Q. An electron is accelerated under a potential difference of 182 V. The maximum velocity of electron will be (Charge of an electron is $1.6 \times 10^{- 19}$ C and its mass is $9.1 \times 10^{- 31}\ $kg)
A) $5.65 \times 10^{6}$ m/s
B) $4 \times 10^{6}$ m/s
C) $8 \times 10^{6}$ m/s
D) $16 \times 10^{6}$ m/s
Explanation
(c) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}mv^{2} = eV$
$\frac{1}{2} \times 9 \times 10^{- 31} \times v^{2} = 1.6 \times 10^{- 19} \times 182$
$v^{2} = \frac{1.6 \times 10^{- 19} \times 182 \times 2}{9.1 \times 10^{- 31}}$
$v^{2} = 64 \times 10^{2}$ m/s
$v = 8 \times 10^{6}$ m/s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.