The e/m ratio is maximum for — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Definition of e/m Ratio For comparative purposes, we compute the relative charge-to-mass ratio: $$\text{Relative } \frac{e}{m} \propto \frac{z}{A}$$ Where $z$ = ionic charge (in units of elementary charge) and $A$ = atomic mass (in amu). ### Step 2 - Calculate for Each Ion | Ion | $z$ | $A$ (amu) | $z/A$ | |-----|-----|-----------|-------| | $\ce{Na^+}$ | 1 | 23 | $1/23 \approx 0.043$ | | $\ce{Al^{3+}}$ | 3 | 27 | $3/27 \approx 0.111$ | | $\ce{H^+}$ | 1 | 1 | $1/1 = 1.000$ | | $\ce{Mg^{2+}}$ | 2 | 24 | $2/24 \approx 0.083$ | ### Step 3 - Comparison $$0.043 < 0.083 < 0.111 < 1.000$$ $$\ce{Na^+} < \ce{Mg^{2+}} < \ce{Al^{3+}} \ll \ce{H^+}$$ $\ce{H^+}$ is a bare proton — minimum mass (1 amu), unit charge → maximum e/m ratio. ### Step 4 - Evaluation of Options * **Option (A) $\ce{Na^+}$:** Lowest e/m (0.043). Incorrect. * **Option (B) $\ce{Al^{3+}}$:** e/m = 0.111. Incorrect. * **Option (C) $\ce{H^+}$:** Maximum e/m = 1.000. **Correct.** * **Option (D) $\ce{Mg^{2+}}$:** e/m = 0.083. Incorrect. $$\text{Correct Option: } \boxed{\text{C}}$$