The Thomson experiment

Assume you are in charge of conducting Thomson’s famous experiment to find the value of $q/m$ (i.e. the charge to mass of the electron). You have an apparatus in which a beam of electrons is being transported from left to right, finally stopping at a detector. In between, Thompson has set up a region of length $x1$ where an electric field of strength $E (V/m)$ deflects the electron beam by a vertical distance $y1$. After passing the electric field region the beam passes through free space over an axial distance $x2$ till it reaches the detector during which time it’s vertical deflection increases by $y2$.

 Thomson used a magnetic field of strength $B (T)$ -where, T stands for the unit of Tesla, perpendicular to the motion of the electrons to counteract the electric field force such that the deflection could be overcome; this was done to help find the velocity of the electrons in the horizontal direction. Knowing the above,

(i) Find the velocity of the electron;

(ii) Show how you would evaluate for the value of $q/m$ from knowing: $x_1$, $x_2$, $y1+y2$, and E;

(iii) If $E=4 kV/m$; $B=1.4*10^{-4}T$; $x1=4 cm$; $x2=30 cm$, $y1+y2=1.1cm$, calculate the value of $v$ and, $q/m$. (20 points).

Initially the magnetic force = electric force for the charge to pass undeflected.

$qVB =qE$ that is $V= E/B$

The figure is below (there exists only electric field E)

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In the deflection space

$F =m*a =qE$   that is $a =(qE)/m$

$Y1 =(a*t1^2) /2$

Entering velocity is $V =V0x$ (charge enters parallel to the plates with horizontal speed V)

$X1 =V*t1$   that is $Y1 = (a/2)*(X1/V)^2 = (qE/2m)*(X1/V)^2$

which means that trajectory inside deflection unit is a parabola

Vertical speed achieved at exit of deflection unit is

$Vy =a*t1 =(qE/m)*(X1/V)$

time to travel $X2$ distance $t2 =X2/V$

deflection $Y2$ is $Y2 =Vy*t2 =( qE/m)*(X1/V) *(X2/V)$

Total deflection is

$Y1+Y2 =(qE/m) [X1^2/2 +X1*X2]/V^2$

$(q/m) = [V^2/E] * [(Y1+Y2) / (X1^2/2 +X1*X2)]$

By knowing initial velocity $V=V0x$, electric filed intensity $E$ and deflections $X1$,$X2$ and $Y1+Y2$ you can compute $(q/m)$

With the given values we have:

$V =E/B =(4*10^3)/(1.4*10^{-4}) =2.857*10^7 m/s$

$(q/m) =1.754*10^11 C/kg$    this is specific $q/m$ for ELECTRON