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SF027 is defined as the central core of an atom that is positively charged and contains protons and neutrons. UNIT 15: NUCLEUS SF027 1 15.1 Nuclear Structure { A nucleus of an atom is made up of protons and neutrons that known as nucleons (is defined as the particles found inside the nucleus) nucleus as shown in figure 15.1a. Proton { Proton and neutron are characterised by the following properties in table 15.1a. Proton (p) Neutron Charge (C) Mass (kg) Electron +e (1.60 × 10 −19 ) Neutron (n) 0 (uncharged) 1.672 × 10 −27 1.675 × 10 −27 Table 15.1a { For a neutral atom : The number of protons = the number of electrons orbiting the inside the nucleus nucleus z This is because the magnitude of an electron charge equals to the magnitude of a proton charge but opposite sign. z SF027 2 SF027 { Nuclei are characterised by the number and type of nucleons they contain as shown in table 15.1b. Number Symbol Definition Atomic number Z The number of protons in a nucleus Neutron number N The number of neutrons in a nucleus Mass (nucleon) number A The number of nucleons in a nucleus Table 15.1b Relationship : { { Any nucleus of elements in the periodic table called a nuclide is characterised by its atomic number Z and its mass number A. The nuclide of an element is represented as Mass number Element X SF027 3 Atomic number { The number of protons Z is not necessary equal to the number of neutrons N. 23 e.g. : 11 Na ; 32 16 S ; 195 78 Pt Z = 11 N = A − Z = 12 { Since a nucleus can be modeled as tightly packed sphere where each sphere is a nucleon, thus the average radius of the nucleus is given by femtometre (fermi) fermi) 1 (15.1a) 1 fm = 1 × 10 −15 m 3 R = R0 A where { SF027 R0 : constant = 1.2 × 10 -15 m ( 1.2 fm) A : mass number Example 1 : Based on the periodic table of element, Write down the symbol of nuclide for following cases: a. Z=20 ; A=40 b. Z=17 ; A=35 (exercise) c. 50 nucleons ; 24 electrons d. 106 nucleons ; 48 protons (exercise) e. 214 nucleons ; 131 protons (exercise) 4 SF027 Solution: a. Given Z=20 A Z ; A=40 40 20 X Ca c. Given A=50 and Z=number of protons = number of electrons =24 A Z { X 50 24 Cr Example 2 : What is meant by the symbols below : 1 0 n ; 11 p ; −10 e State the mass number and sign of the charge for each entity above. Solution: 1 Neutron ; A=1 0 Charge : neutral (uncharged) n SF027 { 1 1 p 0 −1 e Proton ; A=1 Charge : positively charged Electron ; A=0 Charge : negatively charged 5 Example 3 : (exercise) Complete the table below : Element nuclide 1 1 9 4 14 7 16 8 23 11 59 27 31 16 133 55 238 92 Number of protons Number of neutrons Total charge in nucleus Number of electrons 11 12 11e 11 H Be N O Na Co S Cs U SF027 6 SF027 15.2 Isotope { { Definition – is defined as the nuclides/elements/atoms that have the same atomic number Z but different in mass number A. From the definition of isotope, thus the number of protons or electrons are equal but different in the number of neutrons N for two isotopes from the same element. For example : z Hydrogen isotopes : 1 1 2 1 3 1 z H : Z=1, A=2, N=1 H : Z=1, A=3, N=2 Oxygen isotopes : 16 8 17 8 18 8 proton (11p) H : Z=1, A=1, N=0 equal deuterium (12D) tritium ( 31T ) not equal O : Z=8, A=16, N=8 O : Z=8, A=17, N=9 O : Z=8, A=18, N=10 not equal equal SF027 7 15.3 Mass-Energy equivalent { { From the theory of relativity leads to the idea that mass is a form of energy. energy Mass and energy can be related by following relation : where (15.3a) E = mc 2 E : amount of energy m : rest mass c : speed of light in vacuum (3.00 × 10 8 m s -1 ) e.g. : The energy for 1 kg of substance is E = mc 2 E = (1)(3.00 × 10 8 ) 2 E = 9.00 × 10 16 J { Unit conversion of mass and energy : z The electron-volt (eV) { is a unit of energy. { is defined as the kinetic energy gained by an electron in being accelerated by