not just described, Explained

Millikan's oil drop experiment theory

The oil drop wasn't normally stationary

Millikan’s experiment would be fairly straightforward if all you needed to do was to suspend an oil drop in an electric field.

If the oil drop is stationary then only two forces act on the drop.  The electric force acts up and the gravitational force acts down.  These forces must be balanced if the drop is not moving.

We can fairly easily come up with an expression for the charge due to the excess electrons, which is what we want to find.

The problem is that you need to know the weight of the oil drop (which is the gravitational force) and for this you need to find the mass.

Getting a value for the mass of the oil drop was the problem

The real problem is getting a value for the mass of the drop.  The oil drops are so tiny that they just appear like pin-pricks of light even under a microscope so you can’t do anything to measure them.  And it’s finding their mass that leads us to endless hassles, which we’ll just give you a flavour of rather than going into all the details.

It turns out that the best way to get a value for the mass of the drop is to allow the drops to move.

The important thing to note is that the drops have a very low terminal velocity.  (It may take 10 seconds just to travel one millimetre.)  In other words they only need to be going very slowly before the air resistance force balances either the weight, or the net force up when the field is on.  The time that the drops take to accelerate to their terminal velocity is of the order of milliseconds so you can’t easily observe it.

As far as the experimenter is concerned the drops move up at a constant velocity or down at a constant velocity.

Considering the forces as the drop goes up and down

Let’s look at the case where the field is off and the drop we’re following falls at a constant velocity.  In this case weight equals drag.  When we turn the field on the drop moves up with a different constant velocity.

The drag always acts in the opposite direction to the movement of the drop and the drag is different because the speed is different.

If we write down the equation for balanced forces now we end up with an equation for the charge on the droplet in terms of the falling velocity (when the electric field is off) and the rising velocity (when the electric field is on).

The experiment is all about finding velocities of the drop

The velocities are what we actually find during the experiment by timing the drops over a given distance as they move up and down.  But we still need to find a value for the mass of the drop.  Everything else we can measure fairly easily.

Calculating the mass of the drop using Stokes' law

We can calculate the mass from its volume and density but how do you get the volume if you can’t measure the size of the drop?

In 1851 George Stokes had found an equation predicting the speed at which small spheres fall through fluids e.g. air.  We can use Stokes’ law to derive an expression for the radius of the oil drop.

If we know the radius of the drop we can calculate its volume and once we know that we can use the density to calculate the mass, which is what we're desperately trying to find.

The radius calculation depends on the viscosity of the air.  Viscosity is a measure of how easily a fluid pours.  Honey is viscous but water isn’t.  Air has a very low viscosity.

Millikan needed to use a fiddle factor for Stokes' law to work

Unfortunately the oil drops are too small for the Stokes equation to work well.  They tend to spend some time falling between the air molecules.  Millikan had to make an educated guess as to how it should be adjusted (by using a fiddle factor) and he came up with this expression for the radius of the oil drop.

The viscosity of air is very sensitive to temperature so Millikan used an oil bath to keep it within 0.02 C and also monitored atmospheric pressure.

Millikan adjusted the fiddle factor so he got the most consistent value for the charge on the electron for drops of different radius.

Millikan's extraordinary achievement

So from the radius of the oil drop he could calculate its volume and mass and at last use this equation to get a value for the charge.

Remember Millikan was trying to detect a charge due to just a few electrons, which is an extraordinary achievement.

back to Lesson 8: Atoms 2: Electrons