The values of charge on the oil droplets experimentally observed were `-1.6 xx10^(-19), -2.4 xx10^(-19) and -4 xx 10^(-19)` coulomb. The value of the electronic charge, indicated by these results is

Options

A. `-1.6 xx 10^(-19)`
B. `-2.4 xx 10^(-19)`
C. `-4 xx 10^(-19)`
D. `-0.8 xx 10^(-19)`

Explanation

To find the value of the electronic charge based on the observed charges on the oil droplets, we can follow these steps:

### Step 1: Identify the charges given
The charges observed on the oil droplets are:
1. \( Q_1 = -1.6 \times 10^{-19} \) coulombs
2. \( Q_2 = -2.4 \times 10^{-19} \) coulombs
3. \( Q_3 = -4.0 \times 10^{-19} \) coulombs

### Step 2: Express the charges in terms of the electronic charge
The charge on the oil droplets can be expressed as:
\[ Q = -nE \]
where \( n \) is an integer (the number of electrons) and \( E \) is the charge of a single electron.

### Step 3: Find the common factor
To find the electronic charge \( E \), we need to determine the integer \( n \) for each observed charge:
1. For \( Q_1 = -1.6 \times 10^{-19} \):
\[ n_1 = \frac{-Q_1}{E} = \frac{-(-1.6 \times 10^{-19})}{E} = \frac{1.6 \times 10^{-19}}{E} \]

2. For \( Q_2 = -2.4 \times 10^{-19} \):
\[ n_2 = \frac{-Q_2}{E} = \frac{-(-2.4 \times 10^{-19})}{E} = \frac{2.4 \times 10^{-19}}{E} \]

3. For \( Q_3 = -4.0 \times 10^{-19} \):
\[ n_3 = \frac{-Q_3}{E} = \frac{-(-4.0 \times 10^{-19})}{E} = \frac{4.0 \times 10^{-19}}{E} \]

### Step 4: Calculate \( E \) using the smallest charge
From the first charge \( Q_1 \):
Assuming \( n_1 = 1 \):
\[ E = 1.6 \times 10^{-19} \text{ C} \]

Now, check if this value of \( E \) satisfies the other charges:
- For \( Q_2 \):
\[ n_2 = \frac{2.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.5 \] (not an integer)

- For \( Q_3 \):
\[ n_3 = \frac{4.0 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.5 \] (not an integer)

### Step 5: Adjust \( E \) to find a consistent integer \( n \)
Next, let's try \( E = 0.8 \times 10^{-19} \text{ C} \):
- For \( Q_1 \):
\[ n_1 = \frac{1.6 \times 10^{-19}}{0.8 \times 10^{-19}} = 2 \] (integer)

- For \( Q_2 \):
\[ n_2 = \frac{2.4 \times 10^{-19}}{0.8 \times 10^{-19}} = 3 \] (integer)

- For \( Q_3 \):
\[ n_3 = \frac{4.0 \times 10^{-19}}{0.8 \times 10^{-19}} = 5 \] (integer)

### Conclusion
The value of the electronic charge \( E \) that satisfies all observed charges is:
\[ E = 0.8 \times 10^{-19} \text{ C} \]