Q. In Thomson’s mass spectrographs, when an electric field of $2 \times 10^{4}\ Vm^{- 1}$ is applied then the deflection produced on the screen is 20 mm. If the length of the plates is 5 cm and the distance of the screen from plates is 21cm and the velocity of positive ions is $10^{6}ms^{- 1}$, then their specific charge will be
A) $10^{7}\ Ckg^{- 1}$
B) $2.59 \times 10^{7}\ Ckg^{- 1}$
C) $5.9 \times 10^{7}\ Ckg^{- 1}$
D) $9.52 \times 10^{7}\ Ckg^{- 1}$
Explanation
(d) Here, $x = \frac{DEql}{mv^{2}}$
or $\frac{q}{m} = \frac{xv^{2}}{DEl} = \frac{0.02 \times \left( 10^{6} \right)^{2}}{0.21 \times \left( 2 \times 10^{4} \right) \times (5 \times 10^{- 2})}$
$= 9.52 \times 10^{7}Ckg^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.