The e/m ratio of anode rays produced in the discharge tube, depends on the — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Origin and Formation of Anode Rays Anode rays (canal rays) are produced when fast-moving electrons (cathode rays) collide with and ionize neutral gas atoms inside the discharge tube: $$\ce{M(g) + e^- -> M^+(g) + 2e^-}$$ The resulting positively charged gaseous ions ($\ce{M^+}$) form the anode rays. --- ### Step 2 - The Charge-to-Mass ($e/m$) Ratio of Anode Rays The specific charge of the positive ions is defined as: $$\text{Specific Charge} = \frac{q}{m}$$ Since anode rays are composed of positive gaseous ions whose mass depends on the atomic mass of the gas used: * For hydrogen gas ($\ce{H2}$): positive ions are protons ($\ce{H^+}$) $$\left(\frac{e}{m}\right)_{\ce{H^+}} \approx \frac{1e}{1.008\text{ amu}} \approx 9.58 \times 10^7\text{ C/kg}$$ * For helium gas ($\ce{He}$): positive ions are $\ce{He^+}$ $$\left(\frac{e}{m}\right)_{\ce{He^+}} \approx \frac{1e}{4.003\text{ amu}} \approx 2.41 \times 10^7\text{ C/kg}$$ Since the mass changes with the gas, the $e/m$ ratio varies with the **nature of the gas filled in the tube**. --- ### Step 3 - Independence from Electrode Materials The electrodes only serve as conductive plates to establish the high-voltage electric field. The anode does not emit positive ions from its metallic surface, and the cathode material does not affect the mass or charge of the gaseous positive ions. Consequently, the $e/m$ ratio is completely independent of electrode materials. --- ### Step 4 - Systematic Analysis of the Options * **Option (A) nature of the gas filled in the tube:** Correct. Different gases yield ions of different masses, so the charge-to-mass ratio changes. * **Option (B) nature of anode material:** Incorrect. The anode material only conducts current and does not affect the positive ions. * **Option (C) nature of cathode material:** Incorrect. The cathode material has no influence on the mass or charge of the positive gaseous ions. * **Option (D) all of these:** Incorrect. Since the ratio is independent of electrode materials, not all of these are correct. $$\text{Correct Option: } \boxed{\text{A}}$$