Calculation of special values:

Which acceleration voltage is needed to accelerate electrons up to 10% of speed of light?$$v_{\text{end}}=0.1\cdot c = \sqrt{2\cdot \frac{e}{m_e}\cdot V_{\text a}}$$Squaring and solve for $V_{\text a}$$$V_{\text a}=\frac {0.01\cdot c^2 \cdot m_e}{2\cdot {e}}$$Plug in $c=3\cdot 10^8~\frac {\text m}{\text s}$, $m_e=9.1\cdot 10^{-31}~\text{kg}$ und $\text e=1.60\cdot 10^{-19} ~\text C$ leads to:$$\bbox[5px,border:2px solid red]{V_{\text a}=2559~\text V}$$An acceleration voltage of 2559 V is needed to accelerate electrons up to 10% of the speed of light. When using higher acceleration voltages a relativistiv calculation is needed.

Differences between classic and relativistic calculation when the acceleration voltage is ${V_{\text a}=2559~\text V}$:$${V_{\text a}=2559~\text V \Rightarrow v_{\text{klassisch}}=0.1\cdot \text c}$$absolute error: $$F=0.1 \cdot c - 0.09962 \cdot c = 0.00038 \cdot c \approx 114000 \frac{\text m}{\text s} = 410400 \frac {\text{km}}{\text h}$$ relative error: $$\frac{0.1}{0.09962}-1=1.0038-1=0.0038=0.38 \% $$ With increasing acceleration voltage this error is getting much bigger:Comparison between electron speed in non-relativistic and relativistic calculation
\[\begin{array}{c|c} & \text{relative error } f\\ \text{Voltage } V_{\rm{a}} & \text{of velocity}\\ \hline 2559\,\rm{V} & 0.38\% \\ 6840\,\rm{V} & 1\% \\ 34600\,\rm{V} & 5\% \\ 70160\,\rm{V} & 10\% \end{array}\]