Relativistic electron

In an electric field of voltage $U = 400 \,\text{kV}$ , the electron has at the cathode a potential energy $E_\text{pot} = eU = 400 \,\text{keV} \approx 6.4\cdot 10^{-14} \,\text{J}$ The rest energy of an electron results according to Einstein $E_0 = m_e c^2 \approx 8.2 \cdot 10^{-14} \,\text{J} \approx 511 \,\text{keV}$ Since the potential energy of the electron is comparable in magnitude to the rest energy, we need relativistic mechanics. The kinetic energy of the electron then is $E_\text{kin} = m c^2 - m_e c^2 = \frac{m_e c^2}{\sqrt{1 - (v/c)^2}} - m_e c^2$ where $m$ denotes the relativistic mass of the electron. In the case of low velocities ( $v/c \ll 1$ ), the classical approximation for the kinetic energy applies: $E_\text{kin}\approx \frac{m_e}{2} v^2 + \mathcal{O}( (v/c)^4)$ In our case, however, we must take into account the relativistic corrections. When the potential energy is converted into kinetic energy ( $E_\text{kin} = E_\text{pot}$ ), it results $\begin{aligned} & & e U &= \frac{m_e c^2}{\sqrt{1 - (v/c)^2}} - m_e c^2 \\ \Rightarrow & & \varepsilon := \frac{eU}{m_e c^2} &= \frac{1}{\sqrt{1 - (v/c)^2}} - 1 \\ \Rightarrow & & \frac{1}{\varepsilon +1} &= \sqrt{1 - (v/c)^2} \\ \Rightarrow & & 1 - \frac{1}{(\varepsilon +1)^2} &= (v/c)^2 \\ \Rightarrow & & v &= c \sqrt{1 - \frac{1}{(\varepsilon +1)^2}} = c \sqrt{1 - \frac{1}{(eU/(m_e c^2) +1)^2}} \end{aligned}$ Insertion of the numerical values yields $v \approx 0.83\cdot c \approx 2.5 \cdot 10^8 \,\text{m/s}$ The graph below compares the velocity for the classical and the relativistic case as a function of kinetic energy