Q.In Milikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $-1.282 \times 10^{-18}\ C$, calculate the number of electrons present on it.

When an electric current is passed through an electrolyte solution, chemical reactions occur at the electrodes. This process is called electrolysis. The amount of substance produced or consumed at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte. This fundamental relationship is described by Faraday's laws of electrolysis.

Specifically, Faraday's first law states that the mass of a substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed. Mathematically, this can be expressed as:

where $m$ is the mass, $Q$ is the charge, and $Z$ is the electrochemical equivalent.

A more practical approach involves understanding the relationship between charge, moles of electrons, and the stoichiometry of the electrode reaction. One mole of electrons carries a charge equal to Faraday's constant ($F = 96500 \text{ C mol}^{-1}$). By determining the number of electrons required for the reduction of one mole of the metal ion, we can directly calculate the moles of metal produced from the total charge passed.

Here, we are electrolysing a $\text{Hg(NO}_3)_2$ solution. Mercury exists as $\text{Hg}^{2+}$ ions in this solution. At the cathode, these $\text{Hg}^{2+}$ ions will gain electrons and be reduced to metallic mercury ($\text{Hg}$).

1. Identify the reduction half-reaction and electron stoichiometry: The mercury ions in $\text{Hg(NO}_3)_2$ are $\text{Hg}^{2+}$. At the cathode, these ions are reduced to metallic mercury ($\text{Hg}$). The half-reaction is:

This equation tells us that 2 moles of electrons are required to produce 1 mole of metallic mercury.

2. Calculate the total charge passed ($Q$):

The current ($I$) is given as 2.0 amp, and the time ($t$) is 3.0 hours. To use the formula $Q = I \times t$, time must be in seconds.

Now, calculate the total charge:

> [!WARNING]
> Always ensure that time is converted to seconds when using current in amperes to calculate charge in coulombs. A common mistake is to use hours directly.

3. Calculate the moles of electrons passed ($n_e$):

We know that 1 mole of electrons carries a charge of 96500 C (Faraday's constant, $F$). We can use this to convert the total charge passed into moles of electrons.

4. Calculate the moles of mercury produced ($n_{Hg}$): From the half-reaction in Step 1, we know that 2 moles of electrons are required to produce 1 mole of Hg. Therefore, the moles of Hg produced will be half the moles of electrons passed. $n_{Hg} = \frac{n_e}{2} = \frac{0.22383 \text{ mol e}^-}{2 \text{ mol e}^- / \text{mol Hg}} \approx 0.111915 \text{ mol Hg}$ …

When electrolytic cells are connected in series, the same current flows through each for the same time. That means the quantity of charge that passes through one cell is identical to the charge that passes through every other cell. This is the central idea: charge is the common thread.

Faraday’s first law says that the mass of a substance deposited at an electrode is directly proportional to the charge passed. But different ions require different amounts of charge to deposit one mole — that’s where the equivalent weight comes in. The mass deposited is given by:

where $E$ is the equivalent weight (molar mass ÷ n-factor), $Q$ is the total charge, and $F = 96500$ C/mol is Faraday’s constant. Since $Q$ and $F$ are the same for all three cells, the masses are directly proportional to the equivalent weights.

So the problem reduces to: find the charge that deposits 1.45 g of Ag, then use that charge to find the masses of Al and Cu.

1. Find the equivalent weight of each metal.

   • Silver: Ag⁺ + e⁻ → Ag, so n = 1. $E_{\text{Ag}} = \frac{108}{1} = 108$ g/eq.
   • Aluminium: Al³⁺ + 3e⁻ → Al, so n = 3. $E_{\text{Al}} = \frac{27}{3} = 9$ g/eq.
   • Copper: Cu²⁺ + 2e⁻ → Cu, so n = 2. $E_{\text{Cu}} = \frac{63.5}{2} = 31.75$ g/eq.

2. Calculate the charge that deposits 1.45 g of Ag.

   Using Faraday’s law for Ag:

The key idea here is charge compensation. When a Ca²⁺ ion replaces an Na⁺ ion in the NaCl lattice, it introduces an extra positive charge. To keep the crystal electrically neutral, one Na⁺ site must be left vacant for every Ca²⁺ that enters. This is a classic example of impurity-induced vacancy formation in ionic solids.

Let’s work through the numbers step by step.

1. Interpret the doping concentration. "$10^{-3}$ mol% of CaCl₂" means that for every 100 moles of NaCl, there are $10^{-3}$ moles of CaCl₂. So in one mole of NaCl, the amount of CaCl₂ is:

2. Find the number of Ca²⁺ ions. Each mole of CaCl₂ gives one mole of Ca²⁺ ions. So in one mole of NaCl, the number of Ca²⁺ ions is:

3. Relate Ca²⁺ ions to vacancies.

   Each Ca²⁺ ion replaces two Na⁺ ions (to maintain charge balance: +2 from Ca²⁺ replaces +1 + +1 from two Na⁺). But one of those two Na⁺ sites is taken by Ca²⁺, and the other is left empty — a cationic vacancy. So one Ca²⁺ ion creates exactly one cationic vacancy.

