N at the same time are related by R = λN (13.16) The decay rate of a sample, rather than the number of radioactive 447 nuclei, is a more direct experimentally measurable quantity and is given a specific name: activity. The SI unit for activity is becquerel, named after the discoverer of radioactivity, Henry Becquerel. 2020-21

Physics 1 becquerel is simply equal to 1 disintegration or decay per second. There is also another unit named “curie” that is widely used and is related to the SI unit as: 1 curie = 1 Ci = 3.7 × 1010 decays per second = 3.7 × 1010 Bq Different radionuclides differ greatly in their rate of decay. A common way to characterize this feature is through the notion of half-life. Half-life of a radionuclide MARIE SKLODOWSKA CURIE (1867-1934) (denoted by T1/2) is the time it takes for a sample that has initially, say N0 radionuclei to reduce to N0/2. Putting N = N0/2 and t = T1/2 in Eq. (13.14), we get T1/2 = ln 2 = 0.693 (13.17) λ λ Clearly if N0 reduces to half its value in time T1/2, R0 will also reduce to half its value in the same time according to Eq. (13.16). Another related measure is the average or mean life Marie Sklodowska Curie τ. This again can be obtained from Eq. (13.14). The (1867-1934) Born in Poland. She is recognised number of nuclei which decay in the time interval t to t + both as a physicist and as a chemist. The discovery of ∆t is R(t )∆t (= λN0e–λt∆t ). Each of them has lived for time radioactivity by Henri t. Thus the total life of all these nuclei would be t λN0e–λt Becquerel in 1896 inspired Marie and her husband ∆t. It is clear that some nuclei may live for a short time Pierre Curie in their researches and analyses while others may live longer. Therefore to obtain the mean life, we have to sum (or integrate) this expression over all times from 0 to ∞ , and divide by the total number N0 of nuclei at t = 0. Thus, which led to the isolation of ∞ radium and polonium elements. She was the first ∫λN 0 te –λt dt ∞ person to be awarded two ∫= λ te –λt dt Nobel Prizes- for Physics in τ= 0 1903 and for Chemistry N0 0 in 1911. One can show by performing this integral that τ = 1/λ We summarise these results with the following: T1/2 = ln 2 = τ ln 2 (13.18) λ Radioactive elements (e.g., tritium, plutonium) which are short-lived i.e., have half-lives much less than the age of the universe ( ∼ 15 billion years) have obviously decayed long ago and are not found in nature. They can, however, be produced artificially in nuclear reactions. Example 13.4 The half-life of U238 undergoing α-decay is 4.5 × 109 EXAMPLE 13.4 92 years. What is the activity of 1g sample of U238 ? 92 Solution T1/2 = 4.5 × 109 y 107 = 4.5 × 109 y x 3.16 x s/y 448 = 1.42 × 1017s 2020-21

Nuclei One kmol of any isotope contains Avogadro’s number of atoms, and so 1g of U238 contains 92 10–3 kmol × 6.025 × 1026 atoms/kmol 238 = 25.3 × 1020 atoms. The decay rate R is EXAMPLE 13.4 R = λN = 0.693 N = 0.693 × 25.3 ×1020 s −1 T1/ 2 1.42 ×1017 = 1.23 × 104 s–1 = 1.23 × 104 Bq Example 13.5 Tritium has a half-life of 12.5 y undergoing beta decay. EXAMPLE 13.5 What fraction of a sample of pure tritium will remain undecayed after 25 y. Solution By definition of half-life, half of the initial sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure tritium will remain undecayed. 13.6.2 Alpha decay A well-known example of alpha decay is the decay of uranium 238 U to 92 thorium 234 Th with the emission of a helium nucleus 4 He 90 2 U238 → 234 Th + 4 He (α-decay) (13.19) 90 2 92 In α-decay, the mass number of the product nucleus (daughter nucleus) is four less than that of the decaying nucleus (parent nucleus), while the atomic number decreases by two. In general, α-decay of a parent nucleus A X results in a daughter nucleus YA −4 Z Z −2 A X → YA −4 + 4 He (13.20) Z 2 Z−2 From Einstein’s mass-energy equivalance relation [Eq. (13.6)] and energy conservation, it is clear that this spontaneous decay is possible only when the total mass of the decay products is less than the mass of the initial nucleus. This difference in mass appears as kinetic energy of the products. By referring to a table of nuclear masses, one can check that the total mass of 234 Th and 4 He is indeed less than that of 29328U. 90 2 The disintegration energy or the Q-value of a nuclear reaction is the difference between the initial mass energy and the total mass energy of the decay products. For α-decay Q = (mX – mY – mHe) c2 (13.21) Q is also the net kinetic energy gained in the process or, if the initial 449 nucleus X is at rest, the kinetic energy of the products. Clearly, Q> 0 for exothermic processes such as α-decay. 2020-21

Physics Example 13.6 We are given the following atomic masses: U238 = 238.05079 u 4 He = 4.00260 u 92 2 234 Th = 234.04363 u 1 H = 1.00783 u 90 1 237 Pa = 237.05121 u 91 Here the symbol Pa is for the element protactinium (Z = 91). (a) Calculate the energy released during the alpha decay of 29328U. (b) Show that U238 can not spontaneously emit a proton. 92 Solution (a) The alpha decay of U238 is given by Eq. (13.20). The energy released 92 in this process is given by Q = (mU – mTh – mHe) c 2 Substituting the atomic masses as given in the data, we find Q = (238.05079 – 234.04363 – 4.00260)u× c 2 = (0.00456 u) c 2 = (0.00456 u) (931.5 MeV/u) = 4.25 MeV. (b) If U238 spontaneously emits a proton, the decay process would be 92 U238 → 29317Pa + 1 H 92 1 The Q for this process to happen is EXAMPLE 13.6 = (mU – mPa – mH) c 2 – 1.00783) u × c2 = (238.05079 – 237.05121 = (– 0.00825 u) c 2 = – (0.00825 u)(931.5 MeV/u) = – 7.68 MeV Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a U238 nucleus to make it emit a proton. 92 13.6.3 Beta decay In beta decay, a nucleus spontaneously emits an electron (β − decay) or a positron (β+ decay). A common example of β − decay is 32 P → 32 S + e− + ν (13.22) 15 16 and that of β+ decay is 22 Na → 22 Ne + e+ +ν (13.23) 11 10 The decays are governed by the Eqs. (13.14) and (13.15), so that one can never predict which nucleus will undergo decay, but one can characterize the decay by a half-life T1/2 . For example, T1/2 for the decays above is respectively 14.3 d and 2.6y. The emission of electron in β − decay is accompanied by the emission of an antineutrino (ν ); in β+ decay, instead, a neutrino (ν) is generated. Neutrinos are neutral particles with very small (possiblly, even zero) mass compared to electrons. They have only weak interaction with other particles. They are, therefore, very difficult to detect, since they can penetrate large quantity of matter (even earth) without any 450 interaction. 2020-21

Nuclei In both β − and β+ decay, the mass number A remains unchanged. In β − decay, the atomic number Z of the nucleus goes up by 1, while in β+ decay Z goes down by 1. The basic nuclear process underlying β − decay is the conversion of neutron to proton n → p + e– + ν (13.24) while for β+ decay, it is the conversion of proton into neutron p → n + e+ + ν (13.25) Note that while a free neutron decays to proton, the decay of proton to neutron [Eq. (13.25)] is possible only inside the nucleus, since proton has smaller mass than neutron. 13.6.4 Gamma decay Like an atom, a nucleus also has discrete energy levels - the ground state and excited states. The scale of energy is, however, very different. Atomic energy level spacings are of the order of eV, while the difference in nuclear energy levels is of the order of MeV. When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state), a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. This is the so-called gamma decay. The energy (MeV) corresponds to radiation of extremely short wavelength, shorter than the hard X-ray region. Typically, a gamma ray is emitted when a α or β decay results in a daughter nucleus in an excited state. This then returns to the ground state by a single photon transition or successive transitions involving more than one photon. A familiar example FIGURE 13.4 β-decay of 60 Ni nucleus 28 is the successive emmission of gamma rays of followed by emission of two γ rays energies 1.17 MeV and 1.33 MeV from the from deexcitation of the daughter deexcitation of 60 Ni nuclei formed from β − decay 28 nucleus 60 Ni . 28 of 60 Co . 27 13.7 NUCLEAR ENERGY The curve of binding energy per nucleon Ebn, given in Fig. 13.1, has 451 a long flat middle region between A = 30 and A = 170. In this region the binding energy per nucleon is nearly constant (8.0 MeV). For the lighter nuclei region, A < 30, and for the heavier nuclei region, A > 170, the binding energy per nucleon is less than 8.0 MeV, as we have noted earlier. Now, the greater the binding energy, the less is the total mass of a bound system, such as a nucleus. Consequently, if nuclei with less total binding energy transform to nuclei with greater binding energy, there will be a net energy release. This is what happens when a heavy nucleus decays into two or more intermediate mass fragments (fission) or when light nuclei fuse into a havier nucleus (fusion.) Exothermic chemical reactions underlie conventional energy sources such as coal or petroleum. Here the energies involved are in the range of 2020-21

Physics electron volts. On the other hand, in a nuclear reaction, the energy release is of the order of MeV. Thus for the same quantity of matter, nuclear sources produce a million times more energy than a chemical source. Fission of 1 kg of uranium, for example, generates 1014 J of energy; compare it with burning of 1 kg of coal that gives 107 J. 13.7.1 Fission New possibilities emerge when we go beyond natural radioactive decays and study nuclear reactions by bombarding nuclei with other nuclear particles such as proton, neutron, α-particle, etc. A most important neutron-induced nuclear reaction is fission. An example of fission is when a uranium isotope U235 bombarded with a 92 neutron breaks into two intermediate mass nuclear fragments 1 n +235 U → 236 U →144 Ba +8396 Kr + 3 1 n (13.26) 0 92 92 56 0 The same reaction can produce other pairs of intermediate mass fragments 1 n + U235 → U236 → 15313Sb + 99 Nb + 4 1 n (13.27) 0 41 0 92 92 Or, as another example, 1 n + U235 → 140 Xe + 94 Sr + 2 1 n (13.28) 0 54 38 0 92 452 The fragment products are radioactive nuclei; they emit β particles in succession to achieve stable end products. The energy released (the Q value ) in the fission reaction of nuclei like uranium is of the order of 200 MeV per fissioning nucleus. This is estimated as follows: Let us take a nucleus with A = 240 breaking into two fragments each of A = 120. Then Ebn for A = 240 nucleus is about 7.6 MeV, Ebn for the two A = 120 fragment nuclei is about 8.5 MeV. ∴ Gain in binding energy for nucleon is about 0.9 MeV. Hence the total gain in binding energy is 240×0.9 or 216 MeV. The disintegration energy in fission events first appears as the kinetic energy of the fragments and neutrons. Eventually it is transferred to the surrounding matter appearing as heat. The source of energy in nuclear reactors, which produce electricity, is nuclear fission. The enormous energy released in an atom bomb comes from uncontrolled nuclear fission. We discuss some details in the next section how a nuclear reactor functions. 13.7.2 Nuclear reactor Notice one fact of great importance in the fission reactions given in Eqs. (13.26) to (13.28). There is a release of extra neutron (s) in the fission process. Averagely, 2½ neutrons are released per fission of uranium nucleus. It is a fraction since in some fission events 2 neutrons are 2020-21

