Physics-
General
Easy
Question
The following experiment was performed by J.J.Thomson in order to measure the ratio of the charge e to the mass m of an electron. Figure shows a modern version of Thomson's apparatus. Electrons emitted from a hot filament and accelerated by a potential difference V. As the electrons pass through the deflector plates, they encounter both electric and magnetic fields. When the electrons leave the plates they enter a field–free region that extends to the fluorescent screen. The beam of electrons can be observed as a spot of light on the screen. The entire region in which the electrons travel is evacuated with a vacuum pump.
Thomson's procedure was to first set both the electric and magnetic fields to zero, note the position of the undeflected electron beam on the screen, then turn on only the electric field and measure the resulting deflection. The deflection of an electron in an electric field of magnitude E is given by 
, where L is the length of the deflecting plates, and v is the speed of the electron. The deflection d1
can also be calculated from the total deflection of the spot on the screen, 
, and the geometry of the apparatus. In the second part of the experiment Thomson adjusted the magnetic field so as to exactly cancel the force applied by the electric field, leaving the electron beam undeflected. This gives 
By combining this relation with the expression for d1 one can calculate the charge to mass ratio of the electron as a function of the known quantities. The result is 
Why was it important for Thomson to evacuate the air from the apparatus ?
- Electrons travel faster in a vacuum, making the deflection
smaller
- Electromagnetic waves propagate in a vacuum
- The electron collisions with the air molecules cause them to be scattered, and a focused beam will not be produced
- It was not important and could have been avoided
smaller
The correct answer is: The electron collisions with the air molecules cause them to be scattered, and a focused beam will not be produced
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