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ATOMS, , iA INTRODUCTION, , The structure of matter that shapes the world around us was a subject of study, , a long time. Dalton was the first to postulate that matter is made of atoms which, ; invisible. By the end of nineteenth century, enough evidence had accumulated in, favour of atomic hypothesis of matter. In 1897, experiments on electric discharge through, vases carried out by JJ. Thomson revealed that atoms of different elements contain |, negatively charged particles called electrons. However, atoms on a whole are electrically, , neutral. Therefore, an atom must also contain some positive charge to neutralize the, negative charge of the electrons., , The first atom model was rruposed by J.J. Thomson in 1898. According to this, model, the positive charge of atom is uniformly distributed throughout the volume of, the atom, like a pudding. The negatively charged electons are embedded in it like “plums”, in that pudding. Therefore, this atom model is sometimes called “plum-pudding” model., The total positive charge in the atom was evenly and symmetrically balanced by the, negative charges of the electrons., , The Thomson’s atom model failed to satisfy the results of the experiments performed, later by Rutherford and others. The next major step towards the structure of the atom, was Rutherford’s nuclear atom model. According to this model, the electron revolves, round the positively charged nucleus at the centre of the atom, just like the planets, revolve round the sun. Next came the Bohr atom model which was successful in explaining, the spectrum of hydrogen and hydrogen-like atoms, but failed to explain the spectrum, of complex atoms. It was further modified by Sommerfeld’s relativistic atom model and, finally we have the vector atom model which is being used today., , 12.2 ALPHA PARTICLE SCATTERING EXPERIMENT, , In 1911, H. Geiger and E., Marsden performed the alpha, scattering experiment at the, suggestion of E. Rutherford. The, experimental arrangement is as, shown in Fig.12.1. S is a, radioactive source (3Bi2"*) e, Contained in a lead cavity. Alpha >, Particles (of energy 5.5MeV) |] g, cmitted by this source are, collimated into a narrow beam with, , , , , , , , , , , , , , , , , , , , , , , , , , , , we help of a lead slit. The lead, Collimated beam of alpha particles cavity, is allowed, , , , . to fall on a thin gold, , i of thickness 2.1 x 10-7m. The, vila particles scattered in i, , 3 Fig. 12.1, , MMferent directions are observed =, , Detector, (microscope), , , , , , , , Scanned with CamScanner
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968)| unit - 1x @, , through a rotatable detector consisting of a zinc sulphide (208) screen and q, microscope. The alpha particles produce bright flashes or scintillations on the Zng, sereen. These are observed in the microscope and counted at different angles from, the direction of the incident beam. The angle of deviation 0 of the alpha particle, , from its initial direction is called scattering angle., , The graph obtained by plotting, the number of alpha particles, scattered in a given time as a, function of scattering angle is as, shown in Fig. 12.2. The dots in the, figure represent data points in the, actual experiment and the solid curve, is the theoretical prediction based on, the assumption that atom has a small, positively charged nucleus at its iS, centre. It is observed from the graph, that, , (tered, , , , , va, , No. of alpha particles, , (i) many of the alpha particles pass, straight through the gold foil., That means they do not suffer, any collision with gold atoms., , , , 0° Scattering angle 0 180°, , (ii) only about 0.14% of the incident, Fig. 12.2, , alpha particles are scattered by, more than 1°., , (iii) about 1 in 8000 of the incident alpha particles is deflected by more than 90°., , Rutherford argued that, to deflect alpha particles backwards (by more than, 90°), it must experience a large repulsive force. This’ force could be provided if, the greater part of the mass of the atom and its positive charge were concentrated, tightly at its centre in a nucleus. If the incoming alpha particles could get very, close to the positive charge at the centre without penetrating it, then it will experience, a large repulsion. Such a close encounter would result in large deflection of the, alpha particle. Thus, Rutherford’s assumption could explain the large angle scattering, , of alpha particles., , , , , , , , 12.2.1 Rutherford’s atom model :, Following are the essential features of the Rutherford’s nuclear atom model., , (i) The atom has a small, positively charged nucleus. All the positive charge of, the atom and most of the mass of the atom are concentrated in the nucleus., , (ii) The dimensions of the nucleus are negligibly small as compared to the overall, size of the atom so that most of the volume occupied by an atom is actually, , an empty space., , Scanned with CamScanner
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oo, @ ATOMS 569|, , , , , , , , The atomic nucleus is surrounded by cert, is electrically neutral, the total negative, the nucleus is equal to the tot, , ain number of clectrons. As atom, charge on the electrons surrounding, al positive charge on the nucleus., , (iii), , (iv) The electrons in the atom revolve round the nucleus in various circular orbits., The centripetal force required for the revolution of the electron is provided, by the electrostatic force of attraction between the electron and the nucleus., , Rutherford’s experiments suggested the size of the nucleus to be about, yortim to 10-m. The size of the atom is about 10-!m. Therefore most of an, atom is empty space in which electrons revolve round the nucleus. With atom being, largely empty, most of the alpha particles pass through the thin gold foil. However,, when the alpha particle comes near to a nucleus, the intense electric field there, scatters it through a large angle. The atomic electrons