Q.In Thomson's experiment for the determination of $(e/m)$ of electron, electric field of intensity 'E' only is applied. Now the deflection produced by the electron beam is directly proportional to :

The deflection of the electron beam, with only the electric field $E$ applied, is directly proportional to $E$.

In Thomson's $e/m$ experiment, when a transverse electric field of intensity $E$ alone is applied between the deflecting plates, each electron in the beam experiences a constant transverse electric force $F = eE$, producing a transverse acceleration $a = \dfrac{eE}{m}$.

If the electrons enter with a fixed horizontal speed $v$ and traverse plates of length $l$, the time spent between the plates is $t = \dfrac{l}{v}$. The vertical (transverse) deflection acquired while crossing the plates is:

$y = \dfrac12 a t^2 = \dfrac12 \cdot \dfrac{eE}{m}\cdot\left(\dfrac{l}{v}\right)^2$

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Specific charge $q/m$ is largest for the lightest particle (electron) and smallest for the heaviest (deuteron), giving the increasing order deuteron, proton, electron.

Specific charge (charge-to-mass ratio) is defined as $\dfrac{q}{m}$, and it is exactly this ratio that a Bainbridge-type mass spectrometer or a Thomson $e/m$ apparatus measures via the radius of circular deflection in a magnetic field ($r = \dfrac{mv}{qB}$, so $\dfrac{q}{m}=\dfrac{v}{rB}$).

For the electron, the charge magnitude is the elementary charge $e = 1.6\times10^{-19}$ C, and its mass is extremely small, $m_e = 9.11\times10^{-31}$ kg, giving a specific charge $\dfrac{e}{m_e}\approx 1.76\times10^{11}$ C/kg — by far the largest of the three, because the electron is thousands of times lighter than the proton for the same magnitude of charge.

For the proton, the charge is also $+e$, but the mass is $m_p = 1.67\times10^{-27}$ kg (about 1836 times the electron's mass), giving $\dfrac{e}{m_p}\approx 9.58\times10^{7}$ C/kg.

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