Millikan’s Oil Drop Experiment

Electron is a fundamental sub atomic particle which has a charge equal to 1.6×10-19 C.  If ‘e’ is the charge of an electron then the smallest possible charge is ‘e’ and all the charges are equal to the whole number multiple of this charge i.e. e, 2e, 3e etc. At first the charge of an electron was measured by R. A. Millikan using a method which is known today as Millikan’s oil drop experiment.

Millikan’s Oil Drop Experiment

Millikan’s Apparatus

Millikan’s apparatus for the determination of electronic charge consists of a double walled chamber inside which two discs of radius about 20 cm are placed one above another at a separation about 1.5 cm. A hole is drilled above the upper disc through which oil is sprayed by the help of atomizer. The apparatus consists of three windows W1, W2 and W3. Through W1 light passes which illuminates the oil drop charged by X-rays passing through W2. A travelling microscope is fitted on W3 which helps to measure the terminal velocity of the falling drop.

The experiment is carried on two steps:

1. In the absence of electric field: In this method, electric field is switched off and oil is sprayed by atomizer. The oil drop spontaneously attains its terminal velocity due to its small size. Let the terminal velocity is V1. There exists three forces, weight (mg) acting downwards whereas up upward force or thrust (U) and viscous force (Fv) acting upward.

In this motion,

Net downward force = net upward force

i.e. mg = U + Fv

or, vρg = vσg + 6πηrV1

or, v(ρ-σ)g = 6πηrV1………………(1)

Where,

V = volume of oil drop

ρ = density of oil

σ = density of air

η = coefficient of viscosity of air

r = radius of oil drop

From equation (1),

$${\frac{4}{3}\pi \text{r}^3 (\rho-\sigma)\text{g} = 6 \pi \eta r\text{V}_1}$$

$${r = \sqrt{\frac{9\eta{\text{v}_1}}{2(\rho – \sigma){\text{g}}}}}………(ii)$$

Knowing the value of ρ, σ, η, V1 and g, the radius of the drop can be calculated.

2. In the presence of electric force:

Motion of oil drop under electric field

Now, the electric field is switched on. The charged oil drop due to X-rays experience electric force qE acting upward. Since the upper plate is made (+)ve. Now the oil drop moves with different terminal velocity i.e. V2. If the electric force is enough the drop may move upward. Otherwise it will move downward with lesser velocity.

In this case also,

Net downward force = net upward force

i.e., mg = qE + U + FV

or, vρg = qE + vσg + 6πηrV2o

or, V(ρ-σ)g = qE + 6πηrV2

or, 6πηrV1 =  qE + 6πηrV2

or, qE = 6πηr (V­1-V2)

or, $${q = \frac{6 \pi \eta \text{r}(V_1 – V_2)}{E}}$$

Putting the value of r from equation (ii), we get:
$${q =\frac{6 \pi \eta \text(V_1 – V_2)}{E}
\sqrt{\frac{9\eta{\text{v}_1}}{2(\rho – \sigma){\text{g}}}}}$$

If the drop moves upward, then:

$${q =\frac{6 \pi \eta \text(V_1 + V_2)}{E} \sqrt{\frac{9\eta{\text{v}_1}}{2(\rho – \sigma){\text{g}}}}}……(iii)$$

By knowing all the quantities of left hand side, the value of q can be determined.

This experiment is repeated for number of times and it is found that the charge ‘q’ is always the integral multiple of 1.6×10-19. Hence we  conclude that he minimum possible charge is 1.6×10-19 C and it is the charge of the electron.

From Milikan’s oil drop experiment:

  • Quantum Nature of charge (Q =ne) was established.
  • It helps to find the mass of an electron.
  • It help to find the radius of electron.
  • It showed that the electronic charge was the smallest possible charge in charged particles.

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2 Responses to “Millikan’s Oil Drop Experiment”

  1. nareshdevkota

    damiii xa aru hseb ko imp qn post garanu parxa …

  2. Niraj Yadav

    It is very good.