   Therefore, the number of vacancies equals the number of Ca²⁺ ions:

   $$$$

Identify the orbits

Bohr radii scale as $r_n = n^2 a_0$ with $a_0 = 0.0529\ \text{nm}$:

Emitted energy

$\Delta E = 13.6\ \text{eV}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) = 13.6\left(\frac14-\frac1{25}\right) = 13.6\times0.21 = 2.856\ \text{eV}.$ …

The concept here is electrochemical equivalent — it’s the mass of a substance liberated by one coulomb of charge. Faraday’s laws tell us that the mass deposited is directly proportional to the charge passed: $m = Z Q$, where $Q = I t$. So we just need to compute the total charge and divide the given mass by it.

1. Find the total charge passed.

   Current $I = 0.5$ A, time $t = 60$ minutes = $60 \times 60 = 3600$ seconds.

   Charge $Q = I t = 0.5 \times 3600 = 1800$ C.

2. Apply the relation $m = Z Q$.

   Given mass $m = 0.592$ g, so

3. Calculate $Z$.

4. Match with the options. $3.29 \times 10^{-4}$ is closest to $3.3 \times 10^{-4}$, which is option (B). …

Concept & Intuition

Combustion of methane is a classic stoichiometry problem. The balanced equation tells us exactly how many moles of each reactant and product are involved. Here, the question gives a mass of CO₂, so we first convert that mass to moles (using the molar mass of CO₂), then use the mole ratio from the balanced equation to find the moles of methane required. The trap is forgetting to balance the equation or misreading the ratio.

Step-by-step solution

1. Write and balance the combustion reaction Methane (CH₄) burns in oxygen to produce carbon dioxide and water:

The coefficients show that 1 mole of CH₄ produces 1 mole of CO₂.

2. Convert the given mass of CO₂ to moles Molar mass of CO₂ = 12 g/mol (C) + 2 × 16 g/mol (O) = 44 g/mol.

3. Use the mole ratio to find moles of methane From the balanced equation:

So moles of CH₄ needed = 0.5 mol. …

The problem asks for the ratio of the speed of light in vacuum ($c$) to the speed of an electron in the first Bohr orbit of hydrogen ($v_1$). This is a classic result in atomic physics — the fine-structure constant $\alpha$ is defined as $v_1 / c$, and its reciprocal is about 137. So the ratio $c : v_1$ is roughly $137 : 1$.

Why does this work? In Bohr’s model, the electron in the $n$-th orbit has a quantized angular momentum: $m v r = n \hbar$. For the first orbit ($n=1$), we can find $v_1$ by combining this with the Coulomb force that provides the centripetal acceleration. The result is a speed that depends only on fundamental constants — and it turns out to be about $c/137$.

Let’s derive it step by step.

1. Write the force balance for the electron. The electrostatic attraction between the proton and electron provides the centripetal force:

where $k = \frac{1}{4\pi\epsilon_0}$, $e$ is the elementary charge, $m$ is the electron mass, $v$ is its orbital speed, and $r$ is the orbit radius.

Simplify:

2. Apply Bohr’s quantization condition for the first orbit ($n=1$). Angular momentum is quantized:

where $\hbar = h / (2\pi)$.

3. Eliminate $r$ between (1) and (2). From (2): $r = \frac{\hbar}{m v}$. Substitute into (1):

Cancel $m$ (assuming $v \neq 0$):

So the speed in the first Bohr orbit is:

$v_1 = \frac{k e^2}{\hbar}$ …

The key to this question is remembering that Millikan’s experiment involves three distinct stages where different forces balance. Many students mistakenly think only two forces act, but the viscous drag is essential for the motion and measurement.

Why this approach works:

Millikan’s experiment measures the charge on an electron by observing tiny charged oil droplets moving in an electric field. The droplet experiences:

• Gravitational force (weight) pulling it down.
• Electrostatic force (from the electric field) pushing it up or down depending on the charge.
• Viscous force (air resistance) that opposes motion — crucial because the droplet moves through air, and without it, the droplet would accelerate indefinitely.

The experiment uses the balance of these forces to determine the droplet’s charge.

Step-by-step reasoning:

1. Gravitational force always present

   Every oil drop has mass, so weight $F_g = mg$ acts downward. This is constant.

2. Electrostatic force when field is on

   The drop is charged (by friction or an ion source). When an electric field $E$ is applied between the plates, the drop feels $F_e = qE$. This can be upward or downward depending on the sign of the charge.

3. Viscous force when drop moves

   The drop moves through air, so Stokes’ law gives a drag force $F_v = 6\pi\eta r v$ (where $\eta$ is air viscosity, $r$ is drop radius, $v$ is velocity). This force always opposes motion.

   • When the drop is falling under gravity (no field), $F_v$ balances $F_g$ at terminal velocity.
   • When the field is on and the drop rises or falls at constant speed, $F_v$ balances the net of $F_g$ and $F_e$.

