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FSc Notes Chemistry Part 1 Chapter 5 Atomic Structure Lecture 2

FSc Notes Chemistry Part 1 Chapter 5 Atomic Structure Lecture 2

Charge on an Electron:

Charge on an electron was determined by "Millikan" in 1909. The experiment conducted for determining the charge of an electron is known as "Millikan's" Oil drop method. The apparatus used, consists of a chamber, whose base is a good conductor. Let’s call it as "C" a partitioning is made in the chamber by placing another conduction plate in the middle of the chamber. Let’s call this conducting plate as “B". Both B & C are parallel to each other. A small hole is present in plate “B" ( upper plate ). There are two windows, w1 & w2 in the lower compartment of the chamber. Through w1, X— rays can be passed into the region between the plates. In w2, a telescope is placed, with the help of which the process taking place between the two metallic plates can be observed. A fog of very small droplets of oil is introduced into the upper compartment by an atomizer “A ". The droplets are allowed to fall from plate " B " to "C" through the hole in "B" under the influence of gravity and their velocity "V" is determined through telescope.

I.e V1 = mg or V1 = Kmg -------I

Now X—rays are passed through the air, present in the lower compartment and thus the air is ionized because electrons are brought out of air molecules by x—rays. These electrons are picked up by the oil droplets and thus they become negatively charged. Now the two plates "B" and "C" are connected to a battery such that the upper plate "B" is made as anode while "C" as cathode. The negatively charged oil droplets are thus repelled by plate "C" (cathode) and attracted by plate B (anode). Consequently the oil droplets start moving in the upward direction against the gravity. Thus the upward velocity " V2" of the oil droplets is also determined by telescope. i.e
V2 = (Ee – mg)
V2 = K (Ee – mg) ------------II
Now '/' eq I by II
V1/V2 = Kmg/k (Ee – mg)
V1 (Ee – mg) = V2 mg
V1Ee- V1mg=V2mg.
V1Ee = V2 mg +V1mg
V1Ee = mg (V1 + V2)
E = mg (V1 + V2)/V1E

Values of V1, V2, E & g were known. The mass of droplets was determined by suspending them in between the two plates by varying the strength of electric field frequently. By putting all the values, Millikan, finally succeeded in getting the quantity of charge (e) on an electron. The value of charge (e) of electron was calculated to be "1.06022x10-19 coulombs".

Mass of Electron:

From the values of charge to mass ratio ( e/m ) and charge (e) of electron, the mass of an electron can be calculated as:
As e/m = 1.7588x10(11) C /Kg -------------I
E = 1.6022x10(-19) C
Putting the value of "e" in eq I
1.6022x10(-19) C/m = 1.7588x10(11) C x Kg-1
M x 1.7588x10(11) C x Kg-1 = 1.6022x10(-19) C
M = 1.6022x10(-19) C/1.7588x10(11) C x Kg-1
M= 9.1096x10(-31) Kg
In relative mass unit
( a. m . u ) = 9.1096x10(-27)g/ 1.66x10(-24) = 0.00055 a. m .u

Discovery of Proton (Canal Rays):

The positive rays were first determined by Eugene Goldstein in 1866. He used a special type of discharge tube having a perforated cathode. He observed that some rays were moving from anode towards cathode. They were observed by producing fluorescence on the glass behind cathode. Actually these rays pass through the pore or canals of cathode and strike with the glass producing fluorescence. These rays named as “canal rays " or " positive rays".

Characteristics of Canal Rays:

Some important characteristics of canal rays are as follow.
  1. They travel in straight path.
  2. They are deflected by the electric and magneti8c fields in the opposite direction to that of cathode rays.
  3. The charge to mass ratio of these rays is considerably smaller than that electrons.
  4. The charge to mass ratio of canal rays depends upon the nature of the gas used the highest e/m is observed for hydrogen gas.
  5. The mass of the + ve charge is 1.6726x10(-27) , i.e. 1836 times the mass of electron.


The highest and simplest canal rays are formed when a gas discharge tube contains H2 gas. The positive charge of this hydrogen was found to be 1.6022x10(-19) coulomb, which is just equal to that of an electron. From the charge to mass value of hydrogen the mass of hydrogen was calculated to be 1.6726x10(-27). The positive particles of canal rays of hydrogen are now called as “Proton ".

Written by: Asad Hussain