Chemistry · Structure Of Atom · NEET
It found the value of the charge on one electron: -1.6 x 10^-19 C (the modern accurate value is -1.602 x 10^-19 C). Before this, only the charge-to-mass ratio (e/m) was known from J.J. Thomson. Once Millikan gave the charge (e), scientists could combine it with e/m to also find the mass of the electron. So this experiment completed the picture of the electron.
Water drops evaporate very fast. If the drop changes size or mass during the reading, the calculation goes wrong. Oil evaporates very slowly, so the drop stays almost the same during the experiment. That is why Millikan chose fine oil droplets made by an atomiser (sprayer).
He passed X-rays through the air inside the chamber. The X-rays knocked electrons off gas molecules, making gaseous ions. As the oil drops fell, they collided with these ions and picked up charge. So each oil drop ended up carrying one or more extra electrons.
First he let a drop fall freely and measured its speed of fall through air, which gave the mass of the drop. Then he switched on an electric field between two charged plates. He adjusted the field until the upward electric force exactly balanced the drop's weight, so the drop stayed still (floated). At balance, the electric force = weight of drop. From this he could work out the charge on that drop.
Quantisation means charge only comes in whole-number packets, never in a fraction. Millikan found that the charge on every drop was always a whole-number multiple of one basic value: q = n x e, where n = 1, 2, 3... and e = 1.6 x 10^-19 C. He never got a value smaller than e. This proved that e is the smallest unit of charge and all charge is 'n times e'.
Thomson used a cathode ray (discharge) tube and measured the charge-to-mass ratio (e/m) of the electron, not the charge alone. Millikan used oil drops and X-rays to measure the actual charge (e) of the electron directly. Thomson = e/m ratio; Millikan = e (the charge itself). Together they gave the electron's mass.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
-1.6 x 10^-19 C (Millikan's value). The present accepted value is -1.602 x 10^-19 C. For NEET numericals, use 1.6 x 10^-19 C.
R.A. Millikan performed it between 1906 and 1914.
Use q = n x e. n = 3.2 x 10^-19 / 1.6 x 10^-19 = 2. So the drop carries 2 extra electrons.
It gives the exact charge on the electron and proves charge is quantised (q = ne). NEET often asks 'find the number of electrons on a drop' using q = n x e, so knowing e = 1.6 x 10^-19 C is a must.
By combining Millikan's charge (e) with Thomson's e/m ratio: mass = e divided by (e/m). This gave the electron mass as about 9.1 x 10^-31 kg.
