Q. In order to coincide the parabolas formed by singly ionised ions in one spectrograph and doubly ionized ions in the other Thomson’s mass spectrograph, the electric field and magnetic fields are kept in the ratios 1:2 and 3:2 respectively. Then the ratio of masses of the ions is
A) 3 :4
B) 1 :3
C) 9 :4
D) None of these
Explanation
(c) Using $Z^{2} = k\left( \frac{q}{m} \right)y;$ where$\ k = \frac{B^{2}LD}{E}$
For parabolas to coincide in the two photographs, the $\frac{kq}{m}$ should be same for the two cases
$Thus,\frac{B_{1}^{2}LDe}{E_{1}m_{1}} = \frac{B_{2}^{2}LD(2e)}{E_{2}m_{2}}$
$\Rightarrow \frac{m_{1}}{m_{2}} = \left( \frac{B_{1}}{B_{2}} \right)^{2} \times \left( \frac{E_{2}}{E_{1}} \right) \times \frac{1}{2} = \frac{9}{4} \times \frac{2}{1} \times \frac{1}{2} = \frac{9}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.