Question

(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?

Short answer

(a) 256 kV of potential difference is required for the electron to reach the speed of light.

(b) The electron would attain 0.745c speed through a 256 kV potential difference.

Step-by-step solution

Electron Volt

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.

Classical Approach

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

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

Relativistic approach.

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.