Question

An electron accelerated from rest through a potential difference V acquires a speed of 0.9998c. Find the value of V.

Short answer

The potential energy required to acquire a speed of 0.9998c is25 Mega Volts.

Step-by-step solution

Define relation between kinetic energy and potential difference

The relativistic kinetic energy of a charged particle is equal magnitude to the change in potential energy as it is accelerated through a potential difference.

$\begin{array}{c}KE=\Delta U\\ KE=q\Delta V\\ (\gamma -1)m_e{c}^2=e\Delta V\end{array}$ ...........(1)

Determine the potential difference applied

Here,

• Lorentz factor

  
  

  $\begin{array}{rcl}\gamma & =& \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\\ & =& \frac{1}{\sqrt{1-\frac{{\left(0.9998c\right)}^2}{c^2}}}\\ & =& 50\\ & & \end{array}$
• Mass of electron $m_e=9.1\times {10}^{-31}kg$
• Charge of electron $e=1.6\times {10}^{-19}C$
• 
  

Putting these values in above equation we get,

localid="1659084824833" $\begin{array}{c}(50-1)\times 9.1\times {10}^{-31}kg\times {(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)}^2=(1.6\times {10}^{-19})\times \Delta V\\ \Delta V=25.08MV\end{array}$

So, to get an electron accelerated to a speed of 0.9998c, the required potential difference to be applied is 25 mega volts.