is defined as the central core of an atom that is positively charged and contains protons and neutrons. UNIT 5: NUCLUS SF07 Neutron lectron 5. Nuclear Structure nucleus of an atom is made up of protons and neutrons that known as nucleons (is defined as the particles found inside the nucleus) as shown in figure 5.a. Proton Charge (C)( Mass (kg( kg) Proton and neutron are characterised by the following properties in table 5.a. Proton (p)( Neutron (n)( +e 0 9 (. 60 0 ) (uncharged). 67 0 7 Table 5.a. 675 0 7 For a neutral atom : The number of protons = the number of electrons orbiting the inside the nucleus nucleus This is because the magnitude of an electron charge equals to the magnitude of a proton charge but opposite sign. SF07
Nuclei are characterised by the number and type of nucleons they contain as shown in table 5.b. Number Symbol Definition tomic number Z The number of protons in a nucleus Neutron number N The number of neutrons in a nucleus Mass (nucleon) The number of nucleons in a nucleus number Table 5.b Relationship : ny nucleus of elements in the periodic table called a nuclide is characterised by its atomic number Z and its mass number. The nuclide of an element is represented as Mass number lement X tomic number SF07 The number of protons Z is not necessary equal to the number of neutrons N. 95 e.g. : Na ; 6 S ; 78 Pt Z = N = Z = 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) (5.a) 5 R = R 0-5 where R0 : constant =. 0 m (. fm) : mass number fm = 0 xample : ased on the periodic table of element, Write down the symbol of nuclide for following cases: a. Z=0 ; =40 b. Z=7 ; =5 (exercise) c. 50 nucleons ; 4 electrons d. 06 nucleons ; 48 protons (exercise) e. 4 nucleons ; protons (exercise) SF07 4 m
Solution: a. Given Z=0 ; =40 40 X Ca Z 0 c. Given =50 and Z=number of protons = number of electrons =4 Z X 50 4Cr xample : What is meant by the symbols below : 0 n ; p ; 0 e State the mass number and sign of the charge for each entity above. Solution: 0 n Neutron ; = Charge : neutral (uncharged) p 0 e = Proton ; = Charge : positively charged =0 lectron ; =0 Charge : negatively charged SF07 5 xample : (exercise) Complete the table below : lement nuclide H 9 4 e 4 7 N 6 8O Na 59 7 Co 6 S 55Cs 8 9U Number of protons Number of neutrons Total charge in nucleus Number of electrons e SF07 6
5. Isotope Definition is defined as the nuclides/elements/atoms that have the same atomic number Z but different in mass number. 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 : Hydrogen isotopes : H : Z=, =, N=0 proton ( p) H : Z=, =, N= deuterium ( D) H : Z=, =, N= tritium ( T ) Oxygen isotopes : 6 8O 7 8O 8 8O equal : Z=8, =6, N=8 : Z=8, =7, N=9 : Z=8, =8, N=0 equal not equal not equal SF07 7 5. Mass-nergy equivalent From the theory of relativity leads to the idea that mass is a form of energy. Mass and energy can be related by following relation : where e.g. : The energy for kg of substance is Unit conversion of mass and energy : 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 volt. = mc : amount of m: rest mass : speed of (5.a) energy c light in vacuum (.00 0 8 m s - ) = mc 8 = ( )(. 00 0 ) 6 = 9. 00 0 J 9 ev =. 60 0 J MeV = 0 ev =.60 0 6 - SF07 8 J
The atomic mass unit (u) is a unit of mass. is defined as exactly the mass of a neutral carbon- atom. atomic mass unit (u) can be converted into the unit of energy by using mass-energy equivalent equation (eq. 5.a) : in joule (j): mass of u = u =.66 0 SF07 9 6-7 C kg = mc 7 8 = (. 66 0 )(. 00 0 ) 0 =. 49 0 J u =. 49 0 0 in ev or MeV:. 49 0 6 = = 9. 5 0 ev 9. 60 0 6 u = 9. 5 0 ev or 0 u = 9. 5 MeV J 5.4 Mass Defect and inding nergy 5.4. Mass Defect The mass of a nucleus (M ) is always less than the total mass of its constituent nucleons (Zm p +Nm n ) where M < ( Zmp + Nmn ) with m p : mass of a proton m n : mass of a neutron Hence the difference in this mass is given by m ( Zmp + Nmn ) M = (5.4a) where m is called mass defect. The mass defect is defined as the mass difference between the total mass of the constituent nucleons and the mass of a 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. ny 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. SF07 0