Nuclei INDIA’S ATOMIC ENERGY PROGRAMME The atomic energy programme in India was launched around the time of independence under the leadership of Homi J. Bhabha (1909-1966). An early historic achievement was the design and construction of the first nuclear reactor in India (named Apsara) which went critical on August 4, 1956. It used enriched uranium as fuel and water as moderator. Following this was another notable landmark: the construction of CIRUS (Canada India Research U.S.) reactor in 1960. This 40 MW reactor used natural uranium as fuel and heavy water as moderator. Apsara and CIRUS spurred research in a wide range of areas of basic and applied nuclear science. An important milestone in the first two decades of the programme was the indigenous design and construction of the plutonium plant at Trombay, which ushered in the technology of fuel reprocessing (separating useful fissile and fertile nuclear materials from the spent fuel of a reactor) in India. Research reactors that have been subsequently commissioned include ZERLINA, PURNIMA (I, II and III), DHRUVA and KAMINI. KAMINI is the country’s first large research reactor that uses U-233 as fuel. As the name suggests, the primary objective of a research reactor is not generation of power but to provide a facility for research on different aspects of nuclear science and technology. Research reactors are also an excellent source for production of a variety of radioactive isotopes that find application in diverse fields: industry, medicine and agriculture. The main objectives of the Indian Atomic Energy programme are to provide safe and reliable electric power for the country’s social and economic progress and to be self- reliant in all aspects of nuclear technology. Exploration of atomic minerals in India undertaken since the early fifties has indicated that India has limited reserves of uranium, but fairly abundant reserves of thorium. Accordingly, our country has adopted a three- stage strategy of nuclear power generation. The first stage involves the use of natural uranium as a fuel, with heavy water as moderator. The Plutonium-239 obtained from reprocessing of the discharged fuel from the reactors then serves as a fuel for the second stage — the fast breeder reactors. They are so called because they use fast neutrons for sustaining the chain reaction (hence no moderator is needed) and, besides generating power, also breed more fissile species (plutonium) than they consume. The third stage, most significant in the long term, involves using fast breeder reactors to produce fissile Uranium-233 from Thorium-232 and to build power reactors based on them. India is currently well into the second stage of the programme and considerable work has also been done on the third — the thorium utilisation — stage. The country has mastered the complex technologies of mineral exploration and mining, fuel fabrication, heavy water production, reactor design, construction and operation, fuel reprocessing, etc. Pressurised Heavy Water Reactors (PHWRs) built at different sites in the country mark the accomplishment of the first stage of the programme. India is now more than self-sufficient in heavy water production. Elaborate safety measures both in the design and operation of reactors, as also adhering to stringent standards of radiological protection are the hallmark of the Indian Atomic Energy Programme. produced, in some 3, etc. The extra neutrons in turn can initiate fission 453 processes, producing still more neutrons, and so on. This leads to the possibility of a chain reaction, as was first suggested by Enrico Fermi. If the chain reaction is controlled suitably, we can get a steady energy 2020-21

Physics output. This is what happens in a nuclear reactor. If the chain reaction is uncontrolled, it leads to explosive energy output, as in a nuclear bomb. There is, however, a hurdle in sustaining a chain reaction, as described here. It is known experimentally that slow neutrons (thermal neutrons) are much more likely to cause fission in U235 than fast neutrons. Also 92 fast neutrons liberated in fission would escape instead of causing another Nuclear power plants in India fission reaction. http://www.npcil.nic.in/main/AllProjectOperationDisplay.aspx The average energy of a neutron produced in fission of U235 is 2 MeV. 92 These neutrons unless slowed down will escape from the reactor without interacting with the uranium nuclei, unless a very large amount of fissionable material is used for sustaining the chain reaction. What one needs to do is to slow down the fast neutrons by elastic scattering with light nuclei. In fact, Chadwick’s experiments showed that in an elastic collision with hydrogen the neutron almost comes to rest and proton carries away the energy. This is the same situation as when a marble hits head-on an identical marble at rest. Therefore, in reactors, light nuclei called moderators are provided along with the fissionable nuclei for slowing down fast neutrons. The moderators commonly used are water, heavy water (D2O) and graphite. The Apsara reactor at the Bhabha Atomic Research Centre (BARC), Mumbai, uses water as moderator. The other Indian reactors, which are used for power production, use heavy water as moderator. Because of the use of moderator, it is possible that the ratio, K, of number of fission produced by a given generation of neutrons to the number of fission of the preceeding generation may be greater than one. This ratio is called the multiplication factor; it is the measure of the growth rate of the neutrons in the reactor. For K = 1, the operation of the reactor is said to be critical, which is what we wish it to be for steady power operation. If K becomes greater than one, the reaction rate and the reactor power increases exponentially. Unless the factor K is brought down very close to unity, the reactor will become supercritical and can even explode. The explosion of the Chernobyl reactor in Ukraine in 1986 is a sad reminder that accidents in a nuclear reactor can be catastrophic. The reaction rate is controlled through control-rods made out of neutron-absorbing material such as cadmium. In addition to control rods, reactors are provided with safety rods which, when required, can be inserted into the reactor and K can be reduced rapidly to less than unity. The more abundant isotope U238 in naturally occurring uranium is 92 non-fissionable. When it captures a neutron, it produces the highly radioactive plutonium through these reactions 238 U+ n → U239 → 239 Np +e – + ν 92 93 92 239 Np → 29349Pu + e – +ν (13.29) 93 454 Plutonium undergoes fission with slow neutrons. Figure 13.5 shows the schematic diagram of a nuclear reactor based on thermal neutron fission. The core of the reactor is the site of nuclear 2020-21

Nuclei FIGURE 13.5 Schematic diagram of a nuclear reactor based on A simplified online simulation of a nuclear reactor thermal neutron fission. http://esa21.kennesaw.edu/activities/nukeenergy/nuke.htm fission. It contains the fuel elements in suitably fabricated form. The fuel may be say enriched uranium (i.e., one that has greater abundance of U235 than naturally occurring uranium). The core contains a moderator 92 to slow down the neutrons. The core is surrounded by a reflector to reduce leakage. The energy (heat) released in fission is continuously removed by a suitable coolant. A containment vessel prevents the escape of radioactive fission products. The whole assembly is shielded to check harmful radiation from coming out. The reactor can be shut down by means of rods (made of, for example, cadmium) that have high absorption of neutrons. The coolant transfers heat to a working fluid which in turn may produce stream. The steam drives turbines and generates electricity. Like any power reactor, nuclear reactors generate considerable waste products. But nuclear wastes need special care for treatment since they are radioactive and hazardous. Elaborate safety measures, both for reactor operation as well as handling and reprocessing the spent fuel, are required. These safety measures are a distinguishing feature of the Indian Atomic Energy programme. An appropriate plan is being evolved to study the possibility of converting radioactive waste into less active and short- lived material. 13.7.3 Nuclear fusion – energy generation in stars When two light nuclei fuse to form a larger nucleus, energy is released, since the larger nucleus is more tightly bound, as seen from the binding energy curve in Fig.13.1. Some examples of such energy liberating nuclear fusion reactions are : 11 H + 1 H → 2 H + e+ + ν + 0.42 MeV [13.29(a)] 1 1 2 H + 2 H → 3 He + n + 3.27 MeV [13.29(b)] 1 1 2 2 H + 2 H → 31H + 11H + 4.03 MeV [13.29(c)] 455 1 1 2020-21

Physics In the first reaction, two protons combine to form a deuteron and a positron with a release of 0.42 MeV energy. In reaction [13.29(b)], two deuterons combine to form the light isotope of helium. In reaction (13.29c), two deuterons combine to form a triton and a proton. For fusion to take place, the two nuclei must come close enough so that attractive short-range nuclear force is able to affect them. However, since they are both positively charged particles, they experience coulomb repulsion. They, therefore, must have enough energy to overcome this coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei. It can be shown, for example, that the barrier height for two protons is ~ 400 keV, and is higher for nuclei with higher charges. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough energy to overcome the coulomb barrier: (3/2)k T = K 400 keV, which gives T ~ 3 × 109 K. When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion. Thermonuclear fusion is the source of energy output in the interior of stars. The interior of the sun has a temperature of 1.5×107 K, which is considerably less than the estimated temperature required for fusion of particles of average energy. Clearly, fusion in the sun involves protons whose energies are much above the average energy. The fusion reaction in the sun is a multi-step process in which the hydrogen is burned into helium. Thus, the fuel in the sun is the hydrogen in its core. The proton-proton (p, p) cycle by which this occurs is represented by the following sets of reactions: 1 H + 1 H → 21H + e+ + ν + 0.42 MeV (i) 1 1 e + + e – → γ + γ + 1.02 MeV (ii) 2 H + 1 H → 3 He + γ + 5.49 MeV (iii) 1 1 2 3 He + 3 He → 4 He + 1 H + 11H + 12.86 MeV (iv) (13.30) 2 2 2 1 For the fourth reaction to occur, the first three reactions must occur twice, in which case two light helium nuclei unite to form ordinary helium nucleus. If we consider the combination 2(i) + 2(ii) + 2(iii) +(iv), the net effect is 4 1 H + 2e − → 4 He + 2ν + 6γ + 26.7 MeV 1 2 or (411H + 4e − ) → ( 4 He + 2e − ) + 2ν + 6γ + 26.7 MeV (13.31) 2 Thus, four hydrogen atoms combine to form an 4 He atom with a 2 release of 26.7 MeV of energy. 456 Helium is not the only element that can be synthesized in the interior of a star. As the hydrogen in the core gets depleted and becomes helium, the core starts to cool. The star begins to collapse under its own gravity 2020-21

