In an oil drop experiment, the following charges (in arbitrary units) were found on a series of oil — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Principle of Charge Quantization The fundamental principle underlying Millikan's oil drop experiment is the **quantization of electric charge**. This principle states that the net electrical charge \(Q\) on any physical body must always be an integral multiple of the elementary unit of charge \(e\): \[Q = n \cdot e\] where \(n\) is a positive integer. Consequently, the charge of a single electron must be the **Highest Common Factor (HCF)** of all the observed charges: \[e = \text{HCF}(Q_1, Q_2, Q_3, \dots)\] --- ### Step 2 - Simplify the Given Observed Charges The observed charges on the series of oil droplets (in arbitrary units) are: * \(Q_1 = 4.5 \times 10^{-18}\) * \(Q_2 = 3.0 \times 10^{-18}\) * \(Q_3 = 6.0 \times 10^{-18}\) * \(Q_4 = 7.5 \times 10^{-18}\) * \(Q_5 = 9.0 \times 10^{-18}\) Factoring out \(10^{-18}\), we need HCF of: \(\{4.5, 3.0, 6.0, 7.5, 9.0\}\). --- ### Step 3 - Calculate the Highest Common Factor (HCF) Multiply each coefficient by \(10\) to convert to integers: \(\{45, 30, 60, 75, 90\}\). Prime factorizations: * \(30 = 2 \times 3 \times 5\) * \(45 = 3^2 \times 5\) * \(60 = 2^2 \times 3 \times 5\) * \(75 = 3 \times 5^2\) * \(90 = 2 \times 3^2 \times 5\) Common factors: \(3 \times 5 = 15\). Dividing back by \(10\): \[\text{HCF}(4.5, 3.0, 6.0, 7.5, 9.0) = 1.5\] --- ### Step 4 - Determine the Fundamental Charge of the Electron \[e = 1.5 \times 10^{-18}\text{ arbitrary units}\] Verification (\(Q = n \cdot e\)): * \(Q_1 = 4.5 \times 10^{-18} \implies n_1 = 3\) ✓ * \(Q_2 = 3.0 \times 10^{-18} \implies n_2 = 2\) ✓ * \(Q_3 = 6.0 \times 10^{-18} \implies n_3 = 4\) ✓ * \(Q_4 = 7.5 \times 10^{-18} \implies n_4 = 5\) ✓ * \(Q_5 = 9.0 \times 10^{-18} \implies n_5 = 6\) ✓ \[\text{Charge on electron } (e) = \boxed{1.5 \times 10^{-18}}\] --- ### Step 5 - Evaluation of the Options * **Option (A) \(3.0 \times 10^{-18}\):** Incorrect. The droplet with charge \(4.5 \times 10^{-18}\) would require \(1.5\) electrons — impossible. * **Option (B) \(9.0 \times 10^{-18}\):** Incorrect. This is the maximum observed charge, not the HCF. * **Option (C) \(1.5 \times 10^{-18}\):** Correct. This is the HCF, with each charge being an integer multiple (2, 3, 4, 5, 6). * **Option (D) \(1.6 \times 10^{-19}\):** Incorrect. This is the actual SI value in Coulombs, not in the arbitrary units of this problem. \[\text{Correct Option: } \boxed{\text{C}}\]