Question

The number of electrons moving per second through the filament of a lamp of 60W operating at 120V is nearly (e = 1.6 \(\times\) 10\(^{-19}\) C) \textunderscore\textunderscore\textunderscore\textunderscore\textunderscore\textunderscore\textunderscore

• A. 6.2 \(\times\) 10\(^{18}\)
• B. 6.2 \(\times\) 10\(^{19}\)
• C. 3.1 \(\times\) 10\(^{18}\)
• D. 3.1 \(\times\) 10\(^{19}\)

Hint:

Remember that 1 Ampere corresponds to approximately \( 6.25 \times 10^{18} \) electrons passing per second. Since our current is 0.5 A, the answer must be half of that value, which is \( \approx 3.12 \times 10^{18} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to find the rate of electron flow (electrons per second) in a circuit given the power and voltage of an appliance.
Step 2: Key Formula or Approach:
1. Electric Power: \( P = V \cdot I \Rightarrow I = \frac{P}{V} \).
2. Current and Charge: \( I = \frac{Q}{t} \).
3. Quantization of Charge: \( Q = n \cdot e \).
Combining these, the number of electrons per second (\(n\) for \(t = 1\text{s}\)) is:
\[ n = \frac{I}{e} = \frac{P}{V \cdot e} \]
Step 3: Detailed Explanation:
Given:
- Power, \(P = 60 \text{ W}\).
- Voltage, \(V = 120 \text{ V}\).
- Electronic charge, \(e = 1.6 \times 10^{-19} \text{ C}\).
First, calculate the current (\(I\)):
\[ I = \frac{P}{V} = \frac{60}{120} = 0.5 \text{ A} \]
Now, calculate the number of electrons per second (\(n\)):
\[ n = \frac{I}{e} = \frac{0.5}{1.6 \times 10^{-19}} \]
\[ n = \frac{5}{16} \times 10^{19} \]
\[ n \approx 0.3125 \times 10^{19} = 3.125 \times 10^{18} \]
Rounding to the nearest significant figures as per options: \(3.1 \times 10^{18}\).
Step 4: Final Answer:
The number of electrons moving per second is nearly \(3.1 \times 10^{18}\).