a potential difference (voltage) of 1 volt. volt SF027 1 eV = 1.60 × 10 −19 J 1 MeV = 106 eV = 1.60 × 10 -13 J 8 SF027 z The atomic mass unit (u) { is a unit of mass. { is defined as exactly atom. { 1 the mass of a neutral carboncarbon-12 12 mass of 126C 1u = 12 1 u = 1.66 × 10 - 27 kg 1 atomic mass unit (u) can be converted into the unit of energy by using mass-energy equivalent equation (eq. 15.3a) : (j): E = mc 2 E = (1.66 × 10 −27 )(3.00 × 10 8 ) 2 E = 1.49 × 10 −10 J z in joule z in eV or MeV: 1 u = 1.49 × 10 −10 J 1.49 × 10 −10 = 931.5 × 10 6 eV −19 1.60 × 10 1 u = 931.5 × 106 eV E= or 1 u = 931.5 MeV SF027 9 15.4 Mass Defect and Binding Energy 15.4.1 Mass Defect { The mass of a nucleus (MA) is always less than the total mass of its constituent nucleons (Zmp+Nmn) where M A < (Zm p + Nmn ) with m p : mass of a proton mn : mass of a neutron { { { { SF027 Hence the difference in this mass is given by ∆m = (Zm p + Nmn ) − M A (15.4a) where ∆m is called mass defect. defect The mass defect is defined as the mass difference between the total mass of the constituent nucleons and the mass of a nucleus. nucleus The reduction in mass arises because the act of combining the nucleons to form the nucleus causes some of their mass to be released as energy. Any attempt to separate the nucleons would involve them being given this same amount of energy. This energy is called the binding energy of the nucleus. 10 SF027 15.4.2 Binding Energy { Definition - The binding energy of a nucleus is defined as the energy required to separate completely all the nucleons in the nucleus. nucleus { The binding energy of the nucleus is equal to the energy equivalent of the mass defect. Hence Binding energy in joule EB = (∆m )c 2 (15.4b) Speed of light in vacuum Mass defect in kg { Example 4 : Calculate the binding energy of a lithium nucleus 37 Li in MeV. (Given mass of neutron, mn=1.00867 u ; mass of proton, mp=1.00782 u ; speed of light in vacuum, c=3.00 x 108 m s-1and mass of lithium nucleus, MLi=7.01600 u) ( ) Solution: 7 3 Li Z = 3 and N = A − Z N =4 SF027 11 The mass defect of lithium nucleus is ∆m = (Zm p + Nmn ) − M Li ∆m = [(3 × 1.00782) + (4 × 1.00867 )] − 7.01600 ∆m = 0.04214 u The binding energy of lithium nucleus can be calculated by using two method : 2 1 u = 1.66 × 10 - 27 kg z Use equation 15.4b : E B = ∆m c ( ( ) in kg ∆m = (0.04214 ) 1.66 × 10 −27 ∆m = 6.995 × 10 −29 kg ( SF027 )( ) ) 2 EB = 6.995 × 10 −29 3.00 × 10 8 EB = 6.296 × 10 −12 J 1 MeV = 1.60 × 10 -13 J in MeV : 6.296 × 10 −12 EB = 1.60 × 10 −13 EB = 39.4 MeV 12 SF027 z Using unit conversion ( u⇒ MeV ) 1 u = 931.5 MeV EB = ∆m × 931.5 MeV in u { EB = (0.04214 )× 931.5 EB = 39.3 MeV Example 5 : 35 If the mass of chlorine-35 nucleus 17 Cl is 34.97 u, calculate its binding energy in joule. (Given mass of neutron, mn=1.009 u ; mass of proton, mp=1.007 u and speed of light , c=3.00 x 108 m s-1) Solution: ( 35 17 Cl ) Z = 17and N = A − Z N = 18 The mass defect of chlorince-35 nucleus is ∆m = (Zm p + Nmn ) − M Cl ∆m = [(17 × 1.007 ) + (18 × 1.009 )] − 34.97 ∆m = (0.311 u )× (1.66 × 10 −27 )kg 13 ∆m = 5.16 × 10 −28 kg SF027 The binding energy of chlorince-35 nucleus is EB = (∆m )c 2 ( )( EB = 5.16 × 10 −28 3.00 × 10 8 EB = 4.65 × 10 −11 J or ( ) ) 2 1 u = 1.49 × 10 −10 J EB = ∆m × 1.49 × 10 −10 J EB = (0.311 u )× 1.49 × 10 −10 J EB = 4.63 × 10 −11 J { SF027 ( ) Example 6 : (exercise) Calculate the binding energy in joule of a deuterium nucleus. The mass of a deuterium nucleus is 3.344275 x 10-27 kg. (Given mass of neutron, mn=1.674954 x 10-27 kg ; mass of proton, mp=1.672648 x 10-27 kg and speed of light , c=3.00 x 108 m s-1) Ans. : 2.99 x 10-13 J 14 SF027 15.5 Nucleus Stability { { { { Since the nucleus is viewed as a closed packed of nucleons, thus its stability depends only on the forces exist inside it. The forces involve inside the nucleus are z repulsive