Download Performing the Millikan Oil Drop Experiment

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Transcript
The Millikan Oil Drop
Experiment
An experiment performed by Robert Millikan in 1909 determined the size of the
charge on an electron. He also determined that there was a smallest 'unit' charge, or
that charge is 'quantized'. He received the Nobel Prize for his work. We're going to
explain that experiment here, and show how Millikan was able to determine the size
of a charge on a single electron.
What Millikan did was to put a charge on a tiny drop of oil, and measure how strong
an applied electric field had to be in order to stop the oil drop from falling. Since he
was able to work out the mass of the oil drop, and he could calculate the force of
gravity on one drop, he could then determine the electric charge that the drop must
have. By varying the charge on different drops, he noticed that the charge was always
a multiple of -1.6 x 10 -19 C, the charge on a single electron. This meant that it was
electrons carrying this unit charge.
Here's how it worked. Have a look at the apparatus he used:
An atomizer sprayed a fine mist of oil droplets into the chamber. Some of these tiny
droplets fell through a hole in the upper floor. Millikan first let them fall until they
reached terminal velocity. Using the microscope, he measured their terminal velocity,
and by use of a formula, calculated the mass of each oil drop.
Next, Millikan applied a charge to the falling drops by illuminating the bottom
chamber with x-rays. This caused the air to become ionized, and electrons to attach
themselves to the oil drops.
By attaching a battery to the plates above and below this bottom chamber, he was able
to apply an electric voltage. The electric field produced in the bottom chamber by this
voltage would act on the charged oil drops; if the voltage was just right, the
electromagnetic force would just balance the force of gravity on a drop, and the drop
would hang suspended in mid-air.
Now you try it. Click here to open a simulation of Millikan's chamber. First, allow the
drops to fall. Notice how they accelerate at first, due to gravity. But quickly, air
resistance causes them to reach terminal velocity.
Now focus on a single falling drop, and adjust the electric field upwards until the drop
remains suspended in mid-air. At that instant, for that drop, the electric force on it
exactly equals the force of gravity on it. Some drops have more electrons than others,
so will require a higher voltage to stop.
When you've finished playing with the apparatus, close the window and we'll
continue.
O.K., let's look at the calculation Millikan was now able to do.
When a drop is suspended, its weight m · g is exactly equal to the electric force
applied q · E
http://www.mdclearhills.ab.ca/millikan/experiment.html
The values of E, the applied electric field, m the mass of a drop, and g, the
acceleration due to gravity, are all known values. So you can solve for q, the charge
on the drop:
http://www.mdclearhills.ab.ca/millikan/experiment.html
Millikan determined the charge on a drop. Then he redid the experiment numerous
times, each time varying the strength of the x-rays ionizing the air, so that differing
numbers of electrons would jump onto the oil molecules each time. He obtained
various values for q.
The charge q on a drop was always a multiple of -1.6 x 10 -19 C, the charge on a
single electron.
This number was the one Millikan was looking for, and it also showed that the value
was quantized; the smallest unit of charge was this amount, and it was the charge on a
single electron.
Performing the Millikan Oil Drop Experiment
Initially, drops fall into the space between the parallel plates with no voltage applied. They
fall and achieve terminal velocity. When the voltage is turned on, it is adjusted until some
of the drops start to rise. If a drop rises, it indicates the upward electrical force is greater
than the downward gravitational force. A drop is selected and allowed to fall. Its terminal
velocity in the absence of the electrical field is calculated. The drag on the drop is
calculated using Stokes Law:
Fd = 6πrηv1
where r is the drop radius, η is the viscosity of air and v1 is the terminal velocity of the
drop.
The weight W of the oil drop is the volume V multiplied by the density ρ and the
acceleration due to gravity g.
The apparent weight of the drop in air is the true weight minus the up thrust (equal to the
weight of air displaced by the oil drop). If the drop is assumed to be perfectly spherical
then the apparent weight can be calculated:
W = 4/3 πr3g (ρ - ρair)
The drop is not accelerating at terminal velocity so the total force acting on it must be zero
such that F = W. Under this condition:
r2 = 9ηv1 / 2g(ρ - ρair)
r is calculated so W can be solved. When the voltage is turned on the electric force on the
drop is:
FE = qE
where q is the charge on the oil drop and E is the electric potential across the plates. For
parallel plates:
E = V/d
where V is the voltage and d is the distance between the plates.
The charge on the drop is determined by increasing the voltage slightly so that the oil drop
rises with velocity v2:
qE - W = 6πrηv2
qE - W = Wv2/v1