Q.If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

The key idea is that electric current is the flow of charge, and the total charge is quantised in units of the electron charge.

Step 1 – Total charge passed

Current $I = 0.5\ \text{A}$, time $t = 2\ \text{hours} = 2 \times 3600 = 7200\ \text{s}$.

Charge $Q = I \times t = 0.5 \times 7200 = 3600\ \text{C}$.

Step 2 – Charge per electron

Charge on one electron $e = 1.6 \times 10^{-19}\ \text{C}$.

Step 3 – Number of electrons

Number $n = \frac{Q}{e} = \frac{3600}{1.6 \times 10^{-19}} = 2.25 \times 10^{22}$.

This is a straightforward application of the relation between current, charge, and time — a fundamental idea in electricity. Current is simply the rate of flow of charge: $I = \frac{Q}{t}$. So if you know how much current flows and for how long, you can find the total charge that has passed. Then, since each electron carries a fixed amount of charge (the elementary charge $e$), dividing the total charge by $e$ gives the number of electrons.

Let’s work it out step by step.

1. Convert time to seconds.

   The current is given in amperes (coulombs per second), so time must be in seconds.

   $t = 2 \, \text{hours} = 2 \times 60 \times 60 = 7200 \, \text{s}$.

2. Calculate total charge $Q$.

   Using $Q = I \times t$:

   $Q = 0.5 \, \text{A} \times 7200 \, \text{s} = 3600 \, \text{C}$.

3. Recall the charge of one electron.

   The elementary charge $e = 1.6 \times 10^{-19} \, \text{C}$ (this is a standard value you must remember for exams).

4. Find the number of electrons $n$.

   $n = \frac{Q}{e} = \frac{3600}{1.6 \times 10^{-19}}$.

   Compute:

   $\frac{3600}{1.6} = 2250$, and $2250 \times 10^{19} = 2.25 \times 10^{22}$.

Method: Direct Charge-Quantization Approach

This method uses the fundamental relation between current, time, and the quantized nature of electric charge.

Step 1: Find total charge ($Q$) that flows

Current is charge per unit time:

So:

Given:

• $I = 0.5 \, \text{A}$
• $t = 2 \, \text{hours} = 2 \times 3600 = 7200 \, \text{s}$

Total charge flowing = 3600 C

Step 2: Use charge quantization to find number of electrons

Every electron carries a charge of:

If $n$ is the number of electrons:

So:

Final Answer

Number of electrons = $\mathbf{2.25 \times 10^{22}}$

Key Concept Reminder

• Faraday’s laws deal with electrolysis (chemical change due to current).
• This problem is purely electrical — it uses the quantization of charge (charge is always an integer multiple of $e$).
• The formula $Q = n e$ is the bridge between macroscopic current and microscopic particle count.

Here are the most common mistakes students make on this Faraday’s Laws / Electrolysis type question, along with how to avoid each.

Mistake 1: Forgetting to convert time to seconds

The mistake:

Students directly use time in hours in the formula $Q = I \times t$, getting a wildly wrong charge.

Why it happens:

The formula $Q = I t$ requires time in seconds (SI unit), but the problem gives time in hours.

How to avoid:

Always convert hours → minutes → seconds:

$t = 2 \text{ hours} = 2 \times 60 \times 60 = 7200 \text{ s}$.

Correct step:

$Q = 0.5 \times 7200 = 3600 \text{ C}$.

Mistake 2: Using the wrong value of Faraday constant or electronic charge

The mistake:

Some students use $F = 96500 \text{ C/mol}$ directly without linking it to the number of electrons.

Why it happens:

They confuse the charge per mole of electrons (Faraday) with the charge on a single electron.

How to avoid:

Remember:

• Charge on one electron = $e = 1.6 \times 10^{-19} \text{ C}$
• Number of electrons $n = \frac{Q}{e}$

Correct step:

$n = \frac{3600}{1.6 \times 10^{-19}} = 2.25 \times 10^{22}$ electrons.

Mistake 3: Mixing up Faraday’s laws for electrolysis with this simple current flow

The mistake:

Students try to use $m = \frac{Z I t}{F}$ or involve molar mass, thinking it’s an electrolysis cell.

Why it happens:

The problem mentions “metallic wire” — it’s not an electrolytic cell. It’s just conduction through a metal.

How to avoid:

• Metallic wire → electrons flow directly. Use $Q = I t$ and $n = Q/e$.
• Electrolytic cell → ions carry charge. Use Faraday’s laws.

Mistake 4: Incorrect handling of powers of 10 in division

The mistake:

Students miscalculate $3600 \div (1.6 \times 10^{-19})$ and get $2.25 \times 10^{17}$ or $2.25 \times 10^{21}$.

Why it happens:

Dividing by $10^{-19}$ means multiplying by $10^{19}$, but they forget to adjust the exponent correctly.

How to avoid:

Write it step-by-step:

$\frac{3600}{1.6 \times 10^{-19}} = \frac{3600}{1.6} \times 10^{19} = 2250 \times 10^{19} = 2.25 \times 10^{22}$.

Mistake 5: Not writing the final answer in scientific notation

The mistake:

Leaving the answer as $22500000000000000000000$ or rounding incorrectly.

Why it happens:

They don’t convert to standard form.

How to avoid:

Always express large numbers as $a \times 10^{b}$ where $1 \leq a < 10$.

Final answer:

$\boxed{2.25 \times 10^{22} \text{ electrons}}$

Quick checklist to avoid all mistakes: