ATOMIC PHYSICS V. SUBRAMANIAN, P.G.Teacher(Physics) Government Higher secondary School,Manali , Chennai - 600 068, Tamil Nadu state, India. Presented By 1 weyes57 Dedicated to the owners of all web sites and bloggers from whom I have taken most of the pictorial materials for making this presentation. Thank you .

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Air and gases are poor conductors of electricity at atmospheric pressure. They do not have charged particles. They can be made to conduct by, Applying large potential difference across the gas column at very low pressure. Allowing X - rays to pass through them. The study of discharge of electricity through gases led to the discovery of electrons by J. J.Thomson. 2 weyes57

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At 110 mm of Hg pressure - no discharge. At 100 mm of Hg pressure - irregular streaks and crackling sound. At 10 mm of Hg pressure - positive column. 4) At 0.01 mm of Hg pressure - Crooke's dark space fills the tube and the walls glow with green colour. Potential difference - 50000 V and Pressure - 110 mm of Hg to 0.01 mm of Hg . 3 weyes57

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Different gases glow with different colours 5 weyes57

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Properties of cathode rays Travel in straight line. Possess kinetic energy and momentum. Produce heat, when allowed to fall on matter. Produce fluorescence when they fall on crystals , minerals and salts. 5) When they are suddenly stopped by substances of high atomic weight produce X – rays. 6 weyes57

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6) Ionize the gas through which they pass. 7) Affect photographic plates. 8) Deflected by electric and magnetic fields. The direction of deflection shows that they are negatively charged. 9) Travel nearly with 1/10 times the velocity of light. 10) They are called electrons , fundamental constituents of all atoms. 7 weyes57

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Canal rays Discovered by Goldstein in the year 1886. In a discharge tube at 1 mm pressure with perforated cathode he observed a stream of particles travelling opposite to that of cathode rays. They are called canal rays ,since they emerge from the holes of the cathode. 8 weyes57

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Properties of canal rays 1 ) Canal rays or positive rays are nothing but the positive ions of the gas in the discharge tube. 2) The mass of each ion is almost equal to the mass of the atom. 3) Travel in straight line. 4) The velocity of the canal rays is very small when compared to that of cathode rays. 5) Produce fluorescence. 6) Affect photographic plates. 7) Ionize the gas through which they pass. 8) Deflected by electric and magnetic fields. The direction of deflection shows that they are positively charged. 9 weyes57

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Thomson’s e/m experiment This method is based on the fact that the cathode rays are deflected by electric and magnetic fields. 10 weyes57

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Highly evacuated discharge tube with a cathode and an anode . The cathode and anode are connected to high potential. Fine pencil of electrons are obtained through the pinhole in the anode. The electron beam passes through two parallel plates and strike the flat face of the tube. 11 weyes57

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This flat face is coated with fluorescent material. A spot of light is seen on the flat face. Two coils are placed horizontally to provide uniform magnetic field perpendicular to the plane of the paper. 12 weyes57

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The plates are given a potential difference of V volt and the distance between them is d, then the electric field E = V/d. Due this force the electron beam gets displaced and so the light spot seen displaced on the flat face. The force experienced by the electron is F E = Ee, where e is the charge of the electron . 13 weyes57

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By passing suitable current through the coils a magnetic field is created in such a way that to nullify the effect produced by the electric field. The velocity of the electron beam is v . If B is the magnetic induction, then the force on the electron due to magnetic field is F B = Bev. Force due to electric field = Force due to magnetic field. F E = F B Ee = Bev and v = E / B 14 weyes57

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The magnetic field is switched off. Now only electric field is present and it makes the electron beam to have a downward displacement.. The actual downward displacement inside the plates is Y 1 and that is proportional to the displacement Y , on the flat face . Y = k Y 1 , where k is a constant depends on the geometry of the tube. The downward displacement of the electron beam is given by Newton’s second equation Initial downward velocity u = 0, t = l / v and a = Ee / m. 15 weyes57

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Y - + d The value of e/m was calculated by substituting the known values. The value of e/m = 1.7592 X 10 11 C kg -1 16 weyes57

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Millikan’s oil drop experiment to determine the charge of an electron 17 weyes57

