# Millikan’s Oil Drop Experiment

## Millikan’s Oil Drop Experiment

Millikan’s Oil Drop Experiment or Determination of charge of electron is conduct by an American Scientist R.A.Millikan, who perform an experiment on the charge on oil drops. R.A.Millikan perform several experiments to calculate charge on oil drops and he gets every time its value equal to -1.6×10-19 coulomb.When these results associated with results of cathode rays then conclude that charge present on particle of cathode rays is -1.6×10-19 coulomb.

#### Calculation of mass of electron

As we  know e/m = -1.76×108 coulombs/gram
e = -1.6×10-19 coulomb
then,
(e/m)/e = (-1.76×108)/(-1.6×10-19)
so,
m = 9.102×10-28 gram
m = 9.102×10-31 kilogram

#### Mass of electron in comparison with atom

Mass of electron in comparison with atom is described below-
Mass of 1 mole of Hydrogen = 1.008gms
Number of hydrogen atom in 1 mole = 6.023×1023
Mass of 1 atom of hydrogen = 1.008/6.023×1023
= 1.67×10-27kg
Mass of electron is 9.109×10-31
then,
=  Mass of 1 atom of hydrogen/Mass of electron
= (1.67×10-27)/(9.109×10-31) = 1837
so,
Mass of an electron is 1/1837 th the mass of a hydrogen atom.
for more chemistry notes related to these topics visit at http://chemistrynotesinfo.blogspot.in/2015/01/class-9th-atomic-structure-part-1.html