Oil Drop Experiment by Millikan

Rober Andrews Millikan measured electric charge on an electron in 1909 by his famous Oil Drop Experiment. The apparatus used in this experiment is shown in below.

In this experimental setup we have

  • Cylindrical vessel A
  • Two fixed brass plates C and D, the upper plate C had a pin hole at its centre
  • A sprayer (B) to produce very fine droplets of oil
  • Windows W1 and W2 fitted into opposite walls of vessel A
  • A manometer (pressure measuring instrument) and a vacuum pump

So far we have looked at setup of oil drop experiment. Let’s now see how actually it works.

  • Small oil drops were sprayed with help of sprayer (B)
  • These drops of oil were allowed to fall in between plates C and D
  • Oil drops in between two plates can be observed through window W2 with help of a microscope
  • Gas inside chamber is little bit ionised by using a beam of X-rays
  • Electrons or ions released during ionisation may be captured by falling oil drops
  • Thus some oil drops may become charged
  • Then rate of fall of a drop under the action of gravity is measured
  • The rate of fall of a drop depends upon its size, so from rate of fall we can determine size of a drop
  • When plates C and D are charged to a steady potential, some of drops which carry no charge continue to fall as before but other drops which carry charge (due to capture of electrons or ions) change their speed and may start moving up towards plate C.
  • At a certain strength of electric field, two forces (force of gravity and electric field force) on drops become equal and thus drops stop moving, becoming steady in between two plates.
  • From all these observations, we can calculate charge on an oil drop

Let’s now see maths involved in this experiment.

Rate of fall of a drop under gravity = v1
Then v1 ∝ m g

Where
m = mass of drop
g = acceleration due to gravity

If drop moves upwards against force of gravity with a velocity v2 then

v2 ∝ (Upward force due to electric field – Downward force due to gravity)

Upward force due to electric field = E q
Where E = Strength of electric field and q = charge on oil drop
v2 ∝ E q – mg

So we have v1 ∝ m g
v2 ∝ E q – mg

Dividing both of these equation v2/v1 = E q – mg / mg = E q/mg – 1

v2/v1 = E q/mg – 1

q = mg/E (v2/v1 + 1)

If we assume that drops are spherical then m = 4𝛑 r3p/3
Where
r = Radius of spherical drop
p = Density of oil

Then
q = 4𝛑 r3pg/3E (v2/v1 + 1)

In various experiments with different oil drops charge on drops was found to be in multiple of 1.602 × 10-19 coulomb because one oil drop may contain may electrons. Thus lowest values of the charge on any oil drop was found to be 1.602 × 10-19 C.

From this Millikan concluded that smallest charge on a oil drop should be equal to the charge of electron, hence charge of an electron is 1.602 × 10-19 C.

Related Posts

Leave a Reply

Your email address will not be published.