Nuclear Physics

It exists at the centre of an atom, containing entire positive charge and almost whole of mass. The electron resolve around the neuleus to form an atom. The nucleus consists of protons (+ve charge) and neutrons. A proton has positive charge equal in magnitude to that of an electron (+ 1.6 × 10–19 C) and a mass equal to 1840 C) and a mass equal to 1840 times that of an electron. A neutron has no charge and mass is approximately equal to that of proton. PROPERTIES OF A NUCLEUS (1) Nuclear Mass : As we know that every nucleus contains protons and neutrons and so every nucleus has a define mass. However, since the mass of electron is negligible so atomic mass is roughly equal to nuclear mass. Atomic masses are measured in atomic mass unit (a.m.u.) defiined as  1u = 931.478 MeV/c2 and its energy equivalent is 931.48 MeV Thye number of protons is a nucleus of an atom is called as the atomic number (Z) of that atom. The number of protons plus neutrons (called as Nucleus) in a nucleus of an atom is called as mass number (A) of that atom. A partiicular set of nucleons forming an atom is called as nuclide. It is represented as ZXA . The nuclides having same number of protons (Z), but different number of nucleons (A) are called as isotopes. The nuclide having same number of nucleons (A), but different number of protons (Z) are called as isobars. The nuclide having same number of neutrons (A-Z) are called as isotones. (2) Nuclear Charge : Since nucleus contain +vely charged protons (charge = 1.67 × 10–19 C) and neutrons (neutral) so every nucleus has a net +ve charge. (3) Nuclear Radius : A rough estiimate of nuclear size suggests us that the radius of the nucleus of an atom having mass number ‘A’ is given by R = R0A1/3 Where R0 is a constant found to be equal to R0 = 1.4 × 10–15 m = 1.4 fm. (4) Nuclear Density : In spite of the fact that nuclear radius depends on mass number of the atom but nuclear density is independent of mass number because if neutrons are supposed to be of almost the same mass as that of protons then the total mass of a nucleus is proportional to A. If each nucleon are supposed to have a mass m then nuclear density is given by ρ= (5) mA 4 3 πR 0 A 3 = 3m 4πR 30 (Which is independent of A) Nuclear spin and magnetic moment Like orbial electrons in an atom, nucleons inside nucleus have well defined quantum states. Correspondingly they have angular momentum and hence a magnetic moment. Like electrons nucleons also have intrinsic angular momentum and ‘magnetic’ moment corresponding to their spin. 104 NUCLEAR PHYSICS NUCLEAR FORCES If only the electrostatic and gravitattional forces existed in the nucleus, then it would be impossible to have stable nuclei composed of protons and neutrons. The gravitational forces are much too small to hold the nucleons together compared to the electrostatic forces repelling the protons. Since stable atoms of neutrons and protons do exiist, there must be another attractive force acting within the nucleus. This force is called the nuclear force. Properties of nuclear force (1) They are charge independent. The nuclear force between two proton is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces. (2) They may be repulsive may be attractive (Repulsive at exceedingly small separation between two nucleons appreciably smaller than 10-13 cm. i.e. 10–15 m. (3) It is a short range force. Its radius of action is of the order of 10–13 cm. (4) The nuclear force is of saturation character. Each nucleon in nucleus interacts with a limited number of nucleons. (5) Nuclear force are much stronger than electromagnetic force or gravitational attrqactive forces. It is the strongest of all the forces. This is why it is called strong interaction. (6) Nuclear force is spin dependent. If two interacting nucleons are having parallel spins then nuclear force operative betwee them is comparatively stronger and if their spins are antiparallel, nuclear interaction is comparatively weaker. (7) Nuclear force is a non-central force. They can not be represented as directe along the st. line connecting the centers of the interacting nucleons. Its non central nature is due to the fact that it depends also on the orientation of the nucleon spins. MASS DEFECT It is observed that the mass of a nucleus is slightly less than the sum of the masses of constituent nucleons. Suppose a nucleus consists of ‘Z’ protons and ‘N’ neutrons. Mass of a proton, a neutron and the resulting nucleus are