Title MODERN PHYSICS II.pdf ✅

Modern Physics  Digital www.allendigital.in [ 215 ] Nucleus and Nuclear Force (i) Central core of every atom. (ii) Discovered by Rutherford in -scattering experiment. (iii) The order of nuclear size = 10–15 m or fm while the order of atomic size = 10–10 m or Å (iv) Protons and neutrons, together referred as nucleons. (v) A nuclide is represented by A Z X Z = atomic number = p (no. of protons) A = mass number = total no. of nucleons = n + p (vi) Atomic masses are generally represented by atomic mass unit (u) 1u = 12 mass of C atom 12 = 1.66 × 10–27 kg mp = 1.6726 × 10–27 kg = 1.00727 u ; mn = 1.6749 × 10–27 kg = 1.00866 u me = 9.1 × 10–31 kg = 0.00055 u Types of Nuclei (i) Isotope: same Z (ii) Isobar: same A (iii) Isotone: same (A – Z) Properties of Nuclei Size of Nucleus : (Order is fermi) As the number of nucleons in nucleus increases its size also increases and relation between its radius and mass number is R  A1/3 → R = R0A1/3 Here R0 is a constant and its value R0  1.2 fm. Volume of Nucleus Volume  R3 (But R  A1/3) or volume  A Mass of Nucleus Its mass is quite small compare to gm or kg. Therefore it is measured in another unit – amu (Atomic Mass Unit) Mass of an nucleus of mass number A is ≈ Amp  amu or mass of an nucleus, m  A Density of Nucleus () p p p 17 3 3 3 3 0 0 mass Am Am 3m 2.3 10 kg / m volume 4 4 4 R R R A 3 3  =  = = =     It means  is independent of A. Density of nuclei of all types of element is same and its order is 1017 kg/m3 or 1014 g/cm3 Forces acting inside the nucleus There are three forces interacting between nucleons, these are (i) Gravitational force - weakest force of nature. (ii) Electrostatic repulsive (Colombian) force → only works between proton. This is stronger than gravitational force. (iii) Nuclear force → strongest interaction that holds nucleons together to form nuclei and it is powerful enough to overcome the electric repulsion of proton and proton. Modern Physics-2 [Nuclear Physics & Radioactivity]
NEET : Physics [ 216 ] www.allendigital.in  Digital Features of Nuclear Force (Fn) 1. The strongest force in the universe. 2. Works only between the nucleons. 3. Very short range : only upto size of nucleus (3 or 4 fermi). More than this distance, nuclear force is almost zero. 4. Very much depends upon distance :– Small variation in distance may cause of large change in nuclear force while electrostatic force remains almost unaffected. 5. Independent of charge :– Interacts between n–n as well as between p–p and also between n–p. 6. Spin dependent :– It is stronger between nucleons having same sense of spin than between nucleons having opposite sense of spin. 7. It is not a central force: – Definition of central force (Fc) : Whose line of action always passes through a fixed point and its magnitude depends only on distance, if medium is same. c ( ) n K F rˆ r =  is central force. Electrostatic and gravitational forces are central forces. 8. Nature: – (i) Attractive – If distance is greater than 0.8 fm or above. (ii) Repulsive – If distance is lesser than 0.8 fm. Illustration 1: Calculate mass no. of that nucleus whose radius is half of Ge72. Solution: r  A1/3 → 1/3 r 2 A 1 A A 9 r 72 8 72   =  =  =     Illustration 2: Find the density of 12 6 C. Solution: Mass of nucleus  12mP = 12 × 1.66 × 10–27 kg {mP = mass of proton} –27 17 –3 3 3 –15 1/3 3 M 12 1.66 10 kg 2.4 10 kg m 4 4 R 1.2 10 (12) m 3 3    = = =         Einstein's Mass - Energy Relation According to Einstein, mass can be converted into energy and energy into mass. This relation is given by; E = mc2 Here, E = total energy associated with mass m; c2 = used as a conversion coefficient Mass Defect and Nuclear Binding Energy Mass defect: (i) Mass of a nucleus is always less than the sum of masses of its constituent nucleons. This difference is called mass defect. (ii) If observed mass of nucleus ZXA be M, mass of proton is Mp and mass of neutron is Mn then mass defect = m = [ZMp + (A – Z)Mn] – M. (iii) If M is taken as mass atom of ZXA instead of mass of nucleus then m = [Z(Mp + Me) + (A – Z)Mn] – Matom
Modern Physics  Digital www.allendigital.in [ 217 ] Binding energy (Eb) (i) Binding energy of a nucleus is the energy required to split it into its nucleons (free). (ii) Eb = m.c2 (iii) It is always positive and numerically equal to the energy equivalent of mass defect (or equal to the energy liberated when it was formed) Illustration 3: The mass defect in a nuclear fusion reaction is 0.05%. What amount of energy will be liberated in one kg fusion reaction? Solution: Mass defect = m = 0.05% of 1 kg = 0.05 100 kg = 5 × 10–4 kg Energy liberated = (m)c2 = (5 × 10–4) (9 × 1016)J = 45 ×1012 J Illustration 4: What is energy released by fission of 1 gm U235? Solution: Number of atom in 1 gm of U235 = NA 235 Energy released = NA 235 × 200 MeV = 23 6.023 10 235  × 200 MeV = 5 × 1023 MeV = (5 × 1023) (1.6 × 10–13 J) = 8 × 1010 J = 10 6 8 10 3.6 10   kWh = 2.22 × 104 kWh Illustration 5: What is the power output of 235 92 U reactor if it takes 30 days to use up 2 kg of fuel and if each fission gives 185 MeV of usable energy? Solution: Number of atoms in 2 kg of 23 3 235 24 92 6.02 10 2 10 U 5.12 10 235    = =  Therefore, energy released in 30 day = 5.12 × 1024 × 185 MeV = 1.51 × 1014 J  Energy released per second = 14 1 51 10 30 24 60 60      = 58.4 MW Illustration 6: Obtain the binding energy of a nitrogen nucleus ( ) 14 7 N in MeV from the following data. mH = 1·00783u, mn = 1·00867 u, mN = 14·00307 u Solution: Mass defect, m = 7mp + 7mn – mN = 7 × 1·00783 + 7 × 1·00867 – 14·00307 = 0·11243 amu Binding energy = m × 931 MeV = 0·11243 × 931 MeV = 104·67 MeV
NEET : Physics [ 218 ] www.allendigital.in  Digital Binding Energy per Nucleon and its Curve[ ΔΕb A ] (i) The value of binding energy per nucleon decides the stability of a nucleus. It is obtained by dividing binding energy by the mass number of given nucleus. (ii) The following figure shows the binding energy per nucleon plotted against the mass number of various atoms nuclei Greater the binding energy per nucleon, the more stable the nucleus. (iii) It is maximum for isotope of iron – 56 26Fe and is 8.8 MeV/nucleon. It is the most stable nucleus. (iv) For Uranium, binding energy per nucleon is about 7.7 MeV/nucleon and it is unstable. (v) The medium size nuclei are more stable than light or heavy nuclei. Illustration 7: Explain nuclear fission & fusion on the basis of binding energy of nucleus. Solution: In both cases the net B.E. increases resulting in energy release. Figure: The binding energy per nucleon as a function of mass number. Mass number (A) 0 50 100 150 200 250 0 2 4 6 8 10 2H 3H 6Li 14N 18O 12C 16O 4He 32S 56Fe 100Mo 127I 184W 197An 238U Binding energy per nucleon (MeV) Mass number A C B R Q Binding energy per P nucleon In fission : nucleus A breaks into B & C In fussion : P & Q fuse to result in nucleus R