which have very small mass, compared to that of alpha particle, have negligible effect on the alpha particles., Compared to alpha particle, the nucleus has a large mass. Hence, it can be assumed, that the nucleus remains practically at rest during the scattering process., , 12.2.2 Alpha particle trajectory:, , Alpha particle is'a helium nucleus with charge +2e. When an alpha particle, , is at a distance r from the nucleus of charge +Ze, the electrostatic repulsion due, to the nucleus is ., , 1 (Ze)(2e), , F=- =, , >, 4ne, r, , , , Consider the alpha particle 1, , ne 1, having initial kinetic energy K =F mv?, , (at large distance from the nucleus), , moving directly towards a nucleus. As Me ft Me ag eel te a Nuclevs, , the alpha particle approaches the : s=D, nucleus, its kinetic energy decreases Fig. 12.3, , , , , , , , , , and potential ener, , gy increases. When the alpha particle is at a particular distance, D from the nucleu, , ita S, its kinetic energy becomes zero. This distance D is the distance, , is comple approach. At this distance, the initial kinetic energy of the alpha particle, , ‘i oha ctely Converted into potential energy. Afterwards, because of repulsion the, , a Particle retraces its path. In.other words it is scattered by 180°. The potential, 8Y of the alpha particle at a distance D from the nucleus is, , U < lL (Zey(2e) _ 1 22Ze?, 4ne, iD 4ne, D, , , , Scanned with CamScanner
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SS), \870 UNIT - IX @, , Since loss in kinetic energy is equ, , 1 22 1 2ze sizsteistsssesaete QUIZM), , = o D=e—— se, Amy =D 4aty K, , Consider the alpha particle 2 approaching the nucleus with a small impa, , o . i e, parameter b. Because of electrostatic repulsion, th, . oO, a large angle (scattering angle 0>90°)., , , , , , , , al to the gain in potential energy,, , , , irae ct, alpha particle is scattered through, , hing the nucleus with larger, , ' . ticle 3 approac, Consider the alpha particle pp ler angle (0 < 90°)., , parameter b. It is scattered through a smal, The trajectory of the alpha particles 2 and 3 will be hyperbola,, , The impact parameter (b) is the perpendicular distance from the Centre of, the nucleus to the initial direction of the alpha particle when it is at infinite distance, from the nucleus. Smaller the value of impact parameter, larger is the value of, scattering angle. For direct or head on collision, impact parameter = 0 and Scattering, angle = 180°. It can be shown that,, , impact, , cot =, , D 8, 2 2, , impact parameter b =, , where D is the distance of closest approach and @ is the scattering angle., 12.2.3 Electron orbits :, , In Rutherford’s nuclear model of the atom, the picture of the atom is an, electrically neutral sphere consisting of a very small massive and positively charged, nucleus at the centre surrounded by electrons revolving in their respective dynamically, stable orbits. The centripetal force F_ required to keep the electrons in their stable, orbits is provided by the electrostatic force of attraction F. between the revolving, electrons and the nucleus. If an electron of mass m and charge (-e) is revolving, , round the nucleus with speed v in a stable circular orbit of radius r in a hydrogen, atom, we have, , , , , , , , 2 1 I 2, mv (e) (e) e, f= ™ adr = 7 OO. —#, c r ¢ TE, 2 4a, 2, since’ R=, we tive Melk e, ince =F, we have = Tz or: r= >, . § ro anes r de my?, , The kinetic energy of the electron is, edie (&=} e, , ~ 7M = 3 | dney or 8xeqr*, The potential energy of the electron is, , , , , , tt Oe 2 ., U= a 6 = iiap onsnesaag te eeeeeesengs (12.3), , , , , , Scanned with CamScanner
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rhe total energy of the clectron is, n, , S,. 2 5, ee c__ e¢, , , , gek+U= Saegr Ancgr = Saegr, , , , , , The total enresy of electron is negative. This implies that the electron is bound, \ nucleus and it will move in a closed, stable orbit., the, , ‘ , aToMIC SPECTRUM, 23!, , Each element has a characteristic spectrum of radiation which it emits. When, or vapour in atomic state is excited at low pressure, the emitted radiation, gas spectrum which contains certain specific wavelengths only. A spectrum of, ee * e is called line emission spectrum. It consists of bright lines on a dark, a aid Each spectral line corresponds to a particular wavelength of light. Since, ra line emission spectrum is characteristic of the atoms of the element emitting, ih, it serves as a “finger print” for identification of the element in the gas. When, a continuous spectrum light (light containing continuous range of wavelengths) from, an incandescent source Is passed through a gas and the transmitted light is analysed, using a spectrometer, we observe a some dark lines in the spectrum. These dark, lines correspond pricisely to those wavelengths which are found in the line emission, spectrum of the gas. This is called line absorption spectrum of the gas., , 12.3.1 Spectral series :, It is observed that the frequencies of the light emitted by a particular element, , exhibit some regular pattern. The spacing between lines within certain sets decreases, in a regular way. Each of these sets of lines is called a spectral series., , , , Consider the spectrum of, , , , , , , , , , , , , , , , hydrogen, which is the simplest atom. 5 e 5 E, Inthe visible region, four spectral lines E E 5 5, are clearly observed. These lines show Ss a G 2, a regular pattern (Fig. 12.4). This =f = 8, series is called Balmer series. The line, , with largest wavelength 656.3 nm in, , the red is called H,, line; the next line, , wih wavelength 486.1nm in the blue, iy 's called H. line; the third line, , Wavelength 434.1nm in the blue, , IS ca . : ., , led Hy line; the fourth line with 7 i i, , ‘ave ° r F 8 Y, call length 410.2nm in the violet is Ag oT q Aa, alled i i, , Hq line; and so on. As the Fig. 12.4, , , , , , Wavele, ngth ;, 8th decreases, the lines appear closer together and are weaker in intensity:, , Scanned with CamScanner