J J Thomson and Millikan experiment

Question 5.1:What is J. J. Thomson’s experiment for determining e/m value of electron? Answer:The charge to mass ration(e/m) was determined by J.J. Thomson in 1897. The apparatus consists of a discharge tube shown in fig.

Figure: Measurement of e/m value of electron

The cathode rays can pass through electric and magnetic fields. Electric field is on the plane of paper and magnetic field is perpendicular to paper in the diagram. When both fields are off then beam of cathode rays produce luminous spot at point P1 on the screen. When only magnetic field is on then cathode rays strike at the point p3. When only electric field is applied, cathode rays strike at spot p2. Both electric and magnetic fields are then applied simultaneously, and their strength adjusted in such a way that cathode rays again hit point p1. In this way by comparing the strength of two fields we can determine e/m value of electron. It is 1.7588 x 1011coulombs kg-1.

Question 5.2: Explain Millikan’s oil drop experiment to determine the charge of an electron.
Answer: In 1909, Millikan determined the charge on the electron by oil drop method. The apparatus consists of a metallic chamber filled with air whose pressure can be adjusted by a vacuum pump. There are two electrodes; positively charged A with a hole in it and negatively charged A/. When a droplet of oil created by atomizer pass through the hole of upper plate A/ and fall between illuminated space between A and A/, its velocity (V1) is observed by microscope. Velocity v1 of the droplet of mass m depends upon its weight.


Figure: Millikan oil drop method for determination of charge on electron

After that the droplet takes electron from ionization of air by x-rays and gets charged. Now A and A/ are connected to battery and electric field of strength E is generated between these plates. The droplet moves upwards against the gravity with velocity V2.

Equation Where e is charge on the drop and Ee is the upward driving force due to electric field. Dividing equation (1) by equation (2) we get equation

The values of V1, and V2 are recorded with help of telescope and that of E and g are also known. Hence value of e can be calculated. The smallest value is 1.59 x10-19 which is very close to recent value of 1.6x 10-19. This smallest charge on any droplet is due to addition of single electron to the droplet. So, this is charge on electron. Other drops having more than one electron on them have double or triple amount of this charge.

Question 5.3: Evaluate mass of electron from charge to mass ration and Millikan experiment.
Answer: From Milliken’s experiment, value of charge on electron, e = 1.602 x 10-19 coulombs.

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