electrostatic (Coulomb) forces between protons and z attractive forces that bind all nucleons together in the nucleus. These attractive force is called nuclear force and is responsible for nucleus stability. The general properties of the nuclear force are summarized as follow : z The nuclear force is attractive and is the strongest force in nature. z It is a short range force . It means that a nucleon is attracted only to its nearest neighbours in the nucleus. z It does not depend on charge; neutrons as well as protons are bound and the binding is same for both. e.g. : protonproton-proton (p (p-p) The magnitude of neutronneutron-neutron (n (n-n) nuclear forces are protonproton-neutron (p (p-n) same. same. SF027 15 The nuclear force depends on the binding energy per nucleon. Note that a nucleus is stable if the nuclear force greater than the Coulomb force and vice versa. The binding energy per nucleon of a nucleus is a measure of the nucleus stability where z { { Binding energy per nucleon = { Example 7 : Why is the uranium-238 nucleus 238 less stable than carbon92 U 12 nucleus 12 ? Give an explanation by referring to the C 6 repulsive coulomb force and the binding energy per nucleon. (Given mass of neutron, mn=1.009 u ; mass of proton, mp=1.008 u ; mass of carbon-12 nucleus, MC=12.000 u ; mass of uranium-238 nucleus, MU=238.051 u and 1 u =931.5 MeV) ( ) SF027 Binding energy ( EB ) Nucleon number( A) ( ) 16 SF027 Solution: z From the aspect of repulsive coulomb force : { Uranium-238 nucleus has 92 protons but the carbon-12 nucleus has only 6 protons. { Therefore the coulomb force inside uranium-238 nucleus is 92 or 15.3 times the coulomb force inside carbon-12 6 nucleus. z From the aspect of binding energy per nucleon: 12 Z = 6 and N = 6 { Carbon-12 : 6C The mass defect : ∆m = Zm p + Nmn − M C ( ∆m = 0.102 u ) The binding energy per nucleon: (∆m × 931.5 ) MeV EB = 12 A C EB = 7.92 MeV / nucleon A C SF027 17 { 238 Uranium-238 : 92 U The mass defect : ∆m Z = 92 and N = 146 = (Zm p + Nmn ) − M U ∆m = 1.999 u The binding energy per nucleon: (∆m × 931.5 ) MeV EB = 238 A U EB = 7.82 MeV / nucleon A U { z SF027 From the value of binding energy per nucleon for both nuclei, we obtain that EB E < B A U A C Since the binding energy of uranium-238 nucleus less than the binding energy of carbon-12 and the coulomb force inside uranium-238 nucleus greater than the coulomb force inside carbon-12 nucleus therefore uranium-238 nucleus less stable than carbon-12 nucleus . 18 SF027 { Figure 15.5a shows a graph of the binding energy per nucleon as a function of mass number A. Binding energy per nucleon ((MeV MeV/nucleon) /nucleon) Greatest stability Fig. 15.5a SF027 Mass number A From the graph : { The value of EB/A rises rapidly from 1 MeV/nucleon to 8 MeV/nucleon with increasing mass number A for light nuclei. { For the nuclei with A between 50 and 80, the value of EB/A ranges between 8.0 and 8.9 Mev/nucleon. The nuclei in these range are very stable. The maximum value of the curve occurs in the vicinity of nickel, which has the most stable nucleus. { For A > 62, the values of EB/A decreases slowly, indicating that the nucleons are on average less tightly bound. { For heavy nuclei with A between 200 to 240, the binding energy is between 7.5 and 8.0 MeV/nucleon. These nuclei are unstable and radioactive. Example 8 : (exercise) 20 The mass of neon-20 nucleus 10 Ne is 19.9924 u. Calculate the binding energy per nucleon of neon-20 nucleus in joule per nucleon. (Given mass of neutron, mn=1.009 u ; mass of proton, mp=1.008 u ; 1 u = 1.66 x 10-27 kg 20 Ans. : 1.33 x 10-12 J/nucleon z { SF027 19 ( ) SF027 15.6 Liquid Drop Model { { { { The liquid drop model is one of the successful models that permits us to correlate many facts about nuclear masses and binding energies. The model is proposed by Niels Bohr and later expanded