Atomic Structure from A level chemistry.
- 1. Chapter #2 : Atomic Structure Lecture by: Sir M. Saqib Naveed Shah Victorious Academy G-13 Campus Islamabad.
- 2. Introduction • According to Dalton’s theory, atoms were considered to be ultimate particles which could not be divided any further. • A number of subatomic particles have been discovered. The experiments which led to the discovery of electron, proton and neutron about 100 above. • So Dalton Theory was found controversial.
- 3. Discharge Tube Experiment • A gas discharge tube is fitted with two metallic electrodes acting as cathode and anode. • The tube is filled with a gas, air or vapours of a substance at any desired pressure. • The electrodes are connected to a source of high voltage. The exact voltage required depends upon the length of the tube and the pressure inside the tube. • The tube is attached to a vacuum pump by means of a small side tube so that the conduction of electricity may be studied at any value of low pressure
- 5. • It is observed that current does not flow through the gas at ordinary pressure even at high voltage of 5000 volts. When the pressure inside the tube is reduced and a high voltage of 5000-10000 volts is applied, then an electric discharge takes place through the gas producing a uniform glow inside the tube. • Some rays are produced which create fluorescence on the glass wall opposite to the cathode. These rays are called cathode rays. • J.J Thomson first identified electrons in cathode ray tube in 1887.
- 6. Properties of Cathode Rays Cathode rays are negatively charged. In 1895, J Perrin proved that these rays deflect toward anode showing they have negative charge. They produce a greenish fluorescence on striking the walls of the glass tube. Hiitorf proved that Cathode rays cast a shadow when an opaque object is placed in their path. This proves that they travel q straight line perpendicular to the surface of cathode.
- 7. These rays can drive a small paddle wheel placed in their path. This shows that these rays possess momentum. Cathode rays can produce X-rays when they strike an anode particularly with large atomic mass. Cathode rays can produce heat when they fall on matter. Cathode rays can ionize gases.
- 8. They can cause a chemical change, because they have a reducing effect. Cathode rays can pass through a thin metal foil like aluminum or gold foil. The e/m value of cathode rays shows that they are simply electrons. J.J. Thomson concluded from his experiments that cathode rays consist of streams of negatively charged particles. Stoney named these particles as electrons. Thomson also determined the charge to mass ratio (e/m) of electrons. He found that the e/m value ( 1.7588 x1011 Coloumbs/Kg) of cathode rays remained the same no matter which gas was used in the discharge tube.
- 9. Measurement of e/m Value of Electron In 1897, J.J Thomson devised applied cathode rays to see it’s effect on electric and magnetic fields. • When both the fields are off then a beam of cathode rays, consisted of electrons, produces spot on screen at B. • When only electric field was applied , rays struck floursenece on point A.
- 10. When only magnetic field was applied , rays struck floursenece on point C. When both electric and magnetic field were applied , rays struck floursenece on point B again. In this way by comparing the strengths of the two fields one can determine the e/m value of eIectrons. It comes out to be 1.7588 x 1011 coulombs kg-1. This means that 1 kg of electrons have 1.7588 x 1011 coulombs of charge.
- 11. 5.1.8 Measurement of Charge on Electron - Millikan's Oil Drop Method In 1909, Millikan determined the charge on electron by a simple arrangement. The apparatus consists of a metallic chamber. It has two parts. The upper chamber is filled with air, the pressure of which can be adjusted by a vacuum pump. There are two electrodes A and A’ These electrodes are used- to generate an electrical field in the space between the electrodes. The upper electrode has a hole in it as shown in Fig (5.8). A fine spray of oil droplets is created by an atomizer. A few droplets passes through the hole in the top plate and into the region
- 13. • between the charged plates, where one of them is observed though a microscope. This droplet, when illuminated perpendicularly to the direction of view, appears in the microscope as bright speck against a dark background. The droplet falls under the force of gravity without applying the electric field. The velocity of the droplet is determined. The velocity of the droplet (V1) depends upon its weight, mg. v mg 1a .............. (1) where ’m’ is the mass of the droplet and ‘g’ is the acceleration due to gravity. After that the air between the electrodes is ionized by X-rays. The droplet under observation takes up an electron and gets charged. Now, connect A and A’ to a battery which generates an electric field having a strength, E. The droplet moves upwards against the action of gravity with a velocity (v2). v Ee mg 2a - .............. (2) where ‘e’ is the charge on the electron and Ee is the upward driving force on the droplet due to applied electrical field of strength E. Dividing equation (1) by (2) 1 2 v mg v Ee-mg = .............. (3)
- 14. • The values of v 1 and v2 are recorded with the help of microscope. The factors like g and E are also known. Mass of the droplet can be determined by varying the electric field in such a way that the droplet is suspended in the chamber. Hence ‘e’ can be calculated. By changing the strength of electrical field, Millikan found that the charge on each droplet was different. The smallest charge which he found was 1.59 x 10-19 coulombs, which is very close to the recent value of 1.6022 x 10-19 coulombs. This smallest charge on any droplet is the charge of one electron. The other drops having more than one electron on them, have double or triple the amount of this charge. The charge present on an electron is the smallest charge of electricity that has been measured so far. Mass of Electron The value of charge on electron is 1.602 x 10-19 coulombs, while e/m of electron is 1.7588 x 1011 coulombs kg-1. So, = e 1.6022×10 coulombs -19 = 1.7588x10 coulombs kg 11 -1 = m Mass of electrons Mass of electron= -19 11 -1 1.6022×10 coulombs 1.7588×10 coulombs kg Rearranging Mass of electron= 9.1095x10-31 kg