5.4. inding nergy Definition - The binding energy of a nucleus is defined as the energy required to separate completely all the nucleons in the nucleus. The binding energy of the nucleus is equal to the energy equivalent of the mass defect. Hence inding energy in joule ( ) c m = (5.4b) Speed of light in vacuum Mass defect in kg xample 4 : Calculate the binding energy of a lithium nucleus ( 7 in MeV. Li) (Given mass of neutron, m n =.00867 u ; mass of proton, m p =.0078 u ; speed of light in vacuum, c=.00 x 0 8 m s - and mass of lithium nucleus, M Li =7.0600 u) Solution: 7 Li Z = and N = Z N = 4 SF07 The mass defect of lithium nucleus is m = Zmp + Nmn M m = (. 0078) + ( 4. 00867) 7. m = 0. 044 u The binding energy of lithium nucleus can be calculated by using two method : Use equation 5.4b : = ( m) c -7 u =.66 0 in kg 7 in MeV : ( ) Li [ ] 0600 ( 0. 044)( 66 0 ) m =. 9 m = 6. 995 0 kg 9 8 ( )( ) = 6. 995 0. 00 0 = 6. 96 0 J - 6. 96 0 MeV =.60 0 J =. 60 0 = 9. 4 MeV SF07 kg
Using unit conversion ( u MeV ) = m 9. 5 MeV in u ( 0. 044) 9 5 =. = 9. MeV xample 5 : If the mass of chlorine-5 nucleus ( 5 7 Cl) is 4.97 u, calculate its binding energy in joule. (Given mass of neutron, m n =.009 u ; mass of proton, m p =.007 u and speed of light, c=.00 x 0 8 m s - ) Solution: 5 7 Cl Z = 7 and The mass defect of chlorince-5 nucleus is ( ) Cl = 9. 5 MeV SF07 u N = Z N = 8 m = Zmp + Nmn M m =. m =. 8 m = 5. 6 0 kg [( 7. 007) + ( 8. 009) ] 4 97 7 ( 0. u) ( 66 0 )kg The binding energy of chlorince-5 nucleus is ( ) = m c 8 8 ( )( ) = 5. 6 0. 00 0 = 4. 65 0 J u =. 49 0 or 0 (. 49 0 ) J = m = 0. u. 49 0 = 4. 6 0 J 0 ( ) ( )J xample 6 : (exercise) Calculate the binding energy in joule of a deuterium nucleus. The mass of a deuterium nucleus is.4475 x 0-7 kg. (Given mass of neutron, m n =.674954 x 0-7 kg ; mass of proton, m p =.67648 x 0-7 kg and speed of light, c=.00 x 0 8 m s - ) ns. :.99 x 0 - J 0 J SF07 4
5.5 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 repulsive electrostatic (Coulomb) forces between protons and 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 : The nuclear force is attractive and is the strongest force in nature. It is a short range force. It means that a nucleon is attracted only to its nearest neighbours in the nucleus. It does not depend on charge; neutrons as well as protons are bound and the binding is same for both. e.g. : proton-proton (p-p)( The magnitude of neutron-neutron (n-n)( nuclear forces are proton-neutron neutron (p-n)( same. SF07 5 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 inding energy ( ) inding energy per nucleon = Nucleon number( ) xample 7 : Why is the uranium-8 nucleus ( less stable than carbon- nucleus ( ) 8 9U )? Give an explanation by referring to the 6 C repulsive coulomb force and the binding energy per nucleon. (Given mass of neutron, m n =.009 u ; mass of proton, m p =.008 u ; mass of carbon- nucleus, M C =.000 u ; mass of uranium-8 nucleus, M U =8.05 u and u =9.5 MeV) SF07 6