Nuclei which increases the temperature of the core. If this temperature increases to about 108 K, fusion takes place again, this time of helium nuclei into carbon. This kind of process can generate through fusion higher and higher mass number elements. But elements more massive than those near the peak of the binding energy curve in Fig. 13.1 cannot be so produced. The age of the sun is about 5×109 y and it is estimated that there is enough hydrogen in the sun to keep it going for another 5 billion years. After that, the hydrogen burning will stop and the sun will begin to cool and will start to collapse under gravity, which will raise the core temperature. The outer envelope of the sun will expand, turning it into the so called red giant. NUCLEAR HOLOCAUST In a single uranium fission about 0.9×235 MeV (≈200 MeV) of energy is liberated. If each nucleus of about 50 kg of 235U undergoes fission the amount of energy involved is about 4 × 1015J. This energy is equivalent to about 20,000 tons of TNT, enough for a superexplosion. Uncontrolled release of large nuclear energy is called an atomic explosion. On August 6, 1945 an atomic device was used in warfare for the first time. The US dropped an atom bomb on Hiroshima, Japan. The explosion was equivalent to 20,000 tons of TNT. Instantly the radioactive products devastated 10 sq km of the city which had 3,43,000 inhabitants. Of this number 66,000 were killed and 69,000 were injured; more than 67% of the city’s structures were destroyed. High temperature conditions for fusion reactions can be created by exploding a fission bomb. Super-explosions equivalent to 10 megatons of explosive power of TNT were tested in 1954. Such bombs which involve fusion of isotopes of hydrogen, deuterium and tritium are called hydrogen bombs. It is estimated that a nuclear arsenal sufficient to destroy every form of life on this planet several times over is in position to be triggered by the press of a button. Such a nuclear holocaust will not only destroy the life that exists now but its radioactive fallout will make this planet unfit for life for all times. Scenarios based on theoretical calculations predict a long nuclear winter, as the radioactive waste will hang like a cloud in the earth’s atmosphere and will absorb the sun’s radiation. 13.7.4 Controlled thermonuclear fusion 457 The natural thermonuclear fusion process in a star is replicated in a thermonuclear fusion device. In controlled fusion reactors, the aim is to generate steady power by heating the nuclear fuel to a temperature in the range of 108 K. At these temperatures, the fuel is a mixture of positive ions and electrons (plasma). The challenge is to confine this plasma, since no container can stand such a high temperature. Several countries around the world including India are developing techniques in this connection. If successful, fusion reactors will hopefully supply almost unlimited power to humanity. 2020-21

458 EXAMPLE 13.7Physics Example 13.7 Answer the following questions: (a) Are the equations of nuclear reactions (such as those given in Section 13.7) ‘balanced’ in the sense a chemical equation (e.g., 2H2 + O2→ 2 H2O) is? If not, in what sense are they balanced on both sides? (b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction? (c) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. This is strictly speaking, incorrect. Explain. Solution (a) A chemical equation is balanced in the sense that the number of atoms of each element is the same on both sides of the equation. A chemical reaction merely alters the original combinations of atoms. In a nuclear reaction, elements may be transmuted. Thus, the number of atoms of each element is not necessarily conserved in a nuclear reaction. However, the number of protons and the number of neutrons are both separately conserved in a nuclear reaction. [Actually, even this is not strictly true in the realm of very high energies – what is strictly conserved is the total charge and total ‘baryon number’. We need not pursue this matter here.] In nuclear reactions (e.g., Eq. 13.26), the number of protons and the number of neutrons are the same on the two sides of the equation. (b) We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). Now, since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of a reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in these binding energies appears as energy released or absorbed in a nuclear reaction. Since binding energy contributes to mass, we say that the difference in the total mass of nuclei on the two sides get converted into energy or vice-versa. It is in these sense that a nuclear reaction is an example of mass- energy interconversion. (c) From the point of view of mass-energy interconversion, a chemical reaction is similar to a nuclear reaction in principle. The energy released or absorbed in a chemical reaction can be traced to the difference in chemical (not nuclear) binding energies of atoms and molecules on the two sides of a reaction. Since, strictly speaking, chemical binding energy also gives a negative contribution (mass defect) to the total mass of an atom or molecule, we can equally well say that the difference in the total mass of atoms or molecules, on the two sides of the chemical reaction gets converted into energy or vice-versa. However, the mass defects involved in a chemical reaction are almost a million times smaller than those in a nuclear reaction.This is the reason for the general impression, (which is incorrect ) that mass-energy interconversion does not take place in a chemical reaction. 2020-21

Nuclei SUMMARY 1. An atom has a nucleus. The nucleus is positively charged. The radius of the nucleus is smaller than the radius of an atom by a factor of 104. More than 99.9% mass of the atom is concentrated in the nucleus. 2. On the atomic scale, mass is measured in atomic mass units (u). By definition, 1 atomic mass unit (1u) is 1/12th mass of one atom of 12C; 1u = 1.660563 × 10–27 kg. 3. A nucleus contains a neutral particle called neutron. Its mass is almost the same as that of proton 4. The atomic number Z is the number of protons in the atomic nucleus of an element. The mass number A is the total number of protons and neutrons in the atomic nucleus; A = Z+N; Here N denotes the number of neutrons in the nucleus. A nuclear species or a nuclide is represented as A X , where X is the Z chemical symbol of the species. Nuclides with the same atomic number Z, but different neutron number N are called isotopes. Nuclides with the same A are isobars and those with the same N are isotones. Most elements are mixtures of two or more isotopes. The atomic mass of an element is a weighted average of the masses of its isotopes and calculated in accordance to the relative abundances of the isotopes. 5. A nucleus can be considered to be spherical in shape and assigned a radius. Electron scattering experiments allow determination of the nuclear radius; it is found that radii of nuclei fit the formula R = R0 A1/3, where R0 = a constant = 1.2 fm. This implies that the nuclear density is independent of A. It is of the order of 1017 kg/m3. 6. Neutrons and protons are bound in a nucleus by the short-range strong nuclear force. The nuclear force does not distinguish between neutron and proton. 7. The nuclear mass M is always less than the total mass, Σm, of its constituents. The difference in mass of a nucleus and its constituents is called the mass defect, ∆M = (Z mp + ( A – Z )mn ) – M Using Einstein’s mass energy relation, we express this mass difference in terms of energy as ∆Eb = ∆M c2 The energy ∆Eb represents the binding energy of the nucleus. In the mass number range A = 30 to 170, the binding energy per nucleon is nearly constant, about 8 MeV/nucleon. 8. Energies associated with nuclear processes are about a million times larger than chemical process. 9. The Q-value of a nuclear process is Q = final kinetic energy – initial kinetic energy. Due to conservation of mass-energy, this is also, Q = (sum of initial masses – sum of final masses)c2 10. Radioactivity is the phenomenon in which nuclei of a given species transform by giving out α or β or γ rays; α-rays are helium nuclei; 459 2020-21

Physics β-rays are electrons. γ-rays are electromagnetic radiation of wavelengths shorter than X-rays; 11. Law of radioactive decay : N (t) = N(0) e–λt where λ is the decay constant or disintegration constant. The half-life T1/2 of a radionuclide is the time in which N has been reduced to one-half of its initial value. The mean life τ is the time at which N has been reduced to e–1 of its initial value T1/2 = ln 2 = τ ln 2 λ 12. Energy is released when less tightly bound nuclei are transmuted into more tightly bound nuclei. In fission, a heavy nucleus like U235 breaks 92 into two smaller fragments, e.g., 235 U+10 n →133 Sb + 99 Nb + 4 1 n 92 51 41 0 13. The fact that more neutrons are produced in fission than are consumed gives the possibility of a chain reaction with each neutron that is produced triggering another fission. The chain reaction is uncontrolled and rapid in a nuclear bomb explosion. It is controlled and steady in a nuclear reactor. In a reactor, the value of the neutron multiplication factor k is maintained at 1. 14. In fusion, lighter nuclei combine to form a larger nucleus. Fusion of hydrogen nuclei into helium nuclei is the source of energy of all stars including our sun. Physical Quantity Symbol Dimensions Units Remarks Atomic mass unit [M] u Unit of mass for Disintegration or λ [T –1] s–1 expressing atomic or decay constant T1/2 [T] s nuclear masses. One atomic mass unit equals Half-life [T] s 1/12th of the mass of 12C [ T–1] Bq atom. Mean life τ Time taken for the decay Activity of a radio- R of one-half of the initial active sample number of nuclei present in a radioactive sample. Time at which number of nuclei has been reduced to e–1 of its initial value Measure of the activity of a radioactive source. 460 POINTS TO PONDER 1. The density of nuclear matter is independent of the size of the nucleus. The mass density of the atom does not follow this rule. 2. The radius of a nucleus determined by electron scattering is found to be slightly different from that determined by alpha-particle scattering. 2020-21

Nuclei 461 This is because electron scattering senses the charge distribution of the nucleus, whereas alpha and similar particles sense the nuclear matter. 3. After Einstein showed the equivalence of mass and energy, E = mc 2, we cannot any longer speak of separate laws of conservation of mass and conservation of energy, but we have to speak of a unified law of conservation of mass and energy. The most convincing evidence that this principle operates in nature comes from nuclear physics. It is central to our understanding of nuclear energy and harnessing it as a source of power. Using the principle, Q of a nuclear process (decay or reaction) can be expressed also in terms of initial and final masses. 4. The nature of the binding energy (per nucleon) curve shows that exothermic nuclear reactions are possible, when two light nuclei fuse or when a heavy nucleus undergoes fission into nuclei with intermediate mass. 5. For fusion, the light nuclei must have sufficient initial energy to overcome the coulomb potential barrier. That is why fusion requires very high temperatures. 6. Although the binding energy (per nucleon) curve is smooth and slowly varying, it shows peaks at nuclides like 4He, 16O etc. This is considered as evidence of atom-like shell structure in nuclei. 7. Electrons and positron are a particle-antiparticle pair. They are identical in mass; their charges are equal in magnitude and opposite. ( It is found that when an electron and a positron come together, they annihilate each other giving energy in the form of gamma-ray photons.) 8. In â--decay (electron emission), the particle emitted along with electron is anti-neutrino ( ν ). On the other hand, the particle emitted in β+- decay (positron emission) is neutrino (ν). Neutrino and anti-neutrino are a particle-antiparticle pair. There are anti particles associated with every particle. What should be antiproton which is the anti particle of the proton? 9. A free neutron is unstable ( n → p + e – + ν ). But a similar free proton decay is not possible, since a proton is (slightly) lighter than a neutron. 10. Gamma emission usually follows alpha or beta emission. A nucleus in an excited (higher) state goes to a lower state by emitting a gamma photon. A nucleus may be left in an excited state after alpha or beta emission. Successive emission of gamma rays from the same nucleus (as in case of 60Ni, Fig. 13.4) is a clear proof that nuclei also have discrete energy levels as do the atoms. 11. Radioactivity is an indication of the instability of nuclei. Stability requires the ratio of neutron to proton to be around 1:1 for light nuclei. This ratio increases to about 3:2 for heavy nuclei. (More neutrons are required to overcome the effect of repulsion among the protons.) Nuclei which are away from the stability ratio, i.e., nuclei which have an excess of neutrons or protons are unstable. In fact, only about 10% of knon isotopes (of all elements), are stable. Others have been either artificially produced in the laboratory by bombarding α, p, d, n or other particles on targets of stable nuclear species or identified in astronomical observations of matter in the universe. 2020-21

