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# Notes for Atoms Molecules and Nuclei - Part III

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Notes for Atoms Molecules and Nuclei - Part III

Notes for Atoms Molecules and Nuclei - Part III

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• 1. 9011041155 / 9011031155 • Live Webinars (online lectures) with recordings. • Online Query Solving • Online MCQ tests with detailed solutions • Online Notes and Solved Exercises • Career Counseling 1
• 2. 9011041155 / 9011031155 Atoms, Molecules and Nuclei Q.63 The de-Broglie wavelength of a particle having a momentum of 2×10-28kg m/s is (a.63) 3.3×10-5m (b.63) 6.6×10-6m (c.63) 3.3×10-6m (d.63) 1.65×10-6m Q.64 A proton and an α particle are accelectrated through the same potential difference. The ratio of the de-Broglie wavelength of the proton to the de-Broglie wavelength of the α particle will be (a.64) (b.64) (c.64) (d.64) Composition And Size Of Nucleus Nucleus: - positively charged and high density centre. 99.9% mass of the atom. Protons(+ve) + neutrons(neutral)= nucleons number of protons = atomic number (Z). number of protons & neutrons = mass number(A). 2
• 3. 9011041155 / 9011031155 nucleus symbolically expressed as A X Z For example, gold: - 197 A u and Uranium: 79 238 92 U. Radius of nucleus 1 R = R0 A 3 where R0 is linear constant = 1.2 × 10-15 m. volume of nucleus α A. Density of nucleus is constant = 2.3 × 1017 1kg/m3, radius of carbon nuclei is 1 R C = 1 .2 ´ 1 0 - 15 = 2 .7 4 7 3 ´ 1 0 m ´ (1 2 ) 3 - 15 m Radius of uranium nuclei is 3
• 4. 9011041155 / 9011031155 1 R U = 1 .2 ´ 1 0 - 15 m = 7 .4 3 6 6 ´ 1 0 ´ (238 )3 - 15 m Isotopes, isotones and isobars Isotopes: - The nuclei have same number of protons but different number of neutrons. E.g. deuterium 21 H , tritium 31 H are the isotopes of hydrogen. gold has 32 isotopes. Isotopes have identical chemical behavior and are placed in the same location in the periodic table. Isobars: - The nuclei have same mass number (A). For example, the nuclei 31 H , 3 H e are isobars. 2 4
• 5. 9011041155 / 9011031155 Isotones:- The nuclei having same number of neutrons (A – Z) but different atomic numbers Z. For example 198 80 Hg, 179 79 A u , are isotones. Mass-Energy relation Einstein proved that mass is a form of energy. E = m c2 Energy of electron. Ee = mec2= (9.1 × 10-31) × (3 ×108)2 joule = 9 .1 ´ 9 ´ 1 0 - 31 1 .6 ´ 1 0 ´ 10 - 19 16 eV = 0.511 × 106 eV = 0.511 MeV Similarly, the energies of proton and neutron are Ep = 941.1 MeV and En = 942.2 MeV 5
• 6. 9011041155 / 9011031155 There is one more unit used to express nuclear masses. It is Unified atomic mass unit. It is (1/12)th of the mass of neutral carbon atom in its lowest energy state. Its symbol is ‘u; 1u = 1.66054 × 10-27 = 931 MeV/C2 OR Energy equivalent of mass 1 u is = 931 MeV Mass Defect It is observed that the mass of a nucleus is smaller than the sum of the masses of constituent nucleons in the Free State. The difference between the actual mass of the nucleus and the sum of masses of constituent nucleons is called mass defect. Let M – be the measured mass of nucleus. A – be the mass number (mass of nucleons in free state) Z - atomic number (number of protons) 6
• 7. 9011041155 / 9011031155 Mp - mass of hydrogen atom (i.e. proton) Ma - Mass of free neutron (A – Z) - number of neutrons. The mass defect △m = [Zmp = (A – Z)mn] – M Nuclear Binding Energy Nucleons (protons and neutrons) are bound together in a nucleus with very strong attractive force. Energy must be supplied to the nucleus to separate its constituent nucleons. It is observed that mass of the nucleus is always less than the sum of the masses of its constituent nucleons. The difference in mass is being used as energy that holds nucleons together. 7
• 8. 9011041155 / 9011031155 The amount of energy required to separate all the nucleons from the nucleus is called binding energy of the nucleus. The B.E. of nucleus is very high. For example, it is 2.22 MeV for deuteron nucleus, whereas B.E. for an atom, say hydrogen atom in its ground state is 13.6 eV. That is, B.E. of nucleus is about 10,00,000 times larger than B.E. of atom. The B. E. of nucleus can be expressed in terms of mass defect. B.E. = △m × c2 joule Where △m is mass defect and c is speed of light. But △m = [Zmp + (A – Z)mn] – M ∴ B. E. of nucleus = [Zmp + (A – Z)mn – M]c2 joule The B.E. per nucleon = B. E. o f n u cle u s A 8
• 9. 9011041155 / 9011031155 E = éZ m p + ( A - Z )m n - M ù 2 ê úc ê ú A ë û This is average energy per nucleon to separate a nucleon from the nucleus. B.E. Curve 9
• 10. 9011041155 / 9011031155 The B.E. curve is an indicator of nuclear stability. The higher the B.E. per nucleon, the greater is the stability From B.E curve we can infer as follows: i. The B.E. per nucleon is practically constant and is independent of mass number for nuclei, 30 < A < 170. ii. It is maximum 8.75 MeV, for A = 56 and is 7.6 MeV, for A = 238. 10
• 11. 9011041155 / 9011031155 iii. It is low for both light nuclei (A < 30) and heavy nuclei (A > 170). This means that the nucleons of atoms are loosely bound with nucleus. iv. When heavy nucleus (A = 240) breaks into lighter nuclei (A = 120), B.E. increases i.e. nucleons get more tightly bound. v. When very light nuclei A < 10, join to form a heavier nucleus, B.E. increases, i.e. nucleons get more tightly bound.In both the cases, there is release of energy because, the new nuclei formed have less mass and are more stable. 11
