If in Millikan's oil drop experiment charges on drops are found to be `80 mu C, 12 mu C, 20 mu C`, then quanta of charge is :

Options

A. `8 muC`
B. `4 mu C`
C. `20 muC`
D. `12 mu C`

Answer

To determine the quanta of charge in Millikan's oil drop experiment, we need to find the greatest common divisor (GCD) of the given charges. The charges on the drops are:

- \( q_1 = 80 \, \mu C \)
- \( q_2 = 12 \, \mu C \)
- \( q_3 = 20 \, \mu C \)

### Step 1: Convert the charges to the same unit
Since all the charges are already in microcoulombs (µC), we can proceed to the next step.

### Step 2: Find the GCD of the charges
To find the quanta of charge, we need to calculate the GCD of the three charges.

1. **Find the GCD of \(80\) and \(12\)**:
- The factors of \(80\) are \(1, 2, 4, 5, 8, 10, 16, 20, 40, 80\).
- The factors of \(12\) are \(1, 2, 3, 4, 6, 12\).
- The common factors are \(1, 2, 4\).
- The GCD is \(4\).

2. **Now find the GCD of \(4\) and \(20\)**:
- The factors of \(20\) are \(1, 2, 4, 5, 10, 20\).
- The common factors with \(4\) are \(1, 2, 4\).
- The GCD is \(4\).

Thus, the GCD of \(80 \, \mu C\), \(12 \, \mu C\), and \(20 \, \mu C\) is \(4 \, \mu C\).

### Step 3: Conclusion
The quanta of charge, which is the smallest unit of charge that can be observed in this experiment, is \(4 \, \mu C\).

### Final Answer
The quanta of charge is \(4 \, \mu C\).

---