Physics EXERCISES You may find the following data useful in solving the exercises: e = 1.6×10–19C N = 6.023×1023 per mole 1/(4πε0) = 9 × 109 N m2/C2 k = 1.381×10–23J K–1 1 MeV = 1.6×10–13J 1 u = 931.5 MeV/c2 1 year = 3.154×107 s mH = 1.007825 u mn = 1.008665 u me = 0.000548 u m( 4 He ) = 4.002603 u 2 13.1 (a) Two stable isotopes of lithium 6 Li and 7 Li have respective 3 3 13.2 13.3 abundances of 7.5% and 92.5%. These isotopes have masses 13.4 13.5 6.01512 u and 7.01600 u, respectively. Find the atomic mass 13.6 of lithium. 13.7 13.8 (b) Boron has two stable isotopes, 150B and 151B. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of 150B and 11 B. 5 The three stable isotopes of neon: 20 Ne, 1201Ne and 22 Ne have 10 10 respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. ( )Obtain the binding energy (in MeV) of a nitrogen nucleus 14 N , 7 ( )given m 14 N =14.00307 u 7 Obtain the binding energy of the nuclei 56 Fe and 209 Bi in units of 26 83 MeV from the following data: m ( 56 Fe ) = 55.934939 u m ( 209 Bi ) = 208.980388 u 26 83 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 63 Cu atoms (of mass 62.92960 u). 29 Write nuclear reaction equations for (i) α-decay of 226 Ra (ii) α-decay of 242 Pu 88 94 (iii) β–-decay of 32 P (iv) β–-decay of 210 Bi 15 83 (v) β+-decay of 11 C (vi) β+-decay of 97 Tc 6 43 (vii) Electron capture of 120 Xe 54 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive 14 C present with the 6 stable carbon isotope 12 C . When the organism is dead, its interaction 6 with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of 14 C , and the measured activity, the age of the specimen 6 462 can be approximately estimated. This is the principle of 14 C dating 6 2020-21

Nuclei used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation. 13.9 Obtain the amount of 60 Co necessary to provide a radioactive source 27 of 8.0 mCi strength. The half-life of 60 Co is 5.3 years. 27 13.10 The half-life of 90 Sr is 28 years. What is the disintegration rate of 38 15 mg of this isotope? 13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope 197 Au and the silver isotope 107 Ag . 79 47 13.12 Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) 226 Ra and (b) 220 Rn . 88 86 Given m ( 226 Ra ) = 226.02540 u, m ( 222 Rn ) = 222.01750 u, 88 86 m ( 220 Rn ) = 220.01137 u, m ( 216 Po ) = 216.00189 u. 86 84 13.13 The radionuclide 11C decays according to 11 C →151 B + e++ν : T1/2 =20.3 min 6 The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( 161C ) = 11.011434 u and m ( 11 B ) = 11.009305 u, 6 calculate Q and compare it with the maximum energy of the positron emitted. 13.14 The nucleus 23 Ne decays by β– emission. Write down the β-decay 10 equation and determine the maximum kinetic energy of the electrons emitted. Given that: m ( 23 Ne ) = 22.994466 u 10 m ( 23 Na ) = 22.989770 u. 11 13.15 The Q value of a nuclear reaction A + b → C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 1 H+13H →12 H+12H 1 (ii) 12 C+162 C →1200 Ne+42He 6 Atomic masses are given to be m ( 2 H ) = 2.014102 u 1 m ( 3 H ) = 3.016049 u 1 m ( 12 C ) = 12.000000 u 6 m ( 20 Ne ) = 19.992439 u 10 13.16 Suppose, we think of fission of a 56 Fe nucleus into two equal 26 fragments, 28 Al . Is the fission energetically possible? Argue by 13 working out Q of the process. Given m ( 56 Fe ) = 55.93494 u and 26 m ( 28 Al ) = 27.98191 u. 13 13.17 The fission properties of 239 Pu are very similar to those of 235 U . The 463 94 92 average energy released per fission is 180 MeV. How much energy, 2020-21

Physics in MeV, is released if all the atoms in 1 kg of pure 239 Pu undergo 94 fission? 13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much U235 did it contain initially? Assume that the reactor operates 92 80% of the time, that all the energy generated arises from the fission of U235 and that this nuclide is consumed only by the fission process. 92 13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as 2 H+ 2 H → 3 He +n +3.27 MeV 1 1 2 13.20 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.) 13.21 From the relation R = R0 A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A). 13.22 For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a neutrino is emitted). e+ + A X → Z −A1Y +ν Z Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice–versa. ADDITIONAL EXERCISES 13.23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are 24 Mg 12 (23.98504u), 25 Mg (24.98584u) and 26 Mg (25.98259u). The natural 12 12 abundance of 24 Mg is 78.99% by mass. Calculate the abundances 12 of other two isotopes. 13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 2401Ca and 27 Al from the following data: 13 m ( 40 Ca ) = 39.962591 u 20 m ( 2401Ca ) = 40.962278 u m ( 26 Al ) = 25.986895 u 13 m ( 27 Al ) = 26.981541 u 13 13.25 A source contains two phosphorous radio nuclides 32 P (T1/2 = 14.3d) 15 and 33 P (T1/2 = 25.3d). Initially, 10% of the decays come from 33 P . 15 15 How long one must wait until 90% do so? 13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes: 464 223 Ra → 209 Pb + 14 C 88 82 6 2020-21

Nuclei 223 Ra → 219 Rn + 4 He 88 86 2 Calculate the Q-values for these decays and determine that both are energetically allowed. 13.27 Consider the fission of U238 by fast neutrons. In one fission event, 92 no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 140 Ce and 99 Ru. Calculate Q 58 44 for this fission process. The relevant atomic and particle masses are m ( 238 U ) =238.05079 u 92 m ( 140 Ce ) =139.90543 u 58 m ( 99 Ru ) = 98.90594 u 44 13.28 Consider the D–T reaction (deuterium–tritium fusion) 2 H + 3 H → 4 He + n 1 1 2 (a) Calculate the energy released in MeV in this reaction from the data: m ( 2 H )=2.014102 u 1 m ( 3 H ) =3.016049 u 1 (b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.) 13.29 Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that m (198Au) = 197.968233 u m (198Hg) =197.966760 u FIGURE13.6 465 2020-21

Physics 13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor. 13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV. 466 2020-21

Chapter Fourteen SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS 14.1 INTRODUCTION Devices in which a controlled flow of electrons can be obtained are the basic building blocks of all the electronic circuits. Before the discovery of transistor in 1948, such devices were mostly vacuum tubes (also called valves) like the vacuum diode which has two electrodes, viz., anode (often called plate) and cathode; triode which has three electrodes – cathode, plate and grid; tetrode and pentode (respectively with 4 and 5 electrodes). In a vacuum tube, the electrons are supplied by a heated cathode and the controlled flow of these electrons in vacuum is obtained by varying the voltage between its different electrodes. Vacuum is required in the inter-electrode space; otherwise the moving electrons may lose their energy on collision with the air molecules in their path. In these devices the electrons can flow only from the cathode to the anode (i.e., only in one direction). Therefore, such devices are generally referred to as valves. These vacuum tube devices are bulky, consume high power, operate generally at high voltages (~100 V) and have limited life and low reliability. The seed of the development of modern solid-state semiconductor electronics goes back to 1930’s when it was realised that some solid- state semiconductors and their junctions offer the possibility of controlling the number and the direction of flow of charge carriers through them. Simple excitations like light, heat or small applied voltage can change the number of mobile charges in a semiconductor. Note that the supply 2020-21

