Question

 An electron is to be accelerated from 2.0 x 106m/s to 7.5 x 106m/s. Through what potential difference must the electron pass to accomplish this?

Answer 1

As per the question,

Let the initial velocity be v$_1$ =2$\times10^6$ m/s

and the final velocity be v$_2=7.5\times10^6m/s$

As we know

Mass of electron, m=$9.1\times10^{-31}$ kg

charge of electron , q=$1.6\times10^{-19}$ c

Change in kinetic energy= Charge of electron$\times$ potential applied

$\frac{mv_2^2}{2}-\frac{mv_1^2}{2}=q\times V$

$\frac{m(v_2^2-v_1^2)}{2}= q\times V$

$\frac{9.1\times10^{-31}(7.5^2-2^2)\times10^{12}}{2}=1.6\times10^{-19}\times V$

On solving the above equation we get,

potential V=148.58volts.