Two particles, A and B, having same e/m ratio are projected towards silver nucleus, in different exp — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Energy Conservation at Closest Approach For a charged particle (mass $m_1$, charge $q_1$, speed $v$) projected toward a stationary nucleus (charge $q_2$), at the distance of closest approach $r_0$ all kinetic energy converts to potential energy: $$\frac{1}{2}m_1 v^2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_0}$$ ### Step 2 - Solve for $r_0$ Rearranging: $$r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2 q_1 q_2}{m_1 v^2} = \left(\frac{1}{4\pi\varepsilon_0} \frac{2 q_2}{v^2}\right) \cdot \left(\frac{q_1}{m_1}\right)$$ ### Step 3 - Compare Parameters for A and B In both experiments: * Target: same silver nucleus ($q_2 = 47e$) — constant * Speed: same $v$ — constant * $e/m$ ratio: $\left(\frac{q_1}{m_1}\right)_A = \left(\frac{q_1}{m_1}\right)_B$ — given equal Since all factors in the $r_0$ formula are identical for both particles: $$r_{0,A} = r_{0,B}$$ ### Step 4 - Evaluation of Options * **Option (A) same for both:** Correct — all determining factors are equal. * **Options (B) and (C):** Incorrect — imply different $r_0$ without justification. * **Option (D):** Incorrect in this context — speed is the same for both. $$\text{Correct Option: } \boxed{\text{A}}$$