Physics and flow of charge carriers in the semiconductor devices are within the solid itself, while in the earlier vacuum tubes/valves, the mobile electrons were obtained from a heated cathode and they were made to flow in an evacuated space or vacuum. No external heating or large evacuated space is required by the semiconductor devices. They are small in size, consume low power, operate at low voltages and have long life and high reliability. Even the Cathode Ray Tubes (CRT) used in television and computer monitors which work on the principle of vacuum tubes are being replaced by Liquid Crystal Display (LCD) monitors with supporting solid state electronics. Much before the full implications of the semiconductor devices was formally understood, a naturally occurring crystal of galena (Lead sulphide, PbS) with a metal point contact attached to it was used as detector of radio waves. In the following sections, we will introduce the basic concepts of semiconductor physics and discuss some semiconductor devices like junction diodes (a 2-electrode device) and bipolar junction transistor (a 3-electrode device). A few circuits illustrating their applications will also be described. 14.2 CLASSIFICATION OF METALS, CONDUCTORS AND SEMICONDUCTORS On the basis of conductivity On the basis of the relative values of electrical conductivity (σ ) or resistivity ( ρ = 1/σ ), the solids are broadly classified as: (i) Metals: They possess very low resistivity (or high conductivity). ρ ~ 10–2 – 10–8 Ω m σ ~ 102 – 108 S m–1 (ii) Semiconductors: They have resistivity or conductivity intermediate to metals and insulators. ρ ~ 10–5 – 106 Ω m σ ~ 105 – 10–6 S m–1 (iii)Insulators: They have high resistivity (or low conductivity). ρ ~ 1011 – 1019 Ω m σ ~ 10–11 – 10–19 S m–1 The values of ρ and σ given above are indicative of magnitude and could well go outside the ranges as well. Relative values of the resistivity are not the only criteria for distinguishing metals, insulators and semiconductors from each other. There are some other differences, which will become clear as we go along in this chapter. Our interest in this chapter is in the study of semiconductors which could be: (i) Elemental semiconductors: Si and Ge (ii) Compound semiconductors: Examples are: • Inorganic: CdS, GaAs, CdSe, InP, etc. • Organic: anthracene, doped pthalocyanines, etc. • Organic polymers: polypyrrole, polyaniline, polythiophene, etc. 468 Most of the currently available semiconductor devices are based on elemental semiconductors Si or Ge and compound inorganic 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits semiconductors. However, after 1990, a few semiconductor devices using 469 organic semiconductors and semiconducting polymers have been developed signalling the birth of a futuristic technology of polymer- electronics and molecular-electronics. In this chapter, we will restrict ourselves to the study of inorganic semiconductors, particularly elemental semiconductors Si and Ge. The general concepts introduced here for discussing the elemental semiconductors, by-and-large, apply to most of the compound semiconductors as well. On the basis of energy bands According to the Bohr atomic model, in an isolated atom the energy of any of its electrons is decided by the orbit in which it revolves. But when the atoms come together to form a solid they are close to each other. So the outer orbits of electrons from neighbouring atoms would come very close or could even overlap. This would make the nature of electron motion in a solid very different from that in an isolated atom. Inside the crystal each electron has a unique position and no two electrons see exactly the same pattern of surrounding charges. Because of this, each electron will have a different energy level. These different energy levels with continuous energy variation form what are called energy bands. The energy band which includes the energy levels of the valence electrons is called the valence band. The energy band above the valence band is called the conduction band. With no external energy, all the valence electrons will reside in the valence band. If the lowest level in the conduction band happens to be lower than the highest level of the valence band, the electrons from the valence band can easily move into the conduction band. Normally the conduction band is empty. But when it overlaps on the valence band electrons can move freely into it. This is the case with metallic conductors. If there is some gap between the conduction band and the valence band, electrons in the valence band all remain bound and no free electrons are available in the conduction band. This makes the material an insulator. But some of the electrons from the valence band may gain external energy to cross the gap between the conduction band and the valence band. Then these electrons will move into the conduction band. At the same time they will create vacant energy levels in the valence band where other valence electrons can move. Thus the process creates the possibility of conduction due to electrons in conduction band as well as due to vacancies in the valence band. Let us consider what happens in the case of Si or Ge crystal containing N atoms. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of electrons in the outer orbit is 8 (2s + 6p electrons). So, for the 4N valence electrons there are 8N available energy states. These 8N discrete energy levels can either form a continuous band or they may be grouped in different bands depending upon the distance between the atoms in the crystal (see box on Band Theory of Solids). At the distance between the atoms in the crystal lattices of Si and Ge, the energy band of these 8N states is split apart into two which are separated by an energy gap Eg (Fig. 14.1). The lower band which is 2020-21

Physics completely occupied by the 4N valence electrons at temperature of absolute zero is the valence band. The other band consisting of 4N energy states, called the conduction band, is completely empty at absolute zero. BAND THEORY OF SOLIDS Consider that the Si or Ge crystal contains N atoms. Electrons of each atom will have discrete energies in different orbits. The electron energy will be same if all the atoms are isolated, i.e., separated from each other by a large distance. However, in a crystal, the atoms are close to each other (2 to 3 Å) and therefore the electrons interact with each other and also with the neighbouring atomic cores. The overlap (or interaction) will be more felt by the electrons in the outermost orbit while the inner orbit or core electron energies may remain unaffected. Therefore, for understanding electron energies in Si or Ge crystal, we need to consider the changes in the energies of the electrons in the outermost orbit only. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of outer electrons in the orbit is 8 (2s + 6p electrons). So, out of the 4N electrons, 2N electrons are in the 2N s-states (orbital quantum number l = 0) and 2N electrons are in the available 6N p-states. Obviously, some p-electron states are empty as shown in the extreme right of Figure. This is the case of well separated or isolated atoms [region A of Figure]. Suppose these atoms start coming nearer to each other to form a solid. The energies of these electrons in the outermost orbit may change (both increase and decrease) due to the interaction between the electrons of different atoms. The 6N states for l = 1, which originally had identical energies in the isolated atoms, spread out and form an energy band [region B in Figure]. Similarly, the 2N states for l = 0, having identical energies in the isolated atoms, split into a second band (carefully see the region B of Figure) separated from the first one by an energy gap. At still smaller spacing, however, there comes a region in which the bands merge with each other. The lowest energy state that is a split from the upper atomic level appears to drop below the upper state that has come from the lower atomic level. In this region (region C in Figure), no energy gap exists where the upper and lower energy states get mixed. Finally, if the distance between the atoms further decreases, the energy bands again split apart and are separated hbaysabneeennerreg-yapgpaoprEtiog n(reedgiboentwDeiennFtihguertew).oTbhaentdosta(4l Nnusmtabteesr of available energy states 8N each in the lower and upper energy bands). Here the significant point is that there are exactly as many states in the lower band (4N ) as there are available valence electrons from the atoms (4N ). Therefore, this band ( called the valence band ) is completely filled while the upper band is completely empty. The upper band is called the conduction band. 470 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits The lowest energy level in the hcoignhdeusctteionnergbyanledveilsinshthoewvnalaesncEeCbaanndd tishesrheoawrenaalsarEgVe.nAubmobveerEoCf calnosdelbyeslopwacEedV energy levels, as shown in Fig. 14.1. The gap between the top of the valence band and bottom of the conduction band is called the energy band gap (Energy gap dEeg )p.eIntdminagy be large, small, or zero, upo n the material. These different situations, are depicted in Fig. 14.2 and discussed below: Case I: This refers to a situation, as shown in Fig. 14.2(a). One can have a metal either when the conduction band FIGURE 14.1 The energy band positions in a semiconductor at 0 K. The upper band, called the is partially filled and the balanced band conduction band, consists of infinitely large number is partially empty or when the conduction of closely spaced energy states. The lower band, and valance bands overlap. When there called the valence band, consists of closely spaced is overlap electrons from valence band can completely filled energy states. easily move into the conduction band. This situation makes a large number of electrons available for electrical conduction. When the valence band is partially empty, electrons from its lower level can move to higher level making conduction possible. Therefore, the resistance of such materials is low or the conductivity is high. FIGURE 14.2 Difference between energy bands of (a) metals, 471 (b) insulators and (c) semiconductors. 2020-21

Physics Case II: In this case, as shown in Fig. 14.2(b), a large band gap Eg exists (Eg > 3 eV). There are no electrons in the conduction band, and therefore no electrical conduction is possible. Note that the energy gap is so large that electrons cannot be excited from the valence band to the conduction band by thermal excitation. This is the case of insulators. Case III: This situation is shown in Fig. 14.2(c). Here a finite but small band gap (Eg < 3 eV) exists. Because of the small band gap, at room temperature some electrons from valence band can acquire enough energy to cross the energy gap and enter the conduction band. These electrons (though small in numbers) can move in the conduction band. Hence, the resistance of semiconductors is not as high as that of the insulators. In this section we have made a broad classification of metals, conductors and semiconductors. In the section which follows you will learn the conduction process in semiconductors. 14.3 INTRINSIC SEMICONDUCTOR We shall take the most common case of Ge and Si whose lattice structure is shown in Fig. 14.3. These structures are called the diamond-like structures. Each atom is surrounded by four nearest neighbours. We know that Si and Ge have four valence electrons. In its crystalline structure, every Si or Ge atom tends to share one of its four valence electrons with each of its four nearest neighbour atoms, and also to take share of one electron from each such neighbour. These shared electron pairs are referred to as forming a covalent bond or simply a valence bond. The two shared electrons can be assumed to shuttle back-and- forth between the associated atoms holding them together strongly. Figure 14.4 schematically shows the 2-dimensional representation of Si or Ge structure shown in Fig. 14.3 which overemphasises the covalent bond. It shows an idealised picture in which no bonds are broken (all bonds are intact). Such a situation arises at low temperatures. As the temperature increases, more thermal energy becomes available to these electrons and some of these electrons may break–away (becoming free electrons contributing to conduction). The thermal energy effectively ionises only a few atoms in the crystalline lattice and creates a vacancy in the bond as shown in Fig. 14.5(a). The neighbourhood, from which the free electron (with charge –q ) has come out leaves a vacancy with an effective charge (+q ). This vacancy with the effective positive electronic charge is called a hole. The hole behaves as an apparent free particle with effective positive charge. FIGURE 14.3 Three-dimensional dia- In intrinsic semiconductors, the number of free ewlhecenStrreeeom=nnsiinci,shoncn=eaidsnlulieecqdtuoianrlsttroipntoshsiescencsuarmtrhibeeerruconofnihqcuoelenestpr,arnothip.oenTr(h.1tya4t.i1ins) mond-like crystal structure for Carbon, which, apart from electrons, the holes also move. Silicon or Germanium with respective lattice spacing a equal to 3.56, 5.43 and 5.66 Å. 472 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits Suppose there is a hole at site 1 as shown in Fig. 14.5(a). The movement of holes can be visualised as shown in Fig. 14.5(b). An electron from the covalent bond at site 2 may jump to the vacant site 1 (hole). Thus, after such a jump, the hole is at site 2 and the site 1 has now an electron. Therefore, apparently, the hole has moved from site 1 to site 2. Note that the electron originally set free [Fig. 14.5(a)] is not involved in this process of hole motion. The free electron moves completely independently as conduction eulnecdterroannaanpdpgliiveedserliescetrtiocafnieledle. cRteromnecmubrererntht,aIte the motion of hole is only a convenient way of FIGURE 14.4 Schematic two-dimensional describing the actual motion of bound electrons, representation of Si or Ge structure showing whenever there is an empty bond anywhere in covalent bonds at low temperature the crystal. Under the action of an electric field, (all bonds intact). +4 symbol these holes move towards negative potential indicates inner cores of Si or Ge. tghhiovuliensgctuhthreresehunomtlIeho:cfuthrreenelte,cIht.roTnhecutortraelnctuIreraenndt,tIhies I = Ie + Ih (14.2) It may be noted that apart from the process of generation of conduction electrons and holes, a simultaneous process of recombination occurs in which the electrons recombine with the holes. At equilibrium, the rate of generation is equal to the rate of recombination of charge carriers. The recombination occurs due to an electron colliding with a hole. (a) (b) FIGURE 14.5 (a) Schematic model of generation of hole at site 1 and conduction electron due to thermal energy at moderate temperatures. (b) Simplified representation of possible thermal motion of a hole. The electron from the lower left hand covalent bond (site 2) goes to the earlier hole site1, leaving a hole at its site indicating an 473 apparent movement of the hole from site 1 to site 2. 2020-21

