A body has a charge of `9.6xx10^(-20)` coulomb.It is

Options

A. Possible
B. not possible
C. may (or) may not possible
D. Data not sufficient

Explanation

To determine whether the charge of \(9.6 \times 10^{-20}\) coulombs is possible, we need to analyze it in the context of quantization of electric charge.

### Step-by-Step Solution:

1. **Understanding Charge Quantization**:
- Electric charge is quantized, meaning it can only exist in integer multiples of the elementary charge \(e\). The elementary charge \(e\) is approximately \(1.6 \times 10^{-19}\) coulombs.

2. **Setting Up the Equation**:
- We can express the charge \(Q\) as:
\[ Q = N \times e \]
where \(N\) is an integer (the number of elementary charges).

3. **Substituting the Values**:
- Given \(Q = 9.6 \times 10^{-20}\) coulombs, we can substitute this into the equation:
\[ 9.6 \times 10^{-20} = N \times 1.6 \times 10^{-19} \]

4. **Solving for \(N\)**:
- Rearranging the equation to solve for \(N\):
\[ N = \frac{9.6 \times 10^{-20}}{1.6 \times 10^{-19}} \]
- Performing the division:
\[ N = \frac{9.6}{1.6} \times \frac{10^{-20}}{10^{-19}} = 6 \times 10^{-1} = 0.6 \]

5. **Analyzing the Result**:
- Since \(N = 0.6\) is not an integer, it indicates that the charge \(9.6 \times 10^{-20}\) coulombs cannot be represented as an integer multiple of the elementary charge \(e\).

6. **Conclusion**:
- Therefore, the charge \(9.6 \times 10^{-20}\) coulombs is not possible as it does not conform to the quantization of charge.

### Final Answer:
The charge \(9.6 \times 10^{-20}\) coulombs is not possible as it is not an integer multiple of the elementary charge.