# Thread: Fsc 2nd Year chapter No :1 | ELECTROSTATICS

1. ## Fsc 2nd Year chapter No :1 | ELECTROSTATICS

Fsc 2nd Year chapter No :1 | ELECTROSTATICS

Topic No :12

CHARGE ONA AN ELECTRON BY MILLIKAN'S METHOD

The Millikan Oil Drop Experiment was designed to obtain the charge of an electron which would then allow for the calculation of the electronâ€™s mass. J.J. Thomson had been using an almost identical experiment to that of the Millikan one except he had been using a vapor cloud for the suspension measurement instead of an oil droplet. The water vapor made the experiment too difficult because it evaporated quickly.
So Robert A. Millikan of the University of Chicago used droplets of oil for the suspension measurement.
Experiment
The way this experiment works is that a droplet of oil will come out of an atomizer and go through a tiny slit in an electrode. From the electrode the droplet will pass into a chamber with an electrode parallel from the electrode it had just passed. In this chamber Millikan was able to balance or suspend the droplet by the charge the droplet had picked up when passing through the air. Usining the amount of voltage needed to suspend the droplet he could then calculate the charge the the droplet.

Click on image to view animation
Analysis
Millikan calculated the force gravity would have on the droplet and then equaled that to the observed amount of charge that was required to suspend the droplet to determine the absolute charge of a particular droplet. When Millikan calculated the force gravity had on the droplets he very soon realized that the amount of charge on each minutely affected the velocity of the fall. This was proof of a very small electron mass. Millikan also noticed that the absolute charge of each droplet was a multiple of a smallest quanta of charge, 1.6 X 10-19 coulombs. He assumed this to be the charge of a single electron. Millikan could now calculate the mass of an electron with Thomsonâ€™s ratio of charge to mass (qe/m = 1.76 X 1011 coul/kg)
m = 1.6 X 10-19 coul
1.76 X 1011 coul/kg
m = .091 X 10-30 kg

Conclusion The electron is apprently has a very small mass. The mass of the electron is 1836 times smaller than that of the hydrogen ion. If you multiply 1836 (.091 X 10-30 kg) you get 1.66 X 10-7 kg which is approximately the value of one atomic mass unit.

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