Physics FIGURE 14.6 (a) An intrinsic semiconductor at T = 0 K An intrinsic semiconductor behaves like insulator. (b) At T > 0 K, four thermally generated will behave like an insulator at T = 0 K as shown in Fig. 14.6(a). electron-hole pairs. The filled circles ( ) represent electrons It is the thermal energy at and empty circles ( ) represent holes. higher temperatures (T > 0K), which excites some electrons from the valence band to the conduction band. These thermally excited electrons at T > 0 K, partially occupy the conduction band. Therefore, the energy-band diagram of an intrinsic semiconductor will be as shown in Fig. 14.6(b). Here, some electrons are shown in the conduction band. These have come from the valence band leaving equal number of holes there. EXAMPLE 14.1 Example 14.1 C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors? Solution The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C. 14.4 EXTRINSIC SEMICONDUCTOR The conductivity of an intrinsic semiconductor depends on its temperature, but at room temperature its conductivity is very low. As such, no important electronic devices can be developed using these semiconductors. Hence there is a necessity of improving their conductivity. This can be done by making use of impurities. When a small amount, say, a few parts per million (ppm), of a suitable impurity is added to the pure semiconductor, the conductivity of the semiconductor is increased manifold. Such materials are known as extrinsic semiconductors or impurity semiconductors. The deliberate addition of a desirable impurity is called doping and the impurity atoms are called dopants. Such a material is also called a doped semiconductor. The dopant has to be such that it does not distort the original pure semiconductor lattice. It occupies only a very few of the original semiconductor atom sites in the crystal. A necessary condition to attain this is that the sizes of the dopant and the semiconductor atoms should be nearly the same. There are two types of dopants used in doping the tetravalent Si or Ge: 474 (i) Pentavalent (valency 5); like Arsenic (As), Antimony (Sb), Phosphorous (P), etc. 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits (ii) Trivalent (valency 3); like Indium (In), Boron (B), Aluminium (Al), etc. We shall now discuss how the doping changes the number of charge carriers (and hence the conductivity) of semiconductors. Si or Ge belongs to the fourth group in the Periodic table and, therefore, we choose the dopant element from nearby fifth or third group, expecting and taking care that the size of the dopant atom is nearly the same as that of Si or Ge. Interestingly, the pentavalent and trivalent dopants in Si or Ge give two entirely different types of semiconductors as discussed below. (i) n-type semiconductor Suppose we dope Si or Ge with a pentavalent element as shown in Fig. 14.7. When an atom of +5 valency element occupies the position of an atom in the crystal lattice of Si, four of its electrons bond with the four silicon neighbours while the fifth remains very weakly bound to its parent atom. This is because the four electrons participating in bonding are seen as part of the effective core FIGURE 14.7 (a) Pentavalent donor atom (As, Sb, of the atom by the fifth electron. As a result P, etc.) doped for tetravalent Si or Ge giving n- type semiconductor, and (b) Commonly used the ionisation energy required to set this electron free is very small and even at room schematic representation of n-type material temperature it will be free to move in the which shows only the fixed cores of the lattice of the semiconductor. For example, the energy required is ~ 0.01 eV for germanium, substituent donors with one additional effective positive charge and its associated extra electron. and 0.05 eV for silicon, to separate this electron from its atom. This is in contrast to the energy required to jump the forbidden band (about 0.72 eV for germanium and about 1.1 eV for silicon) at room temperature in the intrinsic semiconductor. Thus, the pentavalent dopant is donating one extra electron for conduction and hence is known as donor impurity. The number of electrons made available for conduction by dopant atoms depends strongly upon the doping level and is independent of any increase in ambient temperature. On the other hand, the number of free electrons (with an equal number of holes) generated by Si atoms, increases weakly with temperature. In a doped semiconductor the total number of conduction electrons ne is due to the electrons contributed by donors and those generated intrinsically, while the total number of holes nh is only due to the holes from the intrinsic source. But the rate of recombination of holes would increase due to the increase in the number of electrons. As a result, the number of holes would get reduced further. Thus, with proper level of doping the number of conduction electrons 475 can be made much larger than the number of holes. Hence in an extrinsic 2020-21

Physics semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority carriers. These semiconductors are, therefore, known as n-type semiconductors. For n-type semiconductors, we have, ne >> nh (14.3) (ii) p-type semiconductor This is obtained when Si or Ge is doped with a trivalent impurity like Al, B, In, etc. The dopant has one valence electron less than Si or Ge and, therefore, this atom can form covalent bonds with neighbouring three Si atoms but does not have any electron to offer to the fourth Si atom. So the bond between the fourth neighbour and the trivalent atom has a vacancy or hole as shown in Fig. 14.8. Since the neighbouring Si atom in the lattice wants an electron in place of a hole, an electron in the outer orbit of an atom in the neighbourhood may jump to fill this vacancy, leaving a vacancy or hole at its own site. Thus the hole is available for conduction. Note that the trivalent foreign atom becomes effectively negatively charged when it shares fourth electron with neighbouring Si atom. Therefore, the dopant atom of p-type material can be treated as core of one negative charge along with its associated hole as shown in Fig. 14.8(b). It is obvious that one acceptor atom gives one hole. These holes are in addition to the intrinsically generated holes while the source of conduction electrons is only intrinsic generation. Thus, for FIGURE 14.8 (a) Trivalent such a material, the holes are the majority carriers and electrons acceptor atom (In, Al, B etc.) doped in tetravalent Si or Ge are minority carriers. Therefore, extrinsic semiconductors doped lattice giving p-type semicon- with trivalent impurity are called p-type semiconductors. For ductor. (b) Commonly used p-type semiconductors, the recombination process will reduce schematic representation of hthaeven,ufomrbpe-rty(pnei)osfeminitcroinnsdiucacltloyrsgenerated electrons to ne. We p-type material which shows nh >> ne (14.4) only the fixed core of the Note that the crystal maintains an overall charge neutrality substituent acceptor with as the charge of additional charge carriers is just equal and one effective additional negative charge and its opposite to that of the ionised cores in the lattice. associated hole. In extrinsic semiconductors, because of the abundance of majority current carriers, the minority carriers produced thermally have more chance of meeting majority carriers and thus getting destroyed. Hence, the dopant, by adding a large number of current carriers of one type, which become the majority carriers, indirectly helps to reduce the intrinsic concentration of minority carriers. The semiconductor’s energy band structure is affected by doping. In the case of extrinsic semiconductors, additional energy states due to donor bdimeialpoguwraritmthieeosbf(noEt-Dtto)yampneEdSCai ocscfeemthpietcoocroninmddupucutcrotiirto,ientshbe(EadnAo)dnaalosnrodeenexleiesrcgtt.yrIonlenvtsheflerEoemnD eistrhgsyilsibglheavntledyl move into the conduction band with very small supply of energy. At room temperature, most of the donor atoms get ionised but very few (~1012) atoms of Si get ionised. So the conduction band will have most electrons 476 coming from the donor impurities, as shown in Fig. 14.9(a). Similarly, 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits tsfohurepppto-ltypyopEfeVesonefemtrhgieycovanandleeunlecccteotrrbo,antnhfderoaamcscstehhpeotwovranleeinnnecFregigby.al1ne4vde.9cl a(EbnA).jiuWsmsitlphigthvoetltryhyeasblmeovavelell tEwhAietahvnavdleerinoyncseimsbeaatlnhl dseu.aEpcplceelcypttoroofrnennseergrisgaeytiuvthepleyah.n(odAlelhteofrrloenmsatfaelellyvl ,edlwoEweAncsawinnhkaeslnsdotohsweayny that into gain external energy.) At room temperature, most of the acceptor atoms get ionised leaving holes in the valence band. Thus at room temperature the density of holes in the valence band is predominantly due to impurity in the extrinsic semiconductor. The electron and hole concentration in a semiconductor in thermal equilibrium is given by nenh = ni2 (14.5) Though the above description is grossly approximate and hypothetical, it helps in understanding the difference between metals, insulators and semiconductors (extrinsic and intrinsic) in a simple manner. The difference in the resistivity of C, Si and Ge depends upon the energy gap between their conduction and valence bands. For C (diamond), Si and Ge, the energy gaps are 5.4 eV, 1.1 eV and 0.7 eV, respectively. Sn also is a group IV element but it is a metal because the energy gap in its case is 0 eV. FIGURE 14.9 Energy bands of (a) n-type semiconductor at T > 0K, (b) p-type semiconductor at T > 0K. Example 14.2 Suppose a pure Si crystal has 5 × 1028 atoms m–3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that ni =1.5 × 1016 m–3. Solution Note that thermally generated electrons (ni ~1016 m–3) are EXAMPLE 14.2 negligibly small as compared to those produced by doping. Therefore, ne n≈i2N, DT.he number of holes Since nenh = nh = (2.25 × 1032)/(5 ×1022) ~ 4.5 × 109 m–3 477 2020-21

Physics 14.5 p-n JUNCTION A p-n junction is the basic building block of many semiconductor devices like diodes, transistor,etc. A clear understanding of the junction behaviour is important to analyse the working of other semiconductor devices. We will now try to understand how a junction is formed and how the junction behaves under the influence of external applied voltage (also called bias). Formation and working of p-n junction diode 14.5.1 p-n junction formation http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding precisely a small quantity of pentavelent impurity, part of the p-Si wafer can be converted into n-Si. There are several processes by which a semiconductor can be formed. The wafer now contains p-region and n-region and a metallurgical junction between p-, and n- region. Two important processes occur during the formation of a p-n junction: diffusion and drift. We know that in an n-type semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons. During the formation of p-n junction, and due to the concentration gradient across p-, and n- sides, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This motion of charge carries gives rise to diffusion current across the junction. When an electron diffuses from n → p, it leaves behind an ionised donor on n-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from n → p, a layer of positive charge (or positive space-charge region) on n-side of the junction is developed. Similarly, when a hole diffuses from p → n due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the p-side of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its free charges (Fig. 14.10). The thickness of depletion region is of the order of one-tenth of a micrometre. Due to the positive space-charge region on n-side of the junction and negative space charge region on p-side of the junction, an electric field directed from positive charge towards negative charge develops. Due to this field, an electron on p-side of the junction moves to n-side and a hole on n- side of the junction moves to p-side. The motion of charge FIGURE 14.10 p-n junction carriers due to the electric field is called drift. Thus a drift current, which is opposite in direction to the diffusion 478 formation process. current (Fig. 14.10) starts. 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits Initially, diffusion current is large and drift current is small. FIGURE 14.11 (a) Diode under As the diffusion process continues, the space-charge regions equilibrium (V = 0), (b) Barrier on either side of the junction extend, thus increasing the electric potential under no bias. field strength and hence drift current. This process continues until the diffusion current equals the drift current. Thus a p-n junction is formed. In a p-n junction under equilibrium there is no net current. The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium exists. Figure 14.11 shows the p-n junction at equilibrium and the potential across the junction. The n-material has lost electrons, and p material has acquired electrons. The n material is thus positive relative to the p material. Since this potential tends to prevent the movement of electron from the n region into the p region, it is often called a barrier potential. Example 14.3 Can we take one slab of p-type semiconductor and EXAMPLE 14.3 physically join it to another n-type semiconductor to get p-n junction? Solution No! Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers. 14.6 SEMICONDUCTOR DIODE pn A semiconductor diode [Fig. 14.12(a)] is basically a FIGURE 14.12 (a) Semiconductor diode, p-n junction with metallic contacts provided at the (b) Symbol for p-n junction diode. ends for the application of an external voltage. It is a two terminal device. A p-n junction diode is symbolically represented as shown in Fig. 14.12(b). The direction of arrow indicates the conventional direction of current (when the diode is under forward bias). The equilibrium barrier potential can be altered by applying an external voltage V across the diode. The situation of p-n junction diode under equilibrium (without bias) is shown in Fig. 14.11(a) and (b). 14.6.1 p-n junction diode under forward bias 479 When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal [Fig. 14.13(a)], it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. (This is because the resistance of the depletion region – a region where there are no charges – is very high compared to the resistance of n-side and p-side.) The direction of the applied voltage (V ) is opposite to the 2020-21

