The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given e=1.6 × 10⁻¹⁹ C)

Solution

Physics sir · 3 steps

Get the current from P = VI first, then count how many electron charges make up that current each second.

1. Find the current. Power, voltage and current are linked by P = VI. So I = P/V = 110/220 = 0.5 A.
2. Current as electron flow. A current of I amperes means I coulombs pass each second. Each electron carries charge e, so the number passing per second is n = I/e.

   n = I/e = 0.5/(1.6 × 10⁻¹⁹) = 3.125 × 10¹⁸ s⁻¹

3. So. 3.125 × 10¹⁸ is the same as 31.25 × 10¹⁷ electrons per second, which is option (b).

The idea to keep. Electrons per second is current divided by e. When a question gives power and voltage, find I = P/V first and then divide by 1.6 × 10⁻¹⁹ C.

Answer from the NTA final answer key. Solution by PrepWiser.