Q. If in a Thomson’s mass spectrograph, the ratio of the electric fields and magnetic fields, in order to obtain coincident parabola of singly ionised and doubly ionised positive ions are 1 : 2 and 3 : 2 respectively, then the ratio of masses of particles will be
A) 3 : 1
B) 2 : 1
C) 9 : 4
D) 9 : 2
Explanation
(c) For similar parabola, $\frac{B^{2}l\ Dq}{E\ m}$ must be same for both the ions.
So, $\frac{m_{1}}{m_{2}} = \left( \frac{B_{1}}{B_{2}} \right)^{2} \times \left( \frac{q_{1}}{q_{2}} \right) \times \frac{E_{2}}{E_{1}} = \left( \frac{3}{2} \right)^{2} \times \left( \frac{1}{2} \right) \times \left( \frac{2}{1} \right) = \frac{9}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.