Electron Charge to Mass Ratio

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Physical Law

The ratio of the charge to the mass of an electron is given by:

$\dfrac e {m_e} \approx 1 \cdotp 75882 \, 017 \times 10^{11} \, \mathrm C \, \mathrm {kg}^{-1}$

This sequence is A081815 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Let it be assumed that the cathode rays consist of particles $\mathrm p$ which have charge $e$ and mass $m_e$.

A beam of cathode rays is passed through a region in which an electric field and a magnetic field are arranged to be perpendicular to each other.

In this arrangement, the forces on the cathode rays caused by the electric field and magnetic field can be made to be equal and opposite.

Let the electric field strength of the electric field be $\mathbf E$.

Let the magnetic field strength of the magnetic field be $\mathbf B$.

Let $\mathbf F_e$ be the force on $\mathrm p$ caused by the electric field.

Let $\mathbf F_m$ be the force on $\mathrm p$ caused by the magnetic field.

Let the speed of the cathode rays be $v$.

From Electric Force on Charged Particle:

$\mathbf F_e = \mathbf E e$

From Magnetic Force on Moving Charged Particle:

$\mathbf F_m = \mathbf B v e$

$\mathbf E$ and $\mathbf B$ are adjusted so as to make sure the beam passes through undeflected.


\(\ds \mathbf F_e\) \(=\) \(\ds \mathbf F_m\)
\(\ds \leadsto \ \ \) \(\ds \mathbf E e\) \(=\) \(\ds \mathbf B v e\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \dfrac {\mathbf E} {\mathbf B}\)

Thus we have determined the speed of the cathode rays.

Next we subject the beam to only one of the fields, and observe the deflection of the beam.

Let the beam be subject only to the electric field.

We arrange that $\mathbf E$ is generated by applying a potential difference between two plates whose dimensions we can determine.

Let the length of the region of space through which $\mathbf E$ acts be $l$.

As before, let $\mathbf F_e$ be the force on $\mathrm p$ caused by the electric field.

Let $t$ be the time during which $\mathrm p$ is between the plates.

$\mathbf F_e$ will act on $\mathrm p$ over $t$.

Let $\mathbf a$ be the acceleration of $\mathrm p$ caused by $\mathbf E$ perpendicular to the direction of the beam.

From Newton's Laws of Motion:

$\mathbf F_e = m_e \mathbf a$
\(\ds \mathbf F_e\) \(=\) \(\ds m_e \mathbf a\)
\(\ds \leadsto \ \ \) \(\ds \mathbf a\) \(=\) \(\ds \dfrac {e \mathbf E} {m_e}\) from $\mathbf F_e = \mathbf E e$

This acceleration acts over time $t = \dfrac l v$.

Hence the component of velocity $v_t$ imparted to $\mathrm p$ perpendicular to the direction of the beam is:

$v_t = \dfrac {e \mathbf E l} {m_e v}$

Hence the change of direction $\theta$ of $\mathrm p$ is given by:

\(\ds \tan \theta\) \(=\) \(\ds \dfrac {v_t} v\)
\(\ds \) \(=\) \(\ds \dfrac {e \mathbf E l / m_e v} v\)
\(\ds \) \(=\) \(\ds \dfrac {e \mathbf E l} {m_e v^2}\)
\(\ds \) \(=\) \(\ds \dfrac {e \mathbf E l} {m_e \paren {\mathbf E / \mathbf B}^2}\) as $v = \dfrac {\mathbf E} {\mathbf B}$ from above
\(\ds \leadsto \ \ \) \(\ds \dfrac e {m_e}\) \(=\) \(\ds \dfrac {\mathbf E} {\mathbf B^2 l} \tan \theta\)

All the values on the right hand side are known.

From these we can calculate the value of $\dfrac e {m_e}$.


Historical Note

When an electrode is charged to a high electric potential in a high vacuum, it is observed to emit rays.

This had been established over the course of several decades of experiments by a number of workers.

These rays were called cathode rays.

It was the work of Joseph John Thomson in $1897$ to attempt to establish some basic facts about these cathode rays.