respectively mp, mn and M then mass defect of the nucleus is given by of the nucleus is given by m  Zm p  Nmn  M If A is the mass number of the nucleus m  [ Zm p ( A  Z )mn  M ] In terms of atomic masses we may also write mass defect as m  [ Zm (11 H ) Nmn  m ( zA X )] Where m (11 H ) = mass of one hydrogen atom. m ( zA X ) = mass of atom having atomic no. Z and mass no. A e.g., mass of 11 H = 1.00784 u mass of neutron = 1.00874 u Expected mass of deuterium = 2.01654 u but measured mass = 2.0141 u mass defect, m = 0.00244 u 104 NUCLEAR STABILIITY Figure shows pot of N vs Z for known nuclides. The stable nuclides are indicated by the black dots. Non-stable nuclides decay by emission of particles, or electromagnetic radiation, in a process called radioactivity. BINDING ENERGY To break a nucleus into its constituent nuclei some energy is required to be supplied. This energy is called Binding Energy of the given nucleus or the energy equivalent of the missing mass of a nucleus is called the binding energy of the nucleus. BE  ( m)c 2  [ Zmp  ( A  Z )mn  M ]c 2 BE = m(on amu) × 931 MeV Where m = mass defect Binding energy per nucleon is a measure of the stability of the nucleus. If there be n nucleons which is equal to A. Binding Energy B.E.  Nucleon A 106 NUCLEAR PHYSICS From the plot of B.E./ nucleons Vs mass number (A), we observe that : (1) Binding energy per nucleon has low value for both heavy and liight nuclei i.e. heavy as well as light nuclei, both are unstable. B.E./nucleons increases on an average and reaches a maximum of about 8.7 MeV for A = 50–80. For more heavy nuclei, B.E./nucleons decreases slowly as A increases. For the heaviest natural element U238 it drops to about 7.5 MeV. From above observation, it follows that nuclei in the region of atomic 50 – 80 are most stable. (2) The intermediate nuclei have large value of binding energy per nucleon so they are more stable. (3) Binding energy per nucleon increases rapidly upto mass number 20 but there are peaks corresponding to 4 2 16 He,12 6 C,8 O which indicates that these nuclei are more stable than neighbours. The reason is that they may be considered to passes magic numbers i.e. their mass number is divisible by 4 and these nuclei may have 42 He as their constituents. (4) The minimum value of the BE/Nucleon is in the case of deutron that is 1.11 MeV. (5) The maximum value of the BE/Nucleon is 8.79 Mev for the nuclei Ex. Using the following plot of BE/nucleon vs mass number, mention the condition for which the energy is absorbed or released for the reaction Z1 P A1 + Z2 Q A2 Z3 56 26 Fe which is therefore the most stable nucleus. P A3 BE/A E3 E1 Sol. Binding energy for reaction is (xEa + yEb) and that for product is zEc Case - I : if (A1E1 + A2E2) > A3E3 Energy is absorbed Case - II : if (A1E1 + A2E2) < A3E3 Energy is released Note (i) If we split a heavy neclues into two medium sized nucleii and total binding energy of new nuclei is greater then parent nuclei, then energy is released (Nuclear fission) (ii) If two nuclei of small mass number combine to form a single medium size nucleus for which biinding energy is greater than the constituent nuclei, then energy is released (Nuclear fusion) NUCLEAR COLLISIONS We can represent a nuclear collision or reaction by the following notation, which means X (a,b) Y a (bombarding particle)  X  Y +b (at rest) 106 We can apply : (i) Conservation of momentum (ii) Conservation of charge (iii) Conservation of mass–energy For any nuclear reaction a + K1 By mass energy conservation X  Y + b K2 K4 K3 (i) K1 + K2 + (m + mx)c2 = K3 + K4 + (mY + mb)c2 (ii) Energy released in any nuclear reaction or collision is called Q value of the reaction (iii) Q = (K3 + K4) – (K1 + K2) = KP –KR = (mR – mP)c2 (iv) (v) If Q is positive, energy is released and products are more stable in comparison to reactants. If Q is negative, energy is absorbed and products are less stable in comparison to reactants. Q = (B.E.)product – (B.E.)reactants Ex. Let us find the Q value of fusion reaction 4 He + 4He  8Be, if B.E. B.E. of He = X and of Be = Y  Q = 8Y – 8X A A Q value for  decay XA  Z–2YA–4 + 2He4 Z Momentum conservation, pY = p p2 K = 2 m 4 Q K    Q = K + KY ....(i) ...