(a) What potential difference would accelerate an electron to speed c according to classical physics? (b) With this potential difference, what speed would the electron actually attain?

The energy required to move a single charge (proton or electron) under a potential difference $\mathbf{∆}\mathit{V}$ of exactly 1 volt is called one electron-volt or 1-eV. The magnitude of this energy is$q∆\mathrm{V}$. The energy relation for a charged particle accelerated through a potential difference is given by

$K=q\Delta V$

Where K is the kinetic energy of the electron.

In the given case, the electron must be accelerated up to speed c through a potential difference to be determined. Solving classically.

$$\begin{gathered} \frac{1}{2}{m}_e{v}^2=q\Delta V \\ \Delta V=\frac{m_e{v}^2}{2q} \\ =\frac{m_e{c}^2}{2e} \end{gathered}$$

For Electron, rest mass is $\text{9.1}\times {10}^{-31}kg$and its energy equivalent is $\approx \text{0.511}MeV$. Therefore substituting $\text{0.511}MeV$for $m_e{c}^2$in the above equation, we get

$$\begin{gathered} \Delta V=\frac{0.511\times {10}^{6}eV}{2e} \\ =0.256\times {10}^{6}V \\ =256kV \end{gathered}$$

In reality, through a 256 kV potential difference, the electron would have achieved a speed less than the speed of light because of relativistic correction. The kinetic energy relation for relativistic speeds is

$K=m_e{c}^2\left(\gamma -1\right)$

Where $m_e$is the electron’s rest mass, and $\gamma$is the Lorentz factor. Using this relation to determine the Lorentz factor,

$$\begin{gathered} q\Delta V=m_e{c}^2\left(\gamma -1\right) \\ \gamma =\frac{q\Delta V}{m_e{c}^2}+1 \end{gathered}$$

For Electron, rest mass is$\text{9.1}\times {10}^{-31}kg$ and its energy equivalent is $\approx \text{0.511}MeV$. Therefore substituting$\text{0.511}MeV$ for$m_e{c}^2$ in the above equation, we get

$$\begin{gathered} \gamma =\frac{e\left(0.256\times {10}^{6}V\right)}{0.511\times {10}^{6}eV}+1 \\ =0.5+1 \\ =1.5 \end{gathered}$$

Now determining the speed parameter $\beta =\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$using the Lorentz factor

$$\begin{gathered} \gamma =1.5 \\ \frac{1}{\sqrt{1-{\beta }^2}}=1.5 \\ \beta =\sqrt{1-\frac{1}{{\left(1.5\right)}^2}} \\ =0.745 \end{gathered}$$

In reality, the electron would attain 0.745c speed through a 256 kV of potential difference.