91Ó°ÊÓ

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

$K=q\mathrm{Δ}V$

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.

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

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

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

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

$K=m_e{c}^2\left(\mathrm{γ}-1\right)$

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

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

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

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

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

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

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