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Found in: Page 67

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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

The potential energy required to acquire a speed of 0.9998c is 25 Mega Volts.

See the step by step solution

## Step 1: 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.

$\begin{array}{c}KE\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\Delta U\\ KE\text{\hspace{0.17em}}=\text{\hspace{0.17em}}q\Delta V\\ \left(\gamma \text{\hspace{0.17em}}-\text{\hspace{0.17em}}1\right){m}_{e}{c}^{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}e\Delta V\end{array}$ ...........(1)

## Step 2: 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 ${m}_{e}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}9.1\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{-31}\text{\hspace{0.17em}}kg$
• Charge of electron $e\text{\hspace{0.17em}}=\text{\hspace{0.17em}}1.6\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{-19}\text{\hspace{0.17em}}C$

Putting these values in above equation we get,

localid="1659084824833" $\begin{array}{c}\left(50\text{\hspace{0.17em}}-\text{\hspace{0.17em}}1\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}}9.1\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{-31}\text{\hspace{0.17em}}kg\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{\left(3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{8}\text{\hspace{0.17em}}m}{s}\right)}^{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(1.6\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{-19}\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\Delta V\\ \Delta V\text{\hspace{0.17em}}=\text{\hspace{0.17em}}25.08\text{\hspace{0.17em}}MV\end{array}$

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