Under the action of a given coulombic force the acceleration of an electron is `2.5 xx 10^(22) ms^(-1)`. Then, the magnitude of the acceleration of a proton under the action of same force is nearly

Options

A. `1.6xx10^(-19)ms^(-2)`
B. `9.1xx10^(31)ms^(-2)`
C. `1.5xx10^(19)ms^(-2)`
D. `1.6xx10^(27)ms^(-2)`

Explanation

To find the magnitude of the acceleration of a proton under the action of the same Coulombic force that causes an electron to accelerate at \(2.5 \times 10^{22} \, \text{m/s}^2\), we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

### Step-by-Step Solution:

1. **Identify the relationship between force, mass, and acceleration**:
\[ F = m_e \cdot a_e = m_p \cdot a_p \]
where:
- \(F\) is the force,
- \(m_e\) is the mass of the electron,
- \(a_e\) is the acceleration of the electron,
- \(m_p\) is the mass of the proton,
- \(a_p\) is the acceleration of the proton.

2. **Rearrange the equation to find the acceleration of the proton**:
\[ a_p = \frac{m_e \cdot a_e}{m_p} \]

3. **Substitute the known values**:
- Given \(a_e = 2.5 \times 10^{22} \, \text{m/s}^2\),
- Mass of the electron \(m_e = 9.1 \times 10^{-31} \, \text{kg}\),
- Mass of the proton \(m_p = 1.66 \times 10^{-27} \, \text{kg}\).

Plugging in these values:
\[ a_p = \frac{(9.1 \times 10^{-31} \, \text{kg}) \cdot (2.5 \times 10^{22} \, \text{m/s}^2)}{1.66 \times 10^{-27} \, \text{kg}} \]

4. **Calculate the numerator**:
\[ 9.1 \times 10^{-31} \times 2.5 \times 10^{22} = 2.275 \times 10^{-8} \, \text{kg m/s}^2 \]

5. **Calculate the acceleration of the proton**:
\[ a_p = \frac{2.275 \times 10^{-8}}{1.66 \times 10^{-27}} \approx 1.37 \times 10^{19} \, \text{m/s}^2 \]

6. **Final result**:
The magnitude of the acceleration of the proton under the action of the same force is approximately:
\[ a_p \approx 1.37 \times 10^{19} \, \text{m/s}^2 \]