Solution: From the aspect of repulsive coulomb force : Uranium-8 nucleus has 9 protons but the carbon- nucleus has only 6 protons. Therefore the coulomb force inside uranium-8 nucleus is 9 or 5. times the coulomb force inside carbon- 6 nucleus. From the aspect of binding energy per nucleon: Carbon- : C Z = 6 and N = 6 6 The mass defect : m = ( Zmp + Nmn ) M C m = 0. 0 u The binding energy per nucleon: C C = ( m 9. 5) SF07 7 MeV = 7. 9 MeV / nucleon 8 Uranium-8 : 9 The mass defect : U Z = 9 m = Zmp + Nm m =. 999 u The binding energy per nucleon: and N = 46 ( n ) MU ( m 9. 5) MeV = U 8 = 7. 8 MeV / nucleon U From the value of binding energy per nucleon for both nuclei, we obtain that U < Since the binding energy of uranium-8 nucleus less than the binding energy of carbon- and the coulomb force inside uranium-8 nucleus greater than the coulomb force inside carbon- nucleus therefore uranium-8 nucleus less stable than carbon- nucleus. C SF07 8
Figure 5.5a shows a graph of the binding energy per nucleon as a function of mass number. Greatest stability inding energy per nucleon ( (MeV MeV/nucleon) Fig. 5.5a SF07 9 Mass number From the graph : The value of / rises rapidly from MeV/nucleon to 8 MeV/nucleon with increasing mass number for light nuclei. For the nuclei with between 50 and 80, the value of / 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 > 6, the values of / decreases slowly, indicating that the nucleons are on average less tightly bound. For heavy nuclei with between 00 to 40, the binding energy is between 7.5 and 8.0 MeV/nucleon. These nuclei are unstable and radioactive. xample 8 : (exercise) The mass of neon-0 nucleus ( 0 is 9.994 u. Calculate the 0 Ne) binding energy per nucleon of neon-0 nucleus in joule per nucleon. (Given mass of neutron, m n =.009 u ; mass of proton, m p =.008 u ; u =.66 x 0-7 kg ns. :. x 0 - J/nucleon SF07 0
5.6 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 ohr and later expanded on by C.F. Von Weiszacker in 95. It is used to estimate the total binding energy of a nucleus by driving semi-empirical empirical mass (binding energy) formula. The liquid drop model based on assumptions below : nucleus likes a drop of incompressible liquid with uniform density and spherical shaped. 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. 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. SF07 Figure 5.6a shows the diagram of liquid drop model. nucleus proton Fig. 5.6a neutron nuclear force There are five major effects (factors) influence the binding energy of the nucleus in the liquid drop model : The volume effect (factor), V. The binding energy of a nucleus is proportional to mass number and therefore proportional to the nucleus volume. The contribution of this effect to the binding energy of the entire nucleus is given by = C (5.6a) where C : constant of volume effect (factor) SF07 : mass number
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 5.6a), surface nucleons reduce the binding energy by an amount to their number. Since the number of surface nucleons is proportional to the surface area 4πR of the nucleus and R /, the surface term can be expressed as Negative sign means the decreasing in binding = C (5.6b) energy of the nucleus where C : constant of surface effect (factor) The Coulomb repulsion effect (factor). very one of the Z protons repels every one of the (Z-) other protons in the nucleus. Therefore The total repulsive Z(Z-) electric potential energy SF07 Nucleus radius or R This repulsive energy decreases the binding energy, so this term is negative and given by CZ( Z ) = (5.6c) where C : 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. For stable light nuclei (small ) N Z ny large asymmetry between N and Z for light nuclei reduces the binding energy and makes the nucleus less stable. For stable heavy nuclei (larger ) N >Z This effect can be described by a binding energy term below : C4 ( Z ) = (5.6d) 4 : constant of symmetry effect (factor) SF07 4 where C 4
