Question

A particle of charge $3.2\times {10}^{-19}C$ travels through a potential difference of ${10}^{-2}$ Volts from rest, the final speed of particle is _______ $(m=1.6\times {10}^{-27}kg)$

• A. $2\times {10}^{3}m/s$
• B. $4.8\times {10}^{3}m/s$
• C. $3.6\times {10}^{3}m/s$
• D. $2.4\times {10}^{3}m/s$

Answer 1

The correct option is A $2\times {10}^{3}m/s$

$\frac{1}{2}m{v}^2=Q\times \Delta V$
$v^2=\frac{2Q\Delta V}{m}$
$v=\sqrt{\frac{2Q\Delta V}{m}}$
=$\sqrt{\frac{2\times 3.2\times {10}^{-19}\times {10}^{-2}}{1.6\times {10}^{-27}}}$
$=2\times {10}^{3}m/s$