Q. In a Thomson set-up for the determination of $e/m,$ electrons accelerated by $2.5\ kV$ enter the region of crossed electric and magnetic fields of strengths $3.6 \times 10^{4}Vm^{- 1}$ and $1.2 \times 10^{- 3}T$ respectively and go through undeflected. The measured value of $e/m$ of the electron is equal to
A) $1.0 \times 10^{11}C$-$kg^{- 1}$
B) $1.76 \times 10^{11}C$-$kg^{- 1}$
C) $1.80 \times 10^{11}C$-$kg^{- 1}$
D) $1.85 \times 10^{11}C$-$kg^{- 1}$
Explanation
(c) $\frac{e}{m} = \frac{E^{2}}{2VB^{2}} = \frac{\left( 3.6 \times 10^{4} \right)^{2}}{2 \times 2.5 \times 10^{3} \times \left( 1.2 \times 10^{- 3} \right)^{2}}$
$= 1.8 \times 10^{11}C/kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.