The cathode ray experiment is conducted in a discharge tube containing four different gases A, B, C, and D. Their molecular weight is $A = B > D > C$. Then the e/m observed will be:
A. A > B > D > C
B. A < D < B < C
C. A = B > D > C
D. A = B = C = D

Hint: Think on what exactly happens in a cathode ray experiment and what environmental factors are taken into consideration. Consider the e/m ratio, what it stands for and also what changes may affect it.

Complete step by step solution:
In the cathode ray experiment, a voltage is applied across two electrodes that are placed in an evacuated glass chamber. The surface behind the anode is seen to glow due to the electrons emitted from the cathode that flow towards the anode due to difference in voltage.
The e/m ratio can be calculated using this experiment.
This e/m ratio is defined as the charge to mass ratio of electrons. It depends solely on the charge on electrons and the mass of electrons and is considered to be a constant.
Thus, no matter what the molecular weights of the gases may be, the e/m ratio does not depend on that and nor is it affected by it.
Hence, the answer is ‘D. $A=B=C=D$’

Note: As the molecular weight of any gas increases, the number of electrons present also increases. This may lead you to believe that since the number of electrons are increasing, the total mass of the electrons will also increase. But the correct answer is not ‘C. $A = B > D > C$’. The total charge on electrons will also increase as the total mass increases. We are effectively finding the unit charge on electrons, per unit mass of electrons.