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Principle The experiment is based on the motion of a uncharged oil drop under free fall due to gravity and charged oil drop in a uniform electric field. By adjusting the field the motion of the drop can be controlled. Experimental arrangement Two horizontal circular metallic plates 22 cm diameter, separated by 16mm. Upper plate has a hole in the middle. The plates are held together by insulating rods of glass or ebonite. The plates are surrounded by a constant temperature bath and the chamber containing dry air. 19 weyes57

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Highly viscous liquid is sprayed by means of an atomizer. The droplet enters the space between the plates. The oil drop is illuminated by an arc lamp. A microscope with micrometer scale is provided to view the oil drop descending due to gravity. As the drop falls freely the viscous force of air increases and the droplet attains constant terminal velocity v. The terminal velocity of the droplet is measured with the microscope. 20 weyes57

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Motion due to gravity Due to the motion in the viscous medium air, The apparent weight of the droplet = weight of the oil drop - the weight of the air displaced. If a is the radius of the droplet , ρ density of oil and σ density of air, then 21 weyes57

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By stokes law, The droplet attains terminal velocity as the apparent weight of the droplet is equal to the viscous force. If η is the coefficient of viscosity of air then, 22 weyes57

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Motion under electric field The air inside the plates is ionized by passing X – rays. The droplets pick up one or more electrons from the ionized air. The charge on the droplet under observation is q. The field is applied in such away that the droplet moves upwards. The droplet starts moving with a terminal velocity This velocity can be measured using the microscope. The upward force on the droplet is Eq. 23 weyes57

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Now again using Stoke’s relation, Adding both the equations Substituting the value of a in the equation, E = V/d , where V is the potential applied and d is the distance between the plates. Millikan determined the value of q for large number of oil drops and found that the value of q is an integral multiple of a least value. The greatest common factor gives the charge e of an electron. The charge of an electron was found to be 1.602X10 -19 coulomb. 24 weyes57

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Calculation of mass of an electron The value of e/m = 1.7592 X 10 11 C kg -1 The value of e = 1.602X10 -19 coulomb. Mass m of the electron = e/(e/m) m = 9.11X10 -31 kg 25 weyes57

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Atom models Dalton model (1803) Matter is made up of extremely small particles called atoms. Atoms are indivisible. Prout model (1815) All atoms are made up of Hydrogen atoms. But the atomic weight of all atoms are not integral multiple of atomic weight of hydrogen. 26 weyes57

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Thomson considered atom as a positive sphere of radius around 10 -10 m. The positive charge is uniformly distributed over the entire sphere and the negative charges are embedded inside like plums in a pudding. 27 weyes57 THOMSON’S ATOM MODEL

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28 weyes57 For an atom of two electrons , they are situated symmetrically with respect to the centre of the sphere. In a 3 electron system, the electrons are placed symmetrically at the corners of an equilateral triangle.

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Draw backs of Thomson’s atom model The vibrating electron must radiate energy. The frequency of emitted spectral line must be same as the frequency of the electron. In the case of Thomson’s model only one spectral line of 1300 A. U alone possible. But there are 5 series of spectral lines available in hydrogen spectrum. Thomson’s model was unable to explain the scattering of alpha particles at larger angles. 29 weyes57

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Rutherford’s alpha particle scattering experiment. 30 weyes57 Rutherford studied the scattering of alpha particles by a thin foil of gold. Alpha particle is a positively charged particle having magnitude twice that of an electron and mass equal to that of a Helium atom. The experiment provided information about the structure of the atom.

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31 weyes57 Polonium is used as alpha emitter and it is placed in a lead box with a narrow opening. A narrow beam of alpha particles incident on the gold foil are scattered through different angles. The scattered alpha particles strike the fluorescent screen , coated with zing sulphide. The tiny flashes of light produced when an alpha particle strikes the screen. Scintillation viewer is a low power microscope used for seeing the tiny flashes.

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32 weyes57 From the graph we find that , more than 90% of the alpha particles go un deflected and below 10% of them get large angle deflection.