Physics built-in potential bVa0.rrAiesr a result, the depletion layer width decreases and the height is reduced [Fig. 14.13(b)]. The effecIftitvheebaaprprileierdhevioglhtatguenidsesrmfoarlwl,atrhdebbiaasrriiser(Vp0o–teVn)t.ial will be reduced only slightly below the equilibrium value, and only a small number of carriers in the material—those that happen to be in the uppermost energy levels—will possess enough energy to cross the junction. So the current will be small. If we increase the applied voltage significantly, the barrier height will be reduced and more number of carriers will have the required energy. Thus the current increases. Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carries). Similarly, holes from p-side cross the junction and reach the n-side (where they are minority carries). This process under FIGURE 14.13 (a) p-n forward bias is known as minority carrier injection. At the junction diode under forward junction boundary, on each side, the minority carrier bias, (b) Barrier potential (1) without battery, (2) Low concentration increases significantly compared to the locations battery voltage, and (3) High far from the junction. voltage battery. Due to this concentration gradient, the injected electrons on p-side diffuse from the junction edge of p-side to the other end of p-side. Likewise, the injected holes on n-side diffuse from the junction edge of n-side to the other end of n-side (Fig. 14.14). This motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion. The magnitude of this current is usually in mA. 14.6.2 p-n junction diode under reverse bias FIGURE 14.14 Forward bias When an external voltage (V ) is applied across the diode such minority carrier injection. that n-side is positive and p-side is negative, it is said to be reverse biased [Fig.14.15(a)]. The applied voltage mostly drops across the depletion region. The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in the electric field. TThheisesffuepcptirveesbsaersrtiehrehfleoiwghotfuenledcetrrorenvsefrrsoembinas→isp(Va0n+dVho),le[Fsigfr.o1m4.p15→(bn)].. Thus, diffusion current, decreases enormously compared to the diode under forward bias. The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close to the junction, they will be swept to its majority zone. This drift of carriers gives rise to current. The drift current is of the order of a few µA. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the junction. The drift current is also there under forward bias but it is negligible (µA) when compared with current due to injected carriers which is usually in mA. The diode reverse current is not very much dependent on the applied 480 voltage. Even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction. The current 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction. The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage, known as ibnrceraekadsoeswsnhvaorlptlayg.eE(vVebnr ). Wslihgehnt iVnc=reVabsr,etihnethdieobdieasrevvoeltrasgeeccuaruresnest a large change in the current. If the reverse current is not limited by an external circuit below the rated value (specified by the manufacturer) the p-n junction will get destroyed. Once it exceeds the rated value, the diode gets destroyed due to overheating. This can happen even for the diode under forward bias, if the forward current exceeds the rated value. The circuit arrangement for studying the V-I characteristics of a diode, (i.e., the variation of current as a function of applied FIGURE 14.15 (a) Diode voltage) are shown in Fig. 14.16(a) and (b). The battery is connected under reverse bias, to the diode through a potentiometer (or reheostat) so that the (b) Barrier potential under applied voltage to the diode can be changed. For different values reverse bias. of voltages, the value of the current is noted. A graph between V and I is obtained as in Fig. 14.16(c). Note that in forward bias measurement, we use a milliammeter since the expected current is large (as explained in the earlier section) while a micrometer is used in reverse bias to measure the current. You can see in Fig. 14.16(c) that in forward FIGURE 14.16 Experimental circuit arrangement for studying V-I characteristics of a p-n junction diode (a) in forward bias, (b) in reverse bias. (c) Typical V-I 481 characteristics of a silicon diode. 2020-21

Physics bias, the current first increases very slowly, almost negligibly, till the voltage across the diode crosses a certain value. After the characteristic voltage, the diode current increases significantly (exponentially), even for a very small increase in the diode bias voltage. This voltage is called the threshold voltage or cut-in voltage (~0.2V for germanium diode and ~0.7 V for silicon diode). For the diode in reverse bias, the current is very small (~µA) and almost remains constant with change in bias. It is called reverse saturation current. However, for special cases, at very high reverse bias (break down voltage), the current suddenly increases. This special action of the diode is discussed later in Section 14.8. The general purpose diode are not used beyond the reverse saturation current region. The above discussion shows that the p-n junction diode primerly allows the flow of current only in one direction (forward bias). The forward bias resistance is low as compared to the reverse bias resistance. This property is used for rectification of ac voltages as discussed in the next section. For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage ∆V to a small change in current ∆I: rd = ∆V (14.6) ∆I Example 14.4 The V-I characteristic of a silicon diode is shown in the Fig. 14.17. Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = –10 V. FIGURE 14.17 Solution Considering the diode characteristics as a straight line EXAMPLE 14.4 between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law. (a) From the curve, at I = 20 mA, V = 0.8 V; I = 10 mA, V = 0.7 V (b) rFfrbo=m∆tVh/e∆cI u=rv0e.1aVt/V10= mA = 10 Ω µA, –10 V, I = –1 Therefore, rrb = 10 V/1µA= 1.0 × 107 Ω 482 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits 14.7 APPLICATION OF JUNCTION DIODE AS A RECTIFIER From the V-I characteristic of a junction diode we see that it allows current to pass only when it is forward biased. So if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier. If an alternating voltage is applied across a diode in series with a load, a pulsating voltage will appear across the load only during the half cycles of the ac input during which the diode is forward biased. Such rectifier circuit, as shown in Fig. 14.18, is called a half-wave rectifier. The secondary of a transformer supplies the desired ac voltage across terminals A and B. When the voltage at A is positive, the diode is forward biased and it conducts. When A is negative, the diode is reverse-biased and it does not conduct. The reverse saturation current of a diode is negligible and can be considered equal to zero for practical purposes. (The reverse breakdown voltage of the diode must be sufficiently higher than the peak ac voltage at the secondary of the transformer to protect the diode from reverse breakdown.) Therefore, in the positive half-cycle of ac there FIGURE 14.18 (a) Half-wave rectifier is a current through the load resistor RL and we circuit, (b) Input ac voltage and output get an output voltage, as shown in Fig. 14.18(b), voltage waveforms from the rectifier circuit. whereas there is no current in the negative half- cycle. In the next positive half-cycle, again we get the output voltage. Thus, the output voltage, though still varying, is restricted to only one direction and is said to be rectified. Since the rectified output of this circuit is only for half of the input ac wave it is called as half-wave rectifier. The circuit using two diodes, shown in Fig. 14.19(a), gives output rectified voltage corresponding to both the positive as well as negative half of the ac cycle. Hence, it is known as full-wave rectifier. Here the p-side of the two diodes are connected to the ends of the secondary of the transformer. The n-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer. So for a full-wave rectifier the secondary of the transformer is provided with a centre tapping and so it is called centre-tap transformer. As can be seen from Fig.14.19(c) the voltage rectified by each diode is only half the total secondary voltage. Each diode rectifies only for half the cycle, but the two do so for alternate cycles. Thus, the output between their common terminals and the centre- tap of the transformer becomes a full-wave rectifier output. (Note that there is another circuit of full wave rectifier which does not need a centre- 483 tap transformer but needs four diodes.) Suppose the input voltage to A 2020-21

Physics with respect to the centre tap at any instant is positive. It is clear that, at that instant, voltage at B being out of phase will be negative as shown in Fig.14.19(b). So, diode D1 gets forward biased and conducts (while D2 being reverse biased is not conducting). Hence, during this positive half cycle we get an output current (and a output voltage across the load resistor RL) as shown in Fig.14.19(c). In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D1 would not conduct but diode D2 would, giving an output current and output voltage (across RL) during the negative half cycle of the input ac. Thus, we get output voltage during both the positive as well as the negative half of the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the half- wave rectifier. The rectified voltage is in the form of pulses of the shape of half sinusoids. Though it is unidirectional it does not have a steady value. To get steady dc output from the pulsating voltage normally a capacitor is connected across the output FIGURE 14.19 (a) A Full-wave rectifier terminals (parallel to the load RL). One can circuit; (b) Input wave forms given to the also use an inductor in series with RL for the same purpose. Since these additional diode D1 at A and to the diode D2 at B; circuits appear to filter out the ac ripple (c) Output waveform across the and give a pure dc voltage, so they are called filters. load RL connected in the full-wave rectifier circuit. Now we shall discuss the role of capacitor in filtering. When the voltage across the capacitor is rising, it gets charged. If there is no external load, it remains charged to the peak voltage of the rectified output. When there is a load, it gets discharged through the load and the voltage across it begins to fall. In the next half-cycle of rectified output it again gets charged to the peak value (Fig. 14.20). The rate of fall of the voltage across the capacitor depends inversely upon the product of capacitance C and the effective resistance RL used in the circuit and is called the time constant. To make the time constant large value of C should be large. So capacitor input filters use large capacitors. The output voltage obtained by using capacitor input filter is nearer to the peak voltage of the rectified voltage. This type of filter is most widely 484 used in power supplies. 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits FIGURE 14.20 (a) A full-wave rectifier with capacitor filter, (b) Input and output voltage of rectifier in (a). 14.8 SPECIAL PURPOSE p-n JUNCTION DIODES In the section, we shall discuss some devices which are basically junction diodes but are developed for different applications. 14.8.1 Zener diode It is a special purpose semiconductor diode, named after its inventor C. Zener. It is designed to operate under reverse bias in the breakdown region and used as a voltage regulator. The symbol for Zener diode is shown in Fig. 14.21(a). Zener diode is fabricated by heavily doping both p-, and n- sides of the junction. Due to this, depletion region formed is very thin (<10–6 m) and the electric field of the junction is extremely high (~5×106 V/m) even for a small reverse bias voltage of about 5V. The I-V characteristics of a Zener diode is shown in Fig. 14.21(b). It is seen that when the applied reverse bias voltage(V ) reaches cthheanbgreeaikndtohwencvuorlrteanget.(VNzo)toef the Zener diode, there is a large that after the breakdown voltage Vz, a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant, even though current through the Zener diode varies over a wide range. This property of the Zener diode is used for regulating supply voltages so that they are constant. Let us understand how reverse current suddenly increases at the breakdown voltage. We know that reverse current is due to the flow of electrons (minority carriers) from p → n and holes from n → p. As the reverse bias voltage is increased, the electric field at the junction becomes significant. When the reverse bias voltage vVal=enVcze, then the electric field strength is high enough to pull electrons from the host atoms on the p-side which are accelerated to n-side. These electrons account for high current observed at the breakdown. The FIGURE 14.21 Zener diode, (a) symbol, (b) I-V emission of electrons from the host atoms due to the high characteristics. electric field is known as internal field emission or field ionisation. The electric field required for field ionisation is of 485 the order of 106 V/m. 2020-21

