[NOTES/EM-01003]-Thomson’s parabola method

In the parabola method the electrons from an electron gun are passed through a uniform electric field, perpendicular to the direction of motion, taken as \(X\)- axis, in the Figure 1. 

While in electric field, the electrons travel a distance \(\ell_1\) in the \(X\)- direction and follow a parabolic path. They suffer a deflection \(\Delta_1\) along the \(Y\)- axis at the end of region II. The electrons then travel through a region having no field and hit a screen at a distance \(\ell_2\). In this region there is no force acting on the electrons and they follow a straight line path. The charge to mass ratio of the electrons is then determined in terms of the electric field and other parameters \(L_1, L_2\) and the upward deflection registered on the screen. We will now obtain an expression for the charge to mass ratio \(e/m\).

Let the the initial velocity of electrons be $v_0$ before the they enter the region I. In region $\text{I}$ the electrons experience an electric force $eE$ and the trajectory is a parabola. While in region \(\text{II}\) they do not experience any force and the trajectory is a straight line .

Motion in region I

\begin{align*} x=v_0t~,~y=\frac{1}{2} at^2\\ ma = F =eE\\ a = \frac{eE}{m} \Rightarrow y=\frac{1}{2}\frac{eE}{m} t^2 \end{align*} If $(x_1,y_1)$ are the coordinates at the end of region I and \(t_1\) denotes the time spent in region I, we have \begin{align*} x_1=v_0t_1\qquad\qquad y_1=\frac{1}{2}\frac{eE}{m} t^2_1\\ y_1 = \frac{eE}{2m} \frac{\ell^2_1}{v_0^2} = \frac{eE\ell_1^2}{4K_0} \equiv \Delta_1 \end{align*} where $K_0$ is the initial $K.E. =\frac{1}{2} mv_0^2$.

Motion in region II

The slope of the linear trajectory in region II is given by \begin{align*} \frac{dy}{dx} = & \left(\frac{dy}{dt}\right)\bigg/ \left(\frac{dx}{dt}\right)\biggl|_{t=t_1} = \left(\frac{eE}{m}t_1\right)/v_0\qquad\quad t_1=\frac{\ell_1}{v_0}\\ =&\frac{eE}{m}\frac{1}{v_0} \frac{\ell_1}{v_0} =\frac{eE}{mv_0^2} \ell_1\\ =&\frac{eE}{2K_0} \ell_1. \end{align*} If the distance of screen from the capacitor plates is \(\ell_2\), then the slope of the trajectory in region II is $\tan\theta={\Delta_2}/{\ell_2}$ and therefore \begin{equation} \frac{eE}{2K_0} \ell_1 = \frac{\Delta_2}{\ell_2}. \end{equation} The electric field is \(E=V/d\) where \(V\) is potential difference between the plates and \(d\) is their separation. Thus the deflection \(\Delta_2\) in region II becomes \begin{align*} \Delta_2 = & L_2\tan\theta = \frac{eE}{2K_0} \ell_1 = \frac{eV}{2K_0} \left(\frac{\ell_1}{d}\right)L_2 \end{align*} Total deflection on the screen, \(\Delta\) is \begin{eqnarray} \Delta &=& \Delta_1+\Delta_2\\ &=& \frac{e V}{4dK_0}\big(\ell_1^2 +2 \ell_2 \ell_1 \big). \end{eqnarray} Substituting \(4K_0=2mv_0^2\), we get \begin{equation} \frac{e}{m}= \frac{2v_0^2}{V}\frac{d\Delta }{(\ell_1^2 + 2\ell_2\ell_1)}. \end{equation}