Chemistry Atomic Structure ETEA 2023
PMDC Verified Question 30 of 95
The maximum \( e/m \) ratio for positive rays is obtained when the discharge tube contains:
A
\( \text{He} \)
B
\( \text{N}_2 \)
C
\( \text{Ne} \)
D
\( \text{H}_2 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( \text{H}_2 \)
Concept: Positive rays (canal rays) are ionized gas molecules. Their charge-to-mass ratio depends entirely on the mass of the gas used in the tube.

Formula: $$ \frac{e}{m} = \frac{\text{Charge}}{\text{Mass}} $$

Solution:
  • To maximize the fraction \( e/m \), the denominator (mass) must be as small as possible.
  • Let's compare the atomic/molecular masses of the gases:
  • Helium (\( \text{He} \)) \( \approx 4 \text{ amu} \)
  • Nitrogen (\( \text{N}_2 \)) \( \approx 28 \text{ amu} \)
  • Neon (\( \text{Ne} \)) \( \approx 20 \text{ amu} \)
  • Hydrogen (\( \text{H}_2 \)) \( \approx 2 \text{ amu} \) (and a single ionized H atom is \( \approx 1 \text{ amu} \)).
  • Hydrogen has the lowest mass of any element, so it yields the absolute maximum \( e/m \) ratio.


Why other options are incorrect: They are much heavier gases, which increases the denominator and significantly lowers the \( e/m \) ratio.

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