Q. A beam of electrons of velocity $3 \times 10^{7}\ ms^{- 1}$ is deflected 1.5 mm is passing 10 cm through an electric field of 1800 $Vm^{- 1}$ perpendicular to their path. The value of $e/m$ for electron is
A) $1.78 \times 10^{11}\ Ckg^{- 1}$
B) $2 \times 10^{11}\ Ckg^{- 1}$
C) $1.5 \times 10^{11}\ Ckg^{- 1}$
D) $3.5 \times 10^{11}\ Ckg^{- 1}$
Explanation
(c) $y = \frac{1}{2}at^{2} = \frac{1}{2}\frac{eE}{m}\frac{x^{2}}{v^{2}}$ or $\frac{e}{m} = \frac{2\ y\ v^{2}}{Ex^{2}}$
$= \frac{2 \times 1.5 \times 10^{- 3} \times \left( 3 \times 10^{7} \right)^{2}}{1800 \times (0.1)^{2}}$
$= 1.5 \times 10^{11}\ Ckg^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.