A proton and a deuteron are projected towards the stationary gold nucleus, in different experiments, — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Principle of Conservation of Energy at Closest Approach When a positively charged subatomic particle is projected directly toward a stationary, positively charged atomic nucleus, at the point of closest approach ($r_{\min}$), all initial kinetic energy is converted into electrostatic potential energy: $$K.E._i = U_f$$ $$\frac{1}{2} m v^2 = \frac{k q_1 q_2}{r_{\min}}$$ Solving for $r_{\min}$: $$r_{\min} = \frac{2 k q_1 q_2}{m v^2}$$ --- ### Step 2 - Proportionality Analysis Under Given Conditions Since the target charge ($q_2$ of the gold nucleus) and initial speed ($v$) are the same for both experiments: $$r_{\min} \propto \frac{q_1}{m}$$ The distance of closest approach is directly proportional to the charge-to-mass ratio of the projected particle. --- ### Step 3 - Compare the Charge-to-Mass Ratios For the **Proton** ($\ce{^{1}_{1}H^+}$): * Charge: $q_p = +1e$ * Mass: $m_p \approx 1\text{ amu} = m$ $$\left(\frac{q}{m}\right)_{\text{proton}} = \frac{e}{m}$$ For the **Deuteron** ($\ce{^{2}_{1}H^+}$): * Charge: $q_d = +1e$ * Mass: $m_d \approx 2\text{ amu} = 2m$ $$\left(\frac{q}{m}\right)_{\text{deuteron}} = \frac{e}{2m}$$ Ratio of closest approach distances: $$\frac{r_{\min, \text{proton}}}{r_{\min, \text{deuteron}}} = \frac{\left(\frac{q_p}{m_p}\right)}{\left(\frac{q_d}{m_d}\right)} = \frac{\frac{e}{m}}{\frac{e}{2m}} = 2$$ Therefore: $r_{\min, \text{proton}} = 2 \cdot r_{\min, \text{deuteron}}$ The distance of closest approach is **greater for the proton**. --- ### Step 4 - Evaluation of Options * **Option (A) same for both:** Incorrect. The deuteron has twice the mass, giving it more momentum and kinetic energy at the same speed, so it can penetrate closer to the nucleus. * **Option (B) greater for proton:** Correct. The proton has a higher charge-to-mass ratio ($e/m$ vs $e/2m$), resulting in twice the distance of closest approach compared to the deuteron. * **Option (C) greater for deuteron:** Incorrect. The deuteron's larger mass gives it more kinetic energy at the same speed, so it gets closer to the nucleus. * **Option (D) depends on speed:** Incorrect. The relative comparison depends solely on intrinsic charge-to-mass ratios since the speed is the same in both experiments. $$\text{Correct Option: } \boxed{\text{B}}$$