# Acceleration in an Electron Gun

If the acceleration voltage is up to 2.7 kV it's often recommended to calculate relativistic because the speed of the electrons reaches 10% of speed of light.
Relativistic calculation of the speed of the electrons: Kinetic energy is total energy less rest energy:$E_{\text{Kin}}=m_{\text{rel}}\cdot c^2 - m_{e}\cdot c^2$ Equal with work done by electric field:$$V_{\text a}\cdot e = m_{\text{rel}}\cdot c^2 - m_{e}\cdot c^2\quad (1)$$$m_{\text{rel}}$ is linked to $m_{e}$ by the Lorenz factor $\gamma$:$m_{\text{rel}}=\gamma \cdot m_{e} = \frac{m_{e}}{\sqrt{1-\frac{v^2}{c^2}}}\quad (2)$ Inserting of (2) into (1), factoring out and dividing by $m_{e}\cdot c^2$ leads to:$$\frac{V_{\text a}\cdot e}{m_{e}\cdot c^2}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1$$ Adding 1 and squaring leads to:$$\left({1+\frac{V_{\text a}\cdot e}{m_{e}\cdot c^2}}\right)^2 = \frac {1}{1-\frac{v^2}{c^2}}$$Inverse equation, miltiply by -1 and adding 1 shows:$$\frac{v^2}{c^2}=1-\frac{1}{\left({1+\frac{V_{\text a}\cdot e}{m_{e}\cdot c^2}}\right)^2}$$Multiply by $c^2$ and take the root shows:$${v_{\text{relativistisch}}=c\cdot \sqrt{1-\frac{1}{\left({1+\frac{V_{\text a}\cdot e}{m_{e}\cdot c^2}}\right)^2}}}$$