• 12. 9011041155 / 9011031155 Radioactivity Becquerel Heavy elements like uranium, radium having A > 82 are unstable and emit highly penetrating radiations. The substances which emit these radiations are known as radioactive substances. The phenomenon of spontaneous emission of radiations from radioactive substance is known as radioactivity. Radioactivity is property of atom and nuclei, hence is unaffected by chemical 12 or physical changes.
• 13. 9011041155 / 9011031155 Radioactivity is nuclear phenomenon in which an unstable nucleus undergoes decay. It is called radioactive decay. There are three types of decay – i. -decay - 4 H e . 2 ii. -decay - electrons or positrons. iii. -decay - high energy photons. 13
• 14. 9011041155 / 9011031155 Properties of –particles 1. +ve. It is helium atom with both electrons removed. Mass:- 6.64×10-27kg & charge +3.2×10-19 coulomb. 2. It is deflected by electric and magnetic field. 3. The speed of emission of -particles depend upon the nature of radioactive element. It varies from (1/10)th to (1/100)th of the speed of light. 4. Affect photographic plate, produce fluorescence. 5. They ionize gas when passed through gas. 6. Range:- through air:- 2.7cm to 8.62cm for thorium. 7. They are scattered when incident on mica, aluminium and gold foil. 8. When an -particle is emitted by an atom, its atomic number decreases by 2 and mass number decreases by 4. e.g. 238 92 U ® 234 90 Th + 4 2 He 14
• 15. 9011041155 / 9011031155 Properties of -particles 1. -rays are fast moving electrons 2. Their speed ranges from 1% to 99% of the speed of light. 3. Being charged particles they are deflected by electric and magnetic field. 4. They can ionize gas but its ionization power is th æ 1 ö ÷ of that of ç ÷ ç ÷ ç1 0 0 ø è -particles. 5. They are more penetrating than -particles. 6. Their range in air depends on their speed. A - particle of 0.5 MeV has a range of 1 m in air. 7. When -particle is radiated, the atomic number increases by 1 and mass number does not change e.g. 32 15 P ® 32 16 S + 1e 0 15
• 16. 9011041155 / 9011031155 Properties of -rays 1. -rays are not particles but they are electromagnetic waves (photons) of very short wavelength. Photons originating from the nucleus are called -rays. 2. They are neutral in charge and not affected by electric and magnetic field. 3. They affect photographic plate and produce fluorescence. th æ 1 ö ÷ 4. They have very low ionization power about ç ç ÷ ç1 0 0 0 ÷ è ø of that of σ a-particles. 5. They have high penetration power and can pas through 25cm thick iron plates. 6. They are diffracted by crystals. 16
• 17. 9011041155 / 9011031155 Radioactive Decay law The spontaneous breaking of nucleus is known as radioactive disintegration. They decay law The number of nuclei undergoing the decay per unit time is proportional to the number of unchanged nuclei present at that instant. Let N be the number of nuclei present at any instant t, dN be the number of nuclei that disintegrated in short interval of time dt. Then according to decay law: - dN dt a N i.e. dN dt = - l N (1 ) Where λ is known as decay constant or disintegration constant. From equation (1) dN N = - l dt 17
• 18. 9011041155 / 9011031155 Integrating both sides ò dN N = ò- l dt loge N = –λt + c where c is constant of integration whose value depends on initial conditions. At t = 0; N = N0 (the number of original nuclei) ∴ loge N0 = 0 + c Substitution the value in above expression loge N = – λt + loge N0 = – λt loge N – loge N0 lo g e æN ö ç ÷ = ç ÷ çN ÷ è 0÷ ø N N0 Or = e l t - l t N = N0 e-λt ...(2) 18
• 19. 9011041155 / 9011031155 This expression shows that number of nuclei of given radioactive substance decreases exponentially with time. Decay constant From equation (1) we have dN l = - dt N The decay constant is defined as ratio of the amount of substance disintegrated per unit time to amount of substance present at that time. We have N = N0e- t 19
• 20. 9011041155 / 9011031155 Let us define ‘t’ as t = 1 1 N N0 e N = N0e-1 N = N = N0 e N0 2 .7 1 8 N = 0.37 N0 1 The decay constant , which is equal to , can be t defined as reciprocal of time duration (t) in which the substance decays to 37% its original quality. Half life period (T) Half life period (T) of radioactive substance is defined as the time in which the half substance is disintegrated. 20
• 21. 9011041155 / 9011031155 We have N = N0eat t = T; N = ∴ or N0 = N0e- 2 1 = e- 2 or e t t N0 2 T T =2 T = loge 2 = 0.693 ∴ T = 0 .6 9 3 l Using this expression, we can determine the half life of radioactive substance if its decay constant is known. 21
• 22. 9011041155 / 9011031155 Nuclear fission Discovered by:- German scientists Otto Hahn and Strassman(1939) 235 92 U + 1 0 n ® 236 92 144 56 U ® 22 Ba + 89 36 Kr + 3 1 0 n
• 23. 9011041155 / 9011031155 Nuclear fusion When two lighter nuclei are fused to form a heavier nucleus, the process is called nuclear fusion. 1 1 H + 2 1 H + 1 1 H ® 2 1 H ® 2 1 4 H + e + 0 .4 2 M e V 4 1 He + 24 M eV 23
• 24. 9011041155 / 9011031155 • Ask Your Doubts • For inquiry and registration, call 9011041155 / 9011031155. 24