Physics Zener diode as a voltage regulator We know that when the ac input voltage of a rectifier fluctuates, its rectified output also fluctuates. To get a constant dc voltage from the dc unregulated output of a rectifier, we use a Zener diode. The circuit diagram of a voltage regulator using a Zener diode is shown in Fig. 14.22. The unregulated dc voltage (filtered output of a rectifier) is connected to the Zener diode through a series rIafenstdhisetZaiennncpeeurRt sdvsioouldtcaehgteahlisanotctrhieneacZsreeensa,esrtehsde.ioTcduheriisrserinentvcetrrhesraeosbueisgahstehRdes. voltage adcrroopssatchreosZsenResr without any change in the voltage diode. This is because in the breakdown region, Zener voltage remains constant even though the current through the Zener diode changes. Similarly, if the input voltage decreases, the current through Rs and Zener diode also decreases. The voltage dvoroltpagaecraocsrsosRss decreases without any change in the FIGURE 14.22 Zener diode as DC the Zener diode. Thus any increase/ voltage regulator decrease in the input voltage results in, increase/ change in voltagedaeccrroesassteheofZethneervdoilotdaeg.eTdhruospthaecrZoesnserRds iowditehaocutst any as a voltage regulator. We have to select the Zener diode according to the required output voltage and accordingly the series resistance Rs. Example 14.5 In a Zener regulated power supply a Zener diode with VthZe=u6n.0reVguilsautesdedinfporutreigsu1la0t.i0onV..TWhehalotasdhcouurlrdenbte is to be 4.0 mA and the value of series resistor RS? EXAMPLE 14.5 Solution The value of RS should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., IvRisZoS1=l=t5a204g0.eΩ0Vm. d/SAr(o2.o,4Tpah×seae1crt0iroe–ot3sa)slrAsecs=uRisr1Stro6ern7itosΩft.h11T5r0oh0.ue0Ωgnhie–saRra6Sep.sip0str,vo=atphlrue4iaer.te0eofo.f rVNcea.o,rt2beT4ohtnhimasrtAegss.ilisiTvgthehosert variation in the value of the resistor does not matter, what is important is that the current IZ should be sufficiently larger than IL. 14.8.2 Optoelectronic junction devices We have seen so far, how a semiconductor diode behaves under applied electrical inputs. In this section, we learn about semiconductor diodes in which carriers are generated by photons (photo-excitation). All these devices are called optoelectronic devices. We shall study the functioning of the following optoelectronic devices: (i) Photodiodes used for detecting optical signal (photodetectors). (ii) Light emitting diodes (LED) which convert electrical energy into light. 486 (iii) Photovoltaic devices which convert optical radiation into electricity (solar cells). 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits (i) Photodiode A Photodiode is again a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on the diode. It is operated under reverse bias. When the photodiode is illuminated with light (photons) with energy (hν) greater than t he e nergy gap (E ) of the semiconductor, then electron-hole gpairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach n-side and holes reach p-side. Electrons are collected on n-side and holes are collected on p-side giving rise to an emf. When an external load is connected, current flows. The magnitude of the photocurrent depends on the intensity of incident light (photocurrent is proportional to incident light intensity). It is easier to observe the change in the current with change in the light intensity, if a reverse bias is applied. Thus photodiode can be used as a photodetector to detect optical signals. The circuit diagram used for the measurement of I-V FIGURE 14.23 (a) An illuminated characteristics of a photodiode is shown in photodiode under reverse bias , (b) I-V characteristics of a photodiode for different Fig. 14.23(a) and a typical I-V characteristics in illumination intensity I4 > I3 > I2 > I1. Fig. 14.23(b). Example 14.6 The current in the forward bias is known to be more EXAMPLE 14.6 (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the photodiodes in reverse bias? Solution Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >> p). On illumination, let the excess electrons and holes generated be ∆n and ∆p, respectively: n′ = n + ∆n p′ = p + ∆p Here n′ and p′ are the electron and hole concentrations* at any particular illumination and n and p are carriers concentration when there is no illumination. Remember ∆n = ∆p and n >> p. Hence, the * Note that, to create an e-h pair, we spend some energy (photoexcitation, thermal 487 excitation, etc.). Therefore when an electron and hole recombine the energy is released in the form of light (radiative recombination) or heat (non-radiative recombination). It depends on semiconductor and the method of fabrication of the p-n junction. For the fabrication of LEDs, semiconductors like GaAs, GaAs- GaP are used in which radiative recombination dominates. 2020-21

EXAMPLE 14.6Physics 488 fractional change in the majority carriers (i.e., ∆n/n ) would be much less than that in the minority carriers (i.e., ∆p/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity. (ii) Light emitting diode It is a heavily doped p-n junction which under forward bias emits spontaneous radiation. The diode is encapsulated with a transparent cover so that emitted light can come out. When the diode is forward biased, electrons are sent from n → p (where they are minority carriers) and holes are sent from p → n (where they are minority carriers). At the junction boundary the concentration of minority carriers increases compared to the equilibrium concentration (i.e., when there is no bias). Thus at the junction boundary on either side of the junction, excess minority carriers are there which recombine with majority carriers near the junction. On recombination, the energy is released in the form of photons. Photons with energy equal to or slightly less than the band gap are emitted. When the forward current of the diode is small, the intensity of light emitted is small. As the forward current increases, intensity of light increases and reaches a maximum. Further increase in the forward current results in decrease of light intensity. LEDs are biased such that the light emitting efficiency is maximum. The V-I characteristics of a LED is similar to that of a Si junction diode. But the threshold voltages are much higher and slightly different for each colour. The reverse breakdown voltages of LEDs are very low, typically around 5V. So care should be taken that high reverse voltages do not appear across them. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV (spectral range of visible light is from about 0.4 µm to 0.7 µm, i.e., from about 3 eV to 1.8 eV). The compound semiconductor Gallium Arsenide – Phosphide (GaAs1–xPx) is used for making LEDs of different colours. GaAs0.6 P0.4 (Eg ~ 1.9 eV) is used for red LED. GaAs (Eg ~ 1.4 eV) is used for making infrared LED. These LEDs find extensive use in remote controls, burglar alarm systems, optical communication, etc. Extensive research is being done for developing white LEDs which can replace incandescent lamps. LEDs have the following advantages over conventional incandescent low power lamps: (i) Low operational voltage and less power. (ii) Fast action and no warm-up time required. (iii) The bandwidth of emitted light is 100 Å to 500 Å or in other words it is nearly (but not exactly) monochromatic. (iv) Long life and ruggedness. (v) Fast on-off switching capability. 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits (iii) Solar cell A solar cell is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. It works on the same principle (photovoltaic effect) as the photodiode, except that no external bias is applied and the junction area is kept much larger for solar radiation to be incident because we are interested in more power. A simple p-n junction solar cell is shown in Fig. 14.24. A p-Si wafer of about 300 µm is taken over which a thin layer (~0.3 µm) of n-Si is grown on one-side by diffusion process. The other side of p-Si is coated with a metal (back contact). On the top of n-Si layer, metal finger electrode (or metallic grid) is deposited. This acts as a front contact. The FIGURE 14.24 (a) Typical p-n junction metallic grid occupies only a very small fraction solar cell; (b) Cross-sectional view. of the cell area (<15%) so that light can be incident on the cell from the top. The generation of emf by a solar cell, when light falls on, it is due to the following three basic processes: generation, separation and collection— (i) generation of e-h pairs due to light e(wleicthtrohnνs>aEngd) close to the junction; (ii) separation of holes due to electric field of the depletion region. Electrons are swept to n-side and holes to p-side; (iii) the electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back contact. Thus p-side becomes positive and n-side becomes negative giving rise to photovoltage. When an external load is connected as shown in the Fig. 14.25(a) cahpahraocttoecruisrtriecns tofILafsloowlasr through the load. A typical I-V cell is shown in the Fig. 14.25(b). Note that the I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. This is because a solar cell does not draw current but supplies the same to the load. Semiconductors with band gap close to 1.5 eV are ideal materials for solar cell fabrication. Solar cells are e(mEVag)d,=ee1tw.c4.it3ThehsVee)m,imCicpdooTnredta(uEncgtt=ocr1rsi.t4lei5rkieaeVSf)o,ir(CEtuhg Ie=n1Sse.e1l2ee(cEVtig)o,=nG1oa.f0A4as material for solar cell fabrication are (i) band gap (~1.0 to 1.8 eV), (ii) high optical absorption (~104 cm–1), (iii) electrical conductivity, (iv) availability of the raw FIGURE 14.25 (a) A typical material, and (v) cost. Note that sunlight is not always illuminated p-n junction solar cell; required for a solar cell. Any light with photon energies (b) I-V characteristics of a solar cell. greater than the bandgap will do. Solar cells are used to power electronic devices in satellites and space vehicles and also as power supply to some calculators. Production of low-cost photovoltaic cells for large-scale solar energy is a topic 489 for research. 2020-21