MICHAEL FARADAYIn 1838, Michael Faraday passed current through the glass tube filled with rarefied air. Conducting the experiment he noticed a strange light arc with its beginning at the anode (the positive electrode) and its end almost at the cathode (the negative electrode). The only place where there was no luminescence was just in front of the cathode. The place is called "cathode dark space", "Faraday dark space", or "Crookes dark space". That was the beginning of the long and "turbulent" time of researches on that luminescence. And the luminescence is called "cathode rays".
In this experiment, Faraday showed that the mass of elements was proportional to the quantity of electricity that passed through them.2. George J. Stoney- he was a scientist who first suggested the term “ Electron” to refer to a negatively charged particle.
3. Joseph John Thomson-was credited for his discovery of the first subatomic particle through his work with the discharge tube.
4. Robert A. Millikan- independently measured the electrons charge through his OIL-DROP Experiment. He successfully attempted to detect and measure the effect of an individual subatomic particle.
Using an atomizer, Millikan sprayed tiny droplets of oil which passed between two charged plates. Given a negative charge(-) by the electrons released from gas particles by x-rays, the oil particles were attracted toward the positive plate. He observed the speed of the droplets as they moved toward the positive plates. The smallest possible charge on a droplet was taken as the charge of an individual atom.
The quantity is considered to be the basic unit charge and is given a value of one minus (1-)
When the electron was discovered , the scientists searched for the positively charged particles.1.Eugene Goldstein-
Goldstein observed rays passing through the hole of a cathode in the cathode tube. These rays are made up of positive particles. Their characteristics depend on the gas inside the cathode tube.The lightest particle was obtained when Hydrogen gas was used.
1. Sir James Chadwick- discovered the neutron in 1932. He found out that the particle was electrically neutral an and its mass was approximately the same as that of the proton.
Proton Symbol p+Position in atom Inside the nucleusRelative charge 1+Relative mass 1Actual mass ( g) 1.673x10-24
Neutron Symbol nPosition in atom Inside the nucleusRelative charge 0Relative mass 1Actual mass 1.675x10 -24
Electron − Symbol ePosition in atom moving around the nucleusRelative charge -1 1 Relative mass ≈0 1840 Actual mass 9.109x10-28
Atomic structure e - - - - - - - - electrone n (s) p (s) - - - - - - - - - - - - - - - nucleus e e e
Atomic content of an atom 11e − no. of protons = 11 no. of electrons =11 12n 11p no. of neutrons = 12 23 Symbol 11 Na
12Atomic Content of 6 C no. of protons = 6 no. of neutrons = 12 - 6 =6 no. of electrons = 6 (An atom is neutral in charge)
14Atomic Content of :7 N no. of protons = 7 no. of electrons = 7 no. of neutrons = 14 – 7 =7
14Atomic Content of 6 C no. of protons = 6 no. of neutrons = 14 – 6 =8 no. of electrons = 6
mass number 14 12 6 C 6 C atomic numberMass number = no. of p + no. of natomic number = no. of protons
Isotopes 12 14 6 C 6 C Name of element : carbon carbon No. of protons : 6 6 No. of neutrons : 8 6 No. of electrons : 6 6Isotopes are atoms of same elements which have the sameatomic number but different mass number
Isotopes 14 12 6 C 6 CBurn in air Carbon dioxide Carbon dioxide Mass 14 units 12 unitsRadioactivity Give radiation Does not give radiation (stable)
Extension : Isotopes ( For reference only ) 8n 6p 7n 7p + e +γ − 0 1 14 14 6 C nucleus 7 N nucleusA neutron in a nucleus of 14 C breaks into 1 p + e −. 1Electrons and energy are emitted to the surroundings Carbon -14 dating
Properties of isotopes Same chemical properties Different physical properties e.g. mass b.p. and m.p. radioactivity
Actual mass VS relative mass 12Mass of 1 6 C atom = 1.993 ×10 −26 kg Or 12Which one is more convenience?Relative mass mass of 1 12C 6Standard reference : 1 is equivalent to 12 = 1.661 × 10 − 27 kg
Simplified rule : We regard Relative mass of 1p = 1 Relative mass of 1n = 1 Relative mass of 1e = very small =0
Relative mass of isotopes 28 29 30 14 Si 14 Si 14 SiRelative mass 28 29 30% abundance 92% 4.7% 3.1%Do the three isotopes have the same chemical properties? Yes
Relative atomic mass of isotopes It is difficult and not necessary to separate the isotopes in most of the reactions. Average relative mass of silicon is used to denote the mass of a silicon atom.
Relative atomic mass of silicon ( P.5 e.g.1 ) 28 29 30 14 Si 14 Si 14 SiRelative mass 28 29 30% abundance 92% 4.7% 3.1% ( Relative ) Atomic Mass of silicon = 28 x 92% + 29 x 4.7% + 30 x 3.1% = 28.053 = 28.05
Ex1. Relative atomic mass of Chlorine 35 37 17 Cl 17 Cl Relative mass 35 37 % abundance 75.5% 24.5%Atomic mass of Chlorine = 35 x 75.5% + 37 x 24.5% = 35.5