on by C.F. Von Weiszacker in 1935. It is used to estimate the total binding energy of a nucleus by driving semisemi-empirical mass (binding energy) formula. formula The liquid drop model based on assumptions below : z A nucleus likes a drop of incompressible liquid with uniform density and spherical shaped. z The force between the nucleons does not depend on the spin and charge of the nucleon. Thus the force between two nucleons either (n-n), (n-p) or (p-p) is the same. z The attractive nuclear force which binds the nucleons is a short range force. It means a nucleon can attract only a few of its nearest neighbours. SF027 21 { Figure 15.6a shows the diagram of liquid drop model. nucleus proton Fig. 15.6a { nuclear force There are five major effects (factors) influence the binding energy of the nucleus in the liquid drop model : z The volume effect (factor), V. { The binding energy of a nucleus is proportional to mass number A and therefore proportional to the nucleus volume. { The contribution of this effect to the binding energy of the entire nucleus is given by where SF027 neutron EB1 = C1 A (15.6a) C1 : constant of volume effect (factor) A : mass number 22 SF027 z The surface effect (factor). { The nucleons on the surface of a nucleus have fewer near neighbours than those in the interior of the nucleus (Figure 15.6a), surface nucleons reduce the binding energy by an amount to their number. { Since the number of surface nucleons is proportional to the surface area 4πR2 of the nucleus and R2 ∝ A2/3, the surface term can be expressed as Negative sign means the decreasing in binding energy of the nucleus z 2 EB 2 = −C2 A 3 where C2 : constant of surface effect (factor) The Coulomb repulsion effect (factor). { { Every one of the Z protons repels every one of the (Z-1) other protons in the nucleus. Therefore The total repulsive Z(ZZ(Z-1) electric potential 1 1 or energy 1 ∝ ∝ Nucleus radius SF027 { R A3 23 This repulsive energy decreases the binding energy, so this term is negative and given by EB 3 = − C3 Z (Z − 1) A z (15.6b) 1 3 (15.6c) where C3 : constant of Coulomb repulsion effect (factor) The symmetry effect (factor). { To be in a stable, low energy state, the nucleus must have a balance between the energies associated with the neutrons and with the protons. z { For stable light nuclei (small A ) ⇒ N ≈ Z Any large asymmetry between N and Z for light nuclei reduces the binding energy and makes the nucleus less stable. z For stable heavy nuclei (larger A) ⇒ N >Z This effect can be described by a binding energy term 2 below : C4 ( A − 2 Z ) (15.6d) A where C4 : constant of symmetry effect (factor) EB 4 = − SF027 24 SF027 z The pairing effect (factor). { The nuclear force favours pairing of protons and of neutrons. { { The binding energy terms positive when both Z and N are even and negative when Z and N are odd. The pairing effect can be written as C (15.6e) EB 5 = ± 54 3 A where C5 : constant of pairing effect (factor) C Z & N are even + 54 3 A EB 5 = 0 C − 54 3 A SF027 { A is odd Z & N are odd 25 As a result, the total estimated binding energy EB is the sum of these five term : 2 C3 Z (Z − 1) C4 ( A − 2 Z ) C5 EB = C1 A − C2 A − − ± 4 1 A 3 A3 A 2 3 (15.6e) where the values of the constants are C1 = 15.75 MeV C2 = 17 .80 MeV C3 = 0.710 MeV C4 = 23.69 MeV C5 = 39.00 MeV { determined from the experimental results The mass of a nucleus for any neutral atom, MA is given by M A = (Zm p + Nmn ) − EB c2 (15.6f) Mass defect (∆m) SemiSemi-empirical mass formula SF027 26 SF027 { Example 9 : Consider the nuclide 37 Li in example 4. Calculate a. the five terms in the binding energy and the total estimated binding energy. b. its neutral atomic mass using the semi-empirical mass formula. (Given mn=1.00867 u ; mp=1.00782 u ; c=3.00 x 108 m s-1; C1=15.75 MeV ; C2=17.80 MeV; C3=0.710 MeV; C4=23.69 MeV; C5=39.00 