# J J Thomson experiment of charge to mass $(e/m)$ ratio

I recently learnt about the 'velocity selector'. It says that this property of charges to pass undeflected in a magnetic and electric field was used by Sir J J Thomson to measure the specific charge $e/m$ of an electron. Now I'm pondering how?

• Google 'Wien filter'... – Jon Custer Oct 17 '16 at 16:41

The velocity selector is commonly used to select particles with a specific velocity from a stream of particles, each having different speeds or to make a particle move with a constant speed. It can be used to find the $q/m$ ratio of a charge. A simple instrument that can be used to find the charge to mass ratio of a particle is the Bainbridge mass spectrograph.

Principle of velocity selector

When a charged particle ($q$) passes through a region of electric field of magnitude $E$, it experiences a force $qE$, which causes the charge to accelerate. If it passes through a crossed magnetic field $B$, it experiences a magnetic force $qvB$, where $v$ is the velocity of the charge. When the magnitude of these two forces become equal (i.e., if we adjust the configuration of the fields such that the electric and magnetic fores become equal and opposite),

$$qE=qvB\implies v=\frac{E}{B}\qquad(1)$$

So, when the two forces are balanced, the charge moves with a constant velocity $v$, undeviated through that region. If there are particles with speed other than $v$ as defined by the magnitude of the fields, then they will feel an unbalanced force on them and they bend off from their trajectory. This way we can filter particles of a definite velocity by choosing appropriate field magnitudes.

How this principle can be used in finding the $e/m$ ratio of a charged particle?

You pass a stream of charged particles through two fine slits and let them to pass through a velocity selector so as to maintain them at a constant velocity given by equation ($1$). Now these constant velocity particles enter into a chamber maintained at a uniform magnetic field $B'$. Due to the effect of this magnetic field, the charges moving with constant velocity $v$ experiences a magnetostatic force $qvB'$ (the field $B'$ acts perpendicular to the plane in which the charge moves).

What is the effect of a charged particle in a magnetic field? Of course the magnetic force bends the particle into a circular trajectory of radius $r$. Let $m$ be the mass of a charge. Then it will experience a centripetal force provided by the magnetic field $B'$:

\begin{align} &\frac{mv^2}{r}=qvB'\\ &\implies mv=qB'r\\ \\ &\implies \bbox[yellow,5px] { \frac{q}{m}=\frac{v}{B'r} \qquad(2) } \end{align}

Hence we determined the charge to mass ratio of a charge. Now, the stream of particles may contain charged particles with different masses (isotopes). Hence how to distinguish them? No worries. According to equation ($2$),

$$r=\frac{mv}{qB'}$$

This means that $r\propto m$. Hence the radius made by charges of different masses are different. So if you place a photographic plate such that the charges that bend in the chamber finally strike the plate, you will get spots corresponding to different masses. By measuring the radii to each spot, you can calculate the mass of the particle. That's why this apparatus is known as a mass spectrograph. This way you can calculate the isotopic masses of particles.

Anyway the main concern of your question is how to determine the $q/m$ ratio of a charge. It is already explained by equation ($2$). Your charged particle is electron. Pass the electron through the velocity selector. The field values define the velocity of the electron. It is then passed through the chamber with constant magnetic field $B'$. The electron experiences a force and it will bend, the radius of which you can find out from a photographic plate as explained above. thus you got $r$. Now, you have the value of the constant field $B'$. the velocity of the particle is found out using equation ($1$). Substitute all these values into equation ($2$). There you go. The $e/m$ ratio of electron is found out.

To determine the ratio of the electron charge to its mass $\frac{e}{m}$, Lorentz used, in essence, a cathode ray tube where electron from an electron gun were shot into a plate capacitor deflector of length $L$ with an electric field $E$ and two coils to create a magnetic field $B$ deflector. He considered the elctron to behave like a classical particle. The deflection was monitored on a luminescent screen. First, a magnetic field was produced by the coils and an electric field was applied so that the electric field exactly canceled the Lorentz force, so that no deflection resulted. Thus the velocity of the electrons was determined from $eE=evB$. Then the B was switched off and the deflection of the electron with speed $v$ was measured in the capacitor where it moved similar to a free fall of a horizontally moving projectile with horizontal speed $v$ yielding the speed $v=\frac{e}{B}$. The "fall distance" in a capacitor of length $L$ would be $s=\frac{1}{2} at^2=\frac{1}{2} \frac{eE}{m}(\frac {L}{v})^2$. From the deflection on the screen s was obtained and from this the charge-to-mass ratio of the electron$\frac{e}{m}$.

A description of the original experiment of J J Thomson can be found here.