4. Why not just gravitational and electrostatic?

   If only those two acted, the drop would accelerate constantly — you couldn’t measure a steady velocity. The viscous force is what allows a terminal velocity measurement, which is the whole basis of the experiment.

5. Eliminate other options …

Concept & Intuition

In any close‑packed arrangement (hcp or ccp), the atoms occupy about 74% of the space; the rest are voids (holes). Two types of voids exist:

• Octahedral voids – one per atom.
• Tetrahedral voids – two per atom.

This ratio (1 octahedral + 2 tetrahedral = 3 voids per atom) is a fixed geometric fact, independent of the specific metal or compound. So if you know the number of atoms, you immediately know the total number of voids.

Step‑by‑step reasoning

1. Identify the number of atoms in 0.5 mole One mole contains Avogadro’s number $N$ of particles.

2. Count voids per atom in hcp

   In hcp (and also in ccp/fcc):

   • Octahedral voids = number of atoms = $1$ per atom.
   • Tetrahedral voids = $2$ per atom. Hence total voids per atom = $1 + 2 = 3$.

3. Total voids in 0.5 mole

   Multiply voids per atom by the number of atoms:

   $\text{Total voids} = 3 \times (0.5 N) = 1.5 N$ …

The problem gives a special condition: the wavelength $\lambda$ of the electron is equal to the distance it travels in one second. That distance is simply its speed $v$ (since distance = speed × time, and time = 1 s). So we have $\lambda = v$.

Now, the de Broglie wavelength for any particle is $\lambda = h/p$, where $p$ is the momentum. For an electron of mass $m$ moving with speed $v$, momentum $p = mv$. So the standard relation is:

But here, the condition says $\lambda = v$. That gives us a way to connect $v$ and the constants $h$ and $m$.

1. Start with the de Broglie relation: $\lambda = \frac{h}{mv}$.
2. Impose the given condition: $\lambda = v$.
3. Substitute $v$ for $\lambda$ in the de Broglie equation: $v = \frac{h}{mv}$.
4. Multiply both sides by $mv$: $mv^2 = h$.
5. Solve for $v$: $v = \sqrt{\frac{h}{m}}$.
6. But the condition also says $\lambda = v$, so $\lambda = \sqrt{\frac{h}{m}}$. …

The key is to first get the formula right. Sodium ferrocyanide is not $\text{NaFe}(\text{CN})_6$ — a common slip. The ferrocyanide ion is $[\text{Fe}(\text{CN})_6]^{4-}$, so it needs four sodium ions to balance the charge. That gives the formula $\text{Na}_4[\text{Fe}(\text{CN})_6]$.

Once you have the formula, the rest is straightforward stoichiometry.

1. Find the number of Na⁺ ions per formula unit.

   In $\text{Na}_4[\text{Fe}(\text{CN})_6]$, the subscript 4 on Na means each molecule contains 4 sodium ions. So 1 mole of sodium ferrocyanide contains 4 moles of Na⁺ ions.

2. Scale to the given amount.

   You have 0.5 mole of the compound.

   Moles of Na⁺ = $0.5 \times 4 = 2$ moles of Na⁺ ions.

3. Convert moles to number of ions.

   Use Avogadro’s number, $N_A = 6.022 \times 10^{23}$ particles per mole.

   Number of Na⁺ ions = $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$.

   That is approximately $12 \times 10^{23}$. …

The key here is to recognise what “exclusive product” means. When a primary amine reacts with excess methyl chloride, the reaction doesn’t stop at the secondary or tertiary stage — it goes all the way to the quaternary ammonium salt. But the problem says a tertiary amine is formed exclusively, which tells us the reaction is controlled: exactly three methyl groups add to the nitrogen, one at a time, and each addition consumes one molecule of methyl chloride.

Ethyl amine is $\text{C}_2\text{H}_5\text{NH}_2$. Its molar mass is $2 \times 12 + 7 \times 1 + 14 = 24 + 7 + 14 = 45$ g/mol. So 90 g corresponds to $90 / 45 = 2$ moles of ethyl amine.

Now, the reaction sequence is:

1. First methylation: $\text{C}_2\text{H}_5\text{NH}_2 + \text{CH}_3\text{Cl} \rightarrow \text{C}_2\text{H}_5\text{NHCH}_3 + \text{HCl}$

   (secondary amine formed)

2. Second methylation: $\text{C}_2\text{H}_5\text{NHCH}_3 + \text{CH}_3\text{Cl} \rightarrow \text{C}_2\text{H}_5\text{N}(\text{CH}_3)_2 + \text{HCl}$

   (tertiary amine formed)

3. Third methylation: $\text{C}_2\text{H}_5\text{N}(\text{CH}_3)_2 + \text{CH}_3\text{Cl} \rightarrow \text{C}_2\text{H}_5\text{N}(\text{CH}_3)_3^+ \text{Cl}^-$

   (quaternary salt — but this is not formed here because the product is exclusively tertiary)

Since the tertiary amine is the exclusive product, the reaction stops after exactly two methylations per molecule of ethyl amine. That means each mole of ethyl amine consumes exactly 2 moles of methyl chloride. …