Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron.

Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated.
Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by `d_(1) = (eEL^(2))/(2mV^(2))` where L = length of deflecting plate and v = speed of electron.
In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives `eE = evB`. Using expression for ` d_(1)` we can find out `(e)/(m) = (2d_(1)E)/(B^(2)L^(2))`
A beam of electron with velocity `3 xx 10^(7) m s^(-1)` is deflected 2 mm while passing through 10 cm in an electric field of ` 1800 V//m` perpendicular to its path. `e//m` for electron is

Options

A. `1.5 xx 10^(11)Ckg^(-1)`
B. `2 xx 10^(11)Ckg^(-1)`
C. `2.5 xx 10^(11)Ckg^(-1)`
D. `3 xx 10^(11)Ckg^(-1)`

Answer

B

Explanation

Since `d_(1)=(eEL^(2))/(2mv^2) or e/m = (2 d_(1)v^(2))/(EL^(2))`
Here `d_(1)= 2 xx 10^(-3)m`
`v= 3xx 10^(7) m//s L=0.1m`
`E=1800 V//m`
`e/m = (2 xx (2 xx 10^(-3)) xx (3 xx 10^(7))^(2))/(1800 xx (0.1)^(2)) = 2 xx 10^(11) C//kg`.
Hence choice (b) is correct.