(ii) p2 4 K KY = 2 m A  4  A  4   4 K A  K A4 A4  K  A4 Q A For  decay A > 210 which means maximum part of released energy is associated with K.E. of . If Q is negative, the reaction is endoergic. The minimum amount of energy that a bombarding particle must have in order to initiate an endoergic reaction is called Threshold energy Eth, given by  m1   1  where m = mass of the projectile. Eth = –Q  1  m2  Eth = minimum kinetic energy of the projectile to initiate the nuclear reaction m2 = mass of the target Ex. Sol. How much energy must a bombarding proton possess to cause the reaction 3Li7 + 1H1  4Be7 + 0n1 (Mass of 3Li7 atom is 7.01600, mass of 1H1 atom is 1.0783, mass of 4Be7 atom is 7.01693) Since the mass of an atom includes the masses of the atomic electrons, the appropriate number of electron masses must be subtracted from the given values. Reactants : Total mass = (7.01600 – 3 me) + (1.0783 – 1 me) = 8.0943 – 4me Products : Total mass = (7.01693 –4me)+ 1.0087 = 8.02563 – 4me The energy is supplied as kinetic energy of the bombarding proton. The incident proton must have more than this energy because the system must possess some kinetic energy even after the reaction, so that momentum is conserved with momentum conservation taken into account, the minimum kinetic energy that the incident particle must possess can be found with the formula. 108 NUCLEAR PHYSICS where, Q = – [(8.02563 – 4me) – (8.0943 –4 me)] 931.5 MeV = – 63.96 MeV m 1   Eth = –  1   Q= –  1   (–63.96) = 73.1 MeV M 7 NUCLEAR FISSION In 1938 by Hahn and Strassmann. By attack of a particle splitting of a heavy nucleus (A > 230) into two or more lighter nuclei. In this process certain mass disappears which is obtained in the form of energy (enormous amount) A+p  excited nucleus  B + C + Q Hahn and Strassmann done the first fision of nucleus of U235). When U235 is bombarded by a neutron it splits into two fragments and 2 or 3 secondary neutrons and releases about 190 MeV ( 200 MeV) energy per fission (or from single nucleus) Fragments are uncertain but each time energy released is almost same. Possible reactions are U235 + 0n1  Ba + Kr + 30n1 + 200 MeV or U235 + 0n1  Xe + Sr + 20n1 + 200 MeV and many other reactions are possible. (i) The average number of secondary neutrons is 2.5. (ii) Nuclear fission can be explained by using "liquid drop model" also. (iii) The mass defect m is about 0.1% of mass of fissioned nucleus (iv) About 93% of released energy (Q) is appear in the form of kinetic energies of products and about 7% part in the form of  – rays. NUCLEAR CHAIN REACTION : The equation of fission of U235 is U235 + 0n1  Ba + Kr + 30n1 + Q. These three secondary neutrons produced in the reaction may cause of fission of three more U235 and give 9 neutrons, which in turn, may cause of nine more fission of U235 and so on. Thus a continuous 'Nuclear Chain reaction' would start. If there is no control on chain reaction then in a short time (10–6 sec.) a huge amount of energy will be released. (This is the principle of 'Atom bomb'). If chain is controlled then produced energy can be used for peaceful purposes. For example nuclear reactor (Based on fission) are generating electricity. NATURAL URANIUM : It is mixture of U235 (0.7%) and U238 (99.3%). U235 is easily fissionable, by slow neutron (or thermal neutrons) having K.E. of the order of 0.03 eV. But U 238 is fissionable with fast neutrons. Note : Chain reaction in natural uranium can't occur. To improve the quality, percentage of U235 is increased to 3%. The proposed uranium is called 'Enriched Uranium' (97% U238 and 3% U235) LOSSES OF SECONDARY NEUTRONS : Leakage of neutrons from the system : Due to their maximum K.E. some neutrons escape from the system. Absorption of neutrons by U238 : Which is not fissionable by these secondary neutrons. CRITICAL SIZE (OR MASS) : In order to sustain chain reaction in a sample of enriched uranium, it is required that the number of lost neutrons should be much smaller than the number of neutrons produced in a fission process. For it the size of uranium block should be equal or greater than a certain size called critical size. 108 REPRODUCTION FACTOR : (K) = rate of production of neutrons rate of loss of neutrons (i) If size of Uranium used is 'Critical' then K = 1 and the chain reaction will be steady or sustained (As in nuclear reaction) (ii) If size of Uranium used is 'Super critical' then K > 1 and chain reaction will accelerate resulting in a explosion (As in atom bomb) (iii) If size of Uranium used is 'Sub Critical' then K < 1 and chain reaction will retard and will stop. NUCLEAR REACTOR (K = 1) : Credit  To Enricho Fermi Construction : (i) Nuclear Fuel : Commonly used are U235 , Pu239. Pu239 is the best. Its critical size