next up previous
Next: Rutherford's experiment and the Up: lecture_3 Previous: lecture_3

J.J. Thomson's experiment and the charge-to-mass ratio of the electron

The first is the experiment of Joseph John Thomson, who first demonstrated that atoms are actually composed of aggregates of charged particles. Prior to his work, it was believed that atoms were the fundamental building blocks of matter. The first evidence contrary to this notion came when people began studying the properties of atoms in large electric fields.



If a gas sample is introduced into the region between two charged plates, a current flow can be observed, suggesting that the atoms have been broken down into charged constituents. The source of these charged particles is a heated cathode that, in fact, causes the atoms of the sample to ionize. These were known as cathode rays. In 1897, Thomson set out to prove that the cathode rays produced from the cathode were actually a stream of negatively charged particles called electrons. (See Figure 1.8 in the textbook for Thomson's experimental setup). From Maxwell's theory, he knew that charged particles could be deflected in a magnetic field. A schematic of the experimental setup is shown below:

Figure 1: Schematic of J.J. Thomson's experiment.
\includegraphics[scale=0.5]{JJThomp.eps}

We now zero in on the field region and set up a coordinate system as shown in the figure below:

Figure 2:
\includegraphics[scale=0.6]{lec3_fig1.eps}

In this coordinate system, electrons enter the region between the plates with an (unknown) velocity $v$ in the $x$-direction. In order to determine this velocity, electric and magnetic fields are both applied, and each gives rise to a force on the electron. These forces are in the $y$-direction. The electric force $F_E = eE$, where $E$ is the magnitude of the electric field, and the magnetic force is $F_H = -evH$, where $H$ is the magnitude of the magnetic field, and is opposed to the force on the electric field.

If these forces balance, then there will be no deflection of the electron in the $y$-direction, i.e. all of the electrons' motion will be along the $x$-direction, which was the initial direction when they entered the field region. If the forces balance, then the total force on the electrons will be zero, that is $F_E + F_H = 0$ or

\begin{displaymath}
eE - evH = 0
\end{displaymath}

from which the unknown velocity can be determined as

\begin{displaymath}
v = {E \over H}
\end{displaymath}



Next, the magnetic field is switched off, so that the total force is due entirely to the electric field. Since the force is non-zero, if the charge carriers can be deflected by the force, this provides evidence for their being fundamental particles. If they are fundamental charged particles, then they should have a well defined mass and charge. In this second part of the experiment, the specific trajectory followed by the particle will be used to determine the ratio of the charge to the mass of the particle.



When there is only an electric field, then there is a nonzero force $F_E = eE$ in the $y$-direction but no force in the $x$-direction. Thus, this problem is exactly the same as that of a projectile in a gravitational field. As can be done in the projectile problem, the $x$ and $y$ motion of the electrons can be analyzed separately and independently.



In the $x$-direction, the motion is very simple because there is no force in this direction. The electrons simply move with a constant velocity $v$, which we already determined has the value $E/H$. Note that this value is correct even though there is no magnetic in this part of the experiment! It is just the velocity we determined from the previous part of the experiment, and this value has not changed. Thus, as a function of time $t$, the $x$-position of the electrons is

\begin{displaymath}
x(t) = vt = {E \over H}t
\end{displaymath}



The force in the $y$-direction is a constant, hence motion in the $y$-direction is analogous the gravitational force. The constant force $F_E$ gives rise to an acceleration $a = F_E/m$, and the $y$-position at time $t$ is then

\begin{displaymath}
y(t) = {1 \over 2}a t^2 = {1 \over 2}{F_E \over m}t^2 = {eE \over 2m}t^2
\end{displaymath}

The electric field is tuned such that the particle traverses the entire plate region in the time required for it to strike the positive plate. Let the total $y$ distance travelled be $s$, as shown in the figure. The time $T$ required to traverse the plate region ($x(T)=l$) is
$\displaystyle l$ $\textstyle =$ $\displaystyle {E \over H}T$  
$\displaystyle T$ $\textstyle =$ $\displaystyle {lH \over E}$  

This is also the time required to move a distance $s$ in the $y$ direction:

\begin{displaymath}
s = {eE \over 2m}\left({lH \over E}\right)^2
\end{displaymath}

Solving the above for the ration $e/m$ gives

\begin{displaymath}
{e \over m} = {2sE \over l^2 H^2}
\end{displaymath}

Thus, using his experimental apparatus, Thomson was able to determine the charge-to-mass ratio of the electron. Today, the accepted value of $e/m$ is 1.7588196$\times$10$^{11}$ C$\cdot$kg$^{-1}$.



In 1906, Robert Millikan was able to determine the value of the charge on the electron in his ``oil drop'' experiment. A schematic of his experiment is shown below:

Figure 3: Schematic of Millikan's experiment.
\includegraphics[scale=0.5]{Milliken.eps}
Then using Thomson's value of $e/m$, he calculated the value of $m$. In his experiment, Millikan used a fine spray of ionized oil droplets, which he allowed to be acted on by gravity but to which he also applied an electric field in the direction opposite gravity, i.e. up. By tuning the electric field, he balanced the force due to the pull of gravity and the electric field force so that the drops remained suspended in space. Thus, if a drop has a positive charge $q$ and a mass $M$, the force balance condition becomes

\begin{displaymath}
Mg = qE
\end{displaymath}

where $E$ is the magnitude of the electric field. Solving for the charge-to-mass ratio, we have $q/M = g/E$. In order to determine the mass, the drops are allowed to fall in the gravitational field without the influence of the electric field. In this case, there are still two forces acting on the drops. One is the gravitational force and the other is the frictional force due to air resistance. Since this force is proportional to the velocity, it vanishes when the drop is stationary, which is why it does not need to be taken into account when the electric field is on. However, when the drops are allowed to fall, the total force is $F=Mg-\gamma v$, since friction opposes gravitational force. Here, $\gamma$ is the coefficient of air resistance. Thus, according to Newton's law of motion
$\displaystyle Ma$ $\textstyle =$ $\displaystyle F$  
$\displaystyle M{dv \over dt} = Mg-\gamma v$      

Assuming the drop starts initially at rest, the equation can be solved for $v(t)$, yielding

\begin{displaymath}
v(t) = {Mg \over \gamma}\left(1-e^{-\gamma t/M}\right)
\end{displaymath}

Since $M$ is very small and $\gamma$ is relatively large, the exponential factor quickly decays to 0, leaving simply a constant velocity $v=Mg/\gamma$ known as the terminal velocity. This velocity can be measured and used to solve for the mass $M$. Once the mass is known, the charge-to-mass ratio determined above can be used to determine the charge $q$. The last remaining technical problem is that it is not known how many electrons are stripped off each drop before the experiment is performed. Thus, each drop will have a different charge $q$ that is a multiple of the fundamental charge unit $e$. By subtracting successive values of $q$ obtained for different drops, however, it is possible to find a ``smallest'' value, which can be assumed to be a prediction of $e$, although the possibility that the result is still a multiple of $e$ cannot be ruled out.

The currently accepted values of $e$ and $m$ are:

$\displaystyle e$ $\textstyle =$ $\displaystyle 1.6021773\;\times\;10^{-19}\;{\rm C}$  
$\displaystyle m$ $\textstyle =$ $\displaystyle 9.109390\;\times\;10^{-31}\;{\rm kg}$  


next up previous
Next: Rutherford's experiment and the Up: lecture_3 Previous: lecture_3
Mark E. Tuckerman 2011-09-13