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Electron Speed
5,930,970
metres per second (m/s)
Fraction of light speed (v/c) 0.019784

What is the Electron Speed Calculator?

This calculator finds how fast an electron travels after being accelerated from rest through an electric potential difference (voltage). It is a staple of physics courses and applications such as cathode-ray tubes, electron microscopes, and particle accelerators. The result is given in metres per second and as a fraction of the speed of light.

How to use it

Enter the accelerating voltage in volts and read off the electron's final speed. For example, a voltage of 100 V produces a speed of roughly 5.93 million m/s — just under 2% of light speed.

The formula explained

An electron carrying charge \(q\) gains kinetic energy equal to the work done on it by the field: \(qV = \tfrac{1}{2}mv^2\). Solving for \(v\) gives $$v = \sqrt{\dfrac{2qV}{m}}$$ Here \(q = 1.602 \times 10^{-19}\ \text{C}\) is the elementary charge and \(m = 9.109 \times 10^{-31}\ \text{kg}\) is the electron rest mass. This is the non-relativistic form; it stays accurate while \(v\) is well below the speed of light (roughly under a few keV of accelerating energy).

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Electron accelerated between two charged plates gaining speed across a voltage gap
An electron accelerated through a potential difference \(V\) gains kinetic energy and exits with speed \(v\).

Worked example

For \(V = 100\ \text{V}\):

$$v = \sqrt{\frac{2 \times 1.602\times10^{-19} \times 100}{9.109\times10^{-31}}} = \sqrt{3.518\times10^{13}} \approx 5{,}930{,}000\ \text{m/s}$$

or about \(0.0198c\).

Energy balance diagram showing electrical work qV converting into kinetic energy half m v squared
The electrical work \(qV\) equals the electron's kinetic energy \(\tfrac{1}{2}mv^2\), which rearranges to the speed formula.

FAQ

Does this account for relativity? No — it uses the classical kinetic energy. At very high voltages (tens of kV+), relativistic corrections matter and this will slightly overestimate the speed.

What if I enter 0 volts? The electron starts and stays at rest, so the speed is \(0\ \text{m/s}\).

Does this work for protons or ions? The formula is identical, but you would substitute the proton/ion charge and mass instead of the electron's values.

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