is less than critical size of U235. But Pu239 is not naturally available and U235 is used in most of the reactors. (ii) Moderator : Its function is to slow down the fast secondary neutrons. Because only slow neutrons can bring the fission of U235. The moderator should be light and it should not absorb the neutrons. Commonly, Heavy water (D2O, molecular weight 20 gm.) Graphite etc. are used. These are rich of protons. Neutrons collide with the protons and interchange their energy. Thus neutrons get slow down. (iiii) Control rods :They have the ability to capture the slow neutrons and can control the chain reaction at any stage. (iv) Coolant : A substance which absorb the produced heat and transfers it to water for further use. Generally coolant Boron and Cadmium are best absorber of neutrons. is water at high pressure FAST BREADER REACTORS The atomic reactor in which fresh fissionable fuel (Pu239) is produced along with energy. The amount of produced fuel (Pu239) is more than consumed fuel (U235) (i) Fuel : Natural Uranium. (ii) Process: During fission of U235, energy and secondary neutrons are produced. These secondary neutrons are absorbed by U238 and U239 is formed. This U239 converts into Pu239 after two beta decay. This Pu239 can be separated, its half life is 2400 years.  92 2  94Pu239 (best fuel of fission) U238 + 0n1  92U239  This Pu239 can be used in nuclear weapons because of its small critical size than U235. (iii) Moderator : Are not used in these reactors. (iv) Coolant : Liquid sodium NUCLEAR FUSION : It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus. A+B  C + Q (Energy) The product (C) is more stable then reactants (A and B) & mC < (ma + mb) and mass defect Energy released is m = [(ma + mb)– mc] amu E = m 931 MeV The total binding energy and binding energy per nucleon C both are more than of A and B. E = Ec – (Ea + Eb) 110 NUCLEAR PHYSICS Fusion of four hydrogen nuclei into helium nucleus : 4(1H1)  2He4 + 2 + 0 + 2 + 26 MeV  Energy released per fission >> Energy released per fusion.   2 00  ; 0 .8 5 MeV  << energy per nucleon in fusion Energy per nucleon in fission    235   24    4 ; 6 MeV    REQUIRED CONDITION FOR NUCLEAR FUSION (i) High temperature : Which provide kinetic energy to nuclei to overcome the repulsive electrostatic force between them. (ii) High Pressure (or density) : Which ensure frequent collision and increases the probability of fusion. The required temperature and pressure at earth (lab) are not possible. These condition exist in the sun and in many other stars. The source of energy in the sun is nuclear fusion, where hydrogen is in plasma state and there protons fuse to form helium nuclei. HYDROGEN BOMB It is based on nuclear fusion and produces more energy than an atom bomb. Ex. In a nuclear reactor, fission is produced in 1 g for U235 (235.0439) in 24 hours by slow neutrons (1.0087 u). Assume that 35Kr92 (91.8973 u) and 56Ba141 (140.9139 amu) are produced in all reactions and no energy is lost. (i) Write the complete reaction Sol. (ii) Calculate the total energy produced in kilowatt hour. Given 1u = 931 MeV. The nuclear fission reaction is 92U235 + 0n1  56 Ba141 + 36Kr92 + 30n1 Mass defect m= [(mu + mn) – (mBa + mKr + 3mn)] = 256.0526 – 235.8373 =0.2153 u Energy released Q = 0.2153 × 931 = 200 MeV. Number of atoms in 1 g = 6 .02  10 23 = 2.56 × 1021 2 35 Energy released in fission of 1 g of U235 is E = 200 × 2.56 × 1021 = 5.12 × 1023 MeV = 5.12 × 1023 × 1.6 × 10–13 = 8.2 × 1010 J = 8.2  10 10 kWh = 2.28 × 104 kWh 3.6  10 6 110 Ex. It is proposed to use the nuclear fusion reaction : 1H2 + 1H2  2He4 in a nuclear reactor of 200 MW rating. If the energy from above reaction is used with at 25% efficiency in the reactor, how many grams of deuterium will be needed per day. (Mass of 1H2 is 2.0141 u and mass of 2He4 is 4.0026 u) Sol. Energy released in the nuclear fusion is Q = mc2 = m(931) MeV (where m is in amu) Q = (2 × 2.0141 – 4.0026) × 931 MeV = 23.834 MeV = 23.834 × 106 eV Since efficiency of reactor is 25% So effective energy used = 25  2 3.8 34  10 6  1.6  10 19 J = 9.534 × 10–13 J 1 00 Since the two deuterium nuclei are involved in a fusion reaction, therefore, energy released per deuterium is 9 .53 4  10 1 3 . 2 For 200 MW power per day, number of deuterium nuclei required = 20 0  1 0 6  8 640 0 = 3.624 × 1025 9.53 4  10 23 2 Since 2g of deuterium constitute 6 × 1023 nuclei, therefore amount of deuterium required is = 2  3.624  10 25 = 120.83 g/day.. 6  10 23 112