Answer to Question #132048 in Mechanics | Relativity for Max

Question

An electron is accelerated from rest to a speed 0.8c by applying potential difference. Find it's value.

Answer 1

solution:

given data:-

rest mass of electron(m₀)=$9.1\times10^{-31}kg$

charge of electron(q)=$1.6\times10^{-19}C$

speed of electrone(v)=0.8c

relation between kinetic energy and potential difrence is

$KE=q\Delta V$ .........eq.1

and

$KE=\frac{m_0c^2}{\sqrt1-\frac{v^2}{c^2}}-m_0c^2 ......eq.2$

using equation 1 and 2 we get

potential difference can be written as

$q\Delta V=\frac{m_0c^2}{\sqrt1-\frac{v^2}{c^2}}-m_0c^2$

$\Delta V=\frac{m_0c^2}{q}(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)$

by putting the value of m_{0 ,}c, v and q

$\Delta V=\frac{9.1\times10^{-31}\times9\times10^{16}}{1.6\times10^{-19}}(\frac{1}{\sqrt{1-\frac{(0.64)c^2}{c^2}}}-1)$

$\Delta V=51.1\times10^4\times\frac{2}{3}$

$=34.06\times10^4$

$\fcolorbox{red}{yellow}{$\Delta V=341\times10^3 volts$}$

therefore potential difference for this electron is 341 KV.