Question

An electron is accelerated through a potential difference of $200$ volts. If $\frac{e}{m}$ ratio for the electron be $1.6\times {10}^{11}$ coulomb/kg, the velocity acquired by the electron will be :

• A. $8\times {10}^{6}m/s$
• B. $8\times {10}^{5}m/s$
• C. $5.9\times {10}^{6}m/s$
• D. $5.9\times {10}^{5}m/s$

Answer 1

The correct option is A $8\times {10}^{6}m/s$

When $e^{-}$ is accelerated through $200$ volts
work done on electron $=q\Delta V$
$=1\cdot 6\times {10}^{-19}\times 200$
All the work done is stored as kinetic energy
$\Rightarrow \frac{1}{2}m{v}^2=q\Delta V$

$\Rightarrow v^2=\frac{200\times e\times 2}{m}$

$\Rightarrow v^2=200\times 2\times 1\cdot 6\times {10}^{11}$
$\Rightarrow v^2=4\times 16\times {10}^{12}$
$\Rightarrow v=2\times 4\times {10}^6$
$\Rightarrow v=8\times {10}^{6}m/s$