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Distance of closest approach - r 0 The α particle projected towards a nucleus , will move close up to a distance r 0 . At this distance the kinetic energy lost gets converted in to electrostatic potential energy according to law of conservation of energy. If ‘m’ be the mass of the α particle and ‘v’ be its velocity. 33 weyes57

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weyes57 34 Rutherford’s atom model Atom may be regarded as a sphere of diameter 10 -10 m . Almost the entire mass and all positive charge of the atom is concentrated in a small central core called Nucleus. The diameter of the nucleus is of the order of 10 -14 m. Rutherford suggested that the electrons are revolving around the nucleus in circular orbits and the electrostatic force of attraction between the nucleus and the electron provides the necessary centripetal force. As the atom is neutral , the total number of positive charges of the nucleus is equal to the total number of electrons having negative charges

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The electron in the circular path experiences a centripetal acceleration. Accelerated electric charge must radiate energy in the form of electromagnetic radiation. As the energy of the electron continuously decreases, it must spiral round and fall in to the nucleus. Thus the atom cannot be stable. 35 weyes57 Drawbacks of Rutherford’s atom model

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As the electron spiral towards the nucleus, the angular velocity tends to become infinity. V = r ω , ω = V / r ω = 2π f V = 2π r f f = V / 2π r As ‘r’ tends to zero, both ‘ω’ and ‘f ‘tends to infinity. The frequency of the emitted light tends to infinity, resulting a continuous spectrum. But in the visible region of hydrogen spectrum , line spectrum is obtained. 36 weyes57

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Bohr’s atom model (1913) Bohr’s postulates Electron cannot revolve around the nucleus in all possible orbits. Electron can revolve in only in certain permitted orbits. In this permitted orbits the angular momentum of the electron is equal to an integral multiple of h / 2 π , where h is Plank’s constant.( h = 6.626X10 -36 Js ) Angular momentum of an electron in the n - th orbit = mv n r n = nh/2π 37 weyes57

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The electron can revolve only in certain permitted orbits where the angular momentum of the electron is equal to integral multiple of h /2 π mv 1 r 1 = h/2π mv 2 r 2 = 2h/2π mv 3 r 3 = 3h/2π mv n r n = nh/2π 38 weyes57

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An electron revolving in a permitted orbit , does not radiate energy. Only when an electron jumps from a higher energy orbit to a lower energy orbit , energy is Given out as electromagnetic radiation. The difference in the energy ∆E , Where h is Plank’s constant and is frequency of emitted radiation. 39 weyes57

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Calculation of radius of the n –th orbit 40 weyes57 The electrostatic force of attraction between a nucleus with charge Ze and an electron of charge e is given by The electron revolving around the nucleus will experience a centripetal force. This electrostatic force supplies the necessary centripetal force to the electron . The centripetal force experienced by the electron revolving in n – th orbit with an angular velocity ω n is F cp = mr n ω n 2 F e = F cp

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From the Bohr’s postulate, the angular momentum of the electron in the n – th orbit is equal to nh/2 π. 41 weyes57

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Substituting for n = 1(First orbit) , Z = 1(Hydrogen atom) e =1.602X10 -19 C, h = 6.626X10 -34 Js, m = 9.11X10 -31 kg and ε 0 = 8.854X10 -12 N -1 C 2 m -2 r 1 = 0.53x10 -10 m m r 2 = 4a 0 , r 3 = 9a 0 , r 4 = 16a 0 etc.. 42 weyes57

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The energy E n of the electron in the n – th orbit is the sum of kinetic and potential energy. F e = F cp 43 weyes57

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Substituting for n = 1 (First orbit) , Z = 1 (Hydrogen atom) e =1.602X10 -19 C, h = 6.626X10 -34 Js, m = 9.11X10 -31 kg and ε 0 = 8.854X10 -12 N -1 C 2 m -2 44 weyes57

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Frequency of spectral lines of hydrogen atom when electron jumps from a higher energy orbit to a lower energy orbit , then the difference In the energy will be given out as electromagnetic radiation. Consider an electron jumping from n 2 - th orbit to n 1 - th orbit, then 45 weyes57

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Substituting and Z = 1(Hydrogen atom) e =1.602X10 -19 C, h = 6.626X10 -34 Js, m = 9.11X10 -31 kg ε 0 = 8.854X10 -12 N -1 C 2 m -2 and c = 3X10 8 m/s R is called as Rydberg’s constant. Substituting the values, R = 1.094x10 7 m -1 46 weyes57