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 where C 5 C5 5 = ± (5.6e) 4 : constant of pairing effect (factor) C5 + Z & N are even 4 5 = 0 is odd C Z & N are odd 5 4 SF07 5 s a result, the total estimated binding energy is the sum of these five term : = C C CZ ( Z ) C ( Z ) where the values of the constants are C = 5. 75 MeV C = 7.80 MeV C = 0. 70 MeV C 4 =.69 MeV C 5 = 9.00 MeV C ± The mass of a nucleus for any neutral atom, M is given by M ( Zmp + Nmn ) 4 determined from the experimental results 5 4 = (5.6f) c Mass defect ( m)( (5.6e) Semi-empirical empirical mass formula SF07 6
xample 9 : Consider the nuclide ( 7 in example 4. Calculate Li) 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 m n =.00867 u ; m p =.0078 u ; c=.00 x 0 8 m s - ; C =5.75 MeV ; C =7.80 MeV; C =0.70 MeV; C 4 =.69 MeV; C 5 =9.00 MeV) Solution: From the nuclide notation : Z=, =7 and N=4 a. The five terms are = C = 0. 5 MeV = C = 65. 4 MeV CZ( Z ) = =. MeV 4 5 C4 ( Z ) = 4 =. 8 MeV 0 is odd SF07 7 = is odd The total estimated binding energy is = 0 5 65. 4. = 9. 5 MeV This value is 0.5% greater than the value in example 4 which is 9. MeV. b. y applying the semi-empirical mass formula, thus ( Zmp + Nmn ) M = c 9. 5 MeV M = ( )(. 0078) + ( 4)(. 00867) 9. 5 MeV / u = 7. 0574 u M (.. 8) MeV [ ] This value is.77 x 0 - % less than the value in example 4 which is 7.0600 u. SF07 8
5.7 ainbridge 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 5.7a shows a schematic diagram of a ainbridge mass spectrometer. Ion source Plate P vacuated chamber - S S Plate P --- +++ - + + r -- ++ S m m r r r r SF07 9 Fig. 5.7a Ions beam Photographic plate Working principle: Ions from an ion source such as a discharge tube are narrowed to a fine beam by the slits S and S. The ions beam then passes through a velocity selector (plates P and P ) which uses a uniform magnetic field and a uniform electric field that are perpendicular to each other. The beam with selected velocity v passes through the velocity selector without deflection and emerge from the slit S. Hence, the force on an ion due to the magnetic field and the electric field are equal in magnitude but opposite in direction (Figure 5.7b). The selected velocity v is given by Plate P Plate P F = F o + F r F r v r + + + + qvsin 90 = q v = SF07 0 Fig. 5.7b Using Fleming s left hand rule. (5.7a)
The ions beam emerging from the slit S enter an evacuated chamber of uniform magnetic field which is perpendicular to the selected velocity v. The force due to the magnetic field causes an ion to move in a semicircle path of radius r given by F = F c o mv qvsin 90 = r mv r = and v = q r = Since the magnetic fields and and the electric field are constants and every ion entering the spectrometer contains the same amount of charge q, therefore r = km with k = : constant q r m SF07 m q (5.7b) If ions of masses m and m strike the photographic plate with radii r and r respectively as shown in figure 5.7a then m r = (5.7c) m r xample 0: beam of singly charged ions of isotopes Ne-0 and Ne- travels straight through the velocity selector of a ainbridge mass spectrometer. The mutually perpendicular electric and magnetic fields in the velocity selector are 0.4 MV m - 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-0 =. x 0-6 kg; mass of Ne- =.65 x 0-6 kg and charge of the beam is.60 x 0-9 C) Solution: m =.x0-6 kg, m =.65x0-6 kg, =0.6 T =0.4x0-6 V m -, q=.60x0-9 C, =0.8 T SF07
a. The selected velocity of the ions is v = v = 6. 67 0 b. The radius of the circular path made by isotope Ne-0 is r r m q = 0. 7 m = The radius of the circular path made by isotope Ne- is r r m q = 0. 90 m = Therefore the separation between them is 5 m s r = r r r = 0. 90 0. 7 r = 0. 07 m SF07 TH ND Next Unit UNIT 6 : Nuclear Reaction SF07 4