MeV) Solution: From the nuclide notation : Z=3, A=7 and N=4 a. The five terms are ( ) EB1 = C1 A SF027 2 3 EB1 = 110.25 MeV EB 2 = −C2 A EB 2 = −65.14 MeV C Z (Z − 1) EB 3 = −2.23 MeV EB 3 = − 3 1 A3 2 C4 ( A − 2 Z ) EB 4 = −3.38 MeV EB 4 = − A 27 A is odd EB 5 = 0 The total estimated binding energy is EB = (110.25 − 65.14 − 2.23 − 3.38 ) MeV EB = 39.5 MeV This value is 0.51% greater than the value in example 4 which is 39.3 MeV. b. By applying the semi-empirical mass formula, thus M A = (Zm p + Nmn ) − EB c2 39.5 MeV M A = [(3)(1.00782) + (4 )(1.00867 )] − 931.5 MeV / u M A = 7.01574 u This value is 3.77 x 10-3 % less than the value in example 4 which is 7.01600 u. SF027 28 SF027 15.7 Bainbridge Mass Spectrometer { { Mass spectrometer is a device that detect the presence of isotopes and determines the mass of the isotope from known mass of the common or stable isotope. Figure 15.7a shows a schematic diagram of a Bainbridge mass spectrometer. Ion source S1 Plate P1 r E - Evacuated chamber SF027 r B Photographic plate × × × × × × × × × × × ×m×1 ×m ×2× × × × r1 × ×r × × × × × × × × ×r× × ×××××× × × × × × × × ×2× × × × × × × × ×B× × × × × × × × × × × × × × × × × × × ×2 × × × × × × × × × × × × × × × × × × ×× Fig. 15.7a { 29 Working principle: z Ions from an ion source such as a discharge tube are narrowed to a fine beam by the slits S1 and S2. z The ions beam then passes through a velocity selector (plates P1 and P2) which uses a uniform magnetic field B1 and a uniform electric field E that are perpendicular to each other. z The beam with selected velocity v passes through the velocity selector without deflection and emerge from the slit S3. Hence, the force on an ion due to the magnetic field B1 and the electric field E are equal in magnitude but opposite in direction (Figure 15.7b). The selected velocity v is given by Plate P1 SF027 Ions beam - ××+ S2 + Plate P2 - ××+ ×× + + - ××+ 1 - ××+ S3 −× ×× × ×× −r + − F×E × × × ×× −× ×× r v Fig. 15.7b Plate P2 × + × r+ ×FB+ × × + FB = FE B1qv sin 90 o = qE v= Using Fleming’ Fleming’s left hand rule. E B1 (15.7a) 30 SF027 z The ions beam emerging from the slit S3 enter an evacuated chamber of uniform magnetic field B2 which is perpendicular to the selected velocity v. The force due to the magnetic field B2 causes an ion to move in a semicircle path of radius r given by FB = Fc 2 mv B2 qv sin 90 o = r E mv and v = r= B1 B2 q mE r= (15.7b) B1 B2 q z Since the magnetic fields B1 and B2 and the electric field E are constants and every ion entering the spectrometer contains the same amount of charge q, therefore r = km r∝m with k= E : constant B1 B2 q SF027 31 z If ions of masses m1 and m2 strike the photographic plate with radii r1 and r2 respectively as shown in figure 15.7a then m1 r1 = m2 r2 { (15.7c) Example 10: A beam of singly charged ions of isotopes Ne-20 and Ne-22 travels straight through the velocity selector of a Bainbridge mass spectrometer. The mutually perpendicular electric and magnetic fields in the velocity selector are 0.4 MV m-1 and 0.6 T respectively. These ions then enter a chamber of uniform magnetic flux density 0.8 T. Calculate a. the selected velocity of the ions. b. the separation between two isotopes on the photographic plate. (Given the mass of Ne-20 = 3.32 x 10-26 kg; mass of Ne-22 = 3.65 x 10-26 kg and charge of the beam is 1.60 x 10-19 C) Solution: m1=3.32x10-26 kg, m2=3.65x10-26 kg, B1=0.6 T E=0.4x10-6 V m-1, q=1.60x10-19 C, B2=0.8 T SF027 32 SF027 a. The selected velocity of the ions is E B1 v = 6.67 × 10 5 m s −1 v= b. The radius of the circular path made by isotope Ne-20 is Em1 B1 B2 q r1 = 0.173 m The radius of the circular path made by isotope Ne-22 is Em2 r2 = B1 B2 q r2 = 0.190 m r1 = Therefore the separation between them is ∆r = r2 − r1 ∆r = 0.190 − 0.173 ∆r = 0.017 m SF027 33 THE END… Next Unit… UNIT 16 : Nuclear Reaction SF027 34