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When an electron in Hydrogen atom makes a transition from a higher energy orbit to a lower energy orbit radiates energy in the form of electromagnetic radiation. The wave number of the emitted radiation is given by the relation, Lyman series : Here the transition is from orbit higher than 1 to first orbit. All the spectral lines are in the ultra violet region. Balmer series : Here the transition is from orbit higher than 2 to second orbit. The first line, n = 3 to 2 is called H α. n = 4 to2, n = 5 to 2 and n = 6 to2 are called H β , H ϒ, H δ respectively. This lines are in visible region of the spectrum. 47 weyes57

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Paschen series : Here the transition is from orbit higher than 3 to third orbit. All the spectral lines are in the infra red region. Brackett series : Here the transition is from orbit higher than 4 to forth orbit. All the spectral lines are in the infra red region. Pfund series : Here the transition is from orbit higher than 5 to fifth orbit. All the spectral lines are in the infra red region. 48 weyes57

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Energy level diagram of hydrogen spectrum 50 weyes57

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Ionization potential energy is the energy required to remove a electron from its ground state to infinity. Excitation Potential energy is the energy required to excite an electron from its ground state to other higher energy states. First excitation energy Second excitation energy Critical potentials are the potentials required for ionization or excitation of electron from its ground state to any higher energy level. Ionization potential of Hydrogen atom is 13.6 V. 51 weyes57

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Spectral lines(emission) of Sodium, Mercury and Hydrogen 52 weyes57

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Bohr’s atom model was unable to explain the following. The spectra of atoms more complex than hydrogen atom. Variation of intensity of spectral lines . The arrangement and distribution of electrons in various orbits. The fine structure of hydrogen spectral lines. The Stark and Zeeman effects. 53 weyes57

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weyes57 54 Sommerfeld atom model Sommerfeld introduced two modifications in Bohr’s theory. 1) The path of an electron around the nucleus is an ellipse, with nucleus at one of its foci. The velocity of the electron in an elliptical orbit varies at different parts of the orbit. This causes the relativistic variation in the mass of the moving electron.

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weyes57 55 In an elliptical orbit, there is a variation of distance of the electron from the nucleus. Also there is a variation of angular position of the electron with respect to the nucleus. To accommodate this two variables two quantum numbers are introduced. The principle quantum number(n) of Bohr’s theory. 2) The orbital or azimuthal quantum number (l ) characterizes the angular momentum of an electron in an orbit. Angular momentum quantum number (l ) can take values fro 0 to n in steps of unity. The orbital quantum number(l) is useful in finding the possible elliptical orbits.

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weyes57 56 If ‘a’ and ‘b’ are the semi major axis and semi minor axis then the possibilities of Elliptical orbits are For any principle quantum number ‘n’ there are n possible orbits of varying eccentricities. These orbits are called sub orbits or sub shells. Out of ‘n’ sub orbits one is circular and ‘n-1’ are elliptical. The energy of the sub orbits are slightly different due to the relativistic variation of mass of the electron.

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weyes57 57 Principle quantum number ( n) 1 2 3 4 5 Orbital quantum number ( 0 to n-1) 0 0 1 0 1 2 0 1 2 3 0 1 2 3 4 Sub orbit designation s s p s p d s p d f s p d f g

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weyes57 58 Fine structure of spectral lines The total energy of an electron in an elliptical orbit is as same as that obtained by Bohr. Introduction of elliptical orbits gives no new energy levels and no new transitions. The attempt to explain fine structure with variation of mass with velocity was effective. The velocity of electron is maximum when it is near the nucleus and minimum when it is away from the nucleus, leading to relativistic variation of mass of the electron.

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weyes57 59 Taking in to consideration the relativistic variation of mass of the electron Sommerfeld modified his theory and showed that the path of electron is not simple ellipse but a precessing ellipse called rosette. Based on this idea the fine structure of spectral lines of Hydrogen atom.

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weyes57 60 Drawbacks of Sommerfeld’s atom model: Sommerfeld’s atom model gave a theoretical background of the fine structure of spectral lines of hydrogen atom. But it was unable to predict and explain, 1) the account for the correct number of fine structure lines observed in Hydrogen atom. 2) the distribution and arrangement of electrons in atoms. 3) the Stark and Zeeman effect. 4) the spectral lines of alkali metals like sodium and potassium. 5) the variation of intensity of the spectral lines.

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weyes57 61 Kindly visit www.freewebs.com/weyes57 and view my other works. Thank you

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