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. `1.6 xx 10^(-19) ms^(-2)`
B. `9.1 xx 10^(31) ms^(-2)`
C. `1.5 xx 10^(19) ms^(-2)`
D. `1.6 xx 10^(27) ms^(-2)`

Answer

To solve the problem, we need to find the acceleration of a proton when it experiences the same Coulombic force that causes an electron to accelerate at \(2.5 \times 10^{22} \, \text{m/s}^2\).

### Step-by-Step Solution:

1. **Understand the relationship between force, mass, and acceleration**:
According to Newton's second law, the force \(F\) acting on an object is given by:
\[ F = m \cdot a \]
where \(m\) is the mass of the object and \(a\) is its acceleration.

2. **Express acceleration in terms of force and mass**:
Rearranging the formula gives us:
\[ a = \frac{F}{m} \]

3. **Identify the relationship between the accelerations of the electron and proton**:
Since the same force \(F\) acts on both the electron and the proton, we can write:
\[ a_e = \frac{F}{m_e} \quad \text{(for the electron)} \]
\[ a_p = \frac{F}{m_p} \quad \text{(for the proton)} \]
where \(a_e\) is the acceleration of the electron, \(a_p\) is the acceleration of the proton, \(m_e\) is the mass of the electron, and \(m_p\) is the mass of the proton.

4. **Relate the two accelerations**:
From the equations for \(a_e\) and \(a_p\), we can derive:
\[ \frac{a_e}{a_p} = \frac{m_p}{m_e} \]
Rearranging gives:
\[ a_p = a_e \cdot \frac{m_e}{m_p} \]

5. **Substitute known values**:
We know:
- \(a_e = 2.5 \times 10^{22} \, \text{m/s}^2\)
- Mass of the electron, \(m_e = 9.1 \times 10^{-31} \, \text{kg}\)
- Mass of the proton, \(m_p = 1.6 \times 10^{-27} \, \text{kg}\)

Now, substituting these values into the equation for \(a_p\):
\[ a_p = 2.5 \times 10^{22} \cdot \frac{9.1 \times 10^{-31}}{1.6 \times 10^{-27}} \]

6. **Calculate the ratio of masses**:
\[ \frac{m_e}{m_p} = \frac{9.1 \times 10^{-31}}{1.6 \times 10^{-27}} = \frac{9.1}{1.6} \times 10^{-31 + 27} = \frac{9.1}{1.6} \times 10^{-4} \]
\[ \frac{9.1}{1.6} \approx 5.6875 \]

7. **Calculate \(a_p\)**:
Now substituting back:
\[ a_p = 2.5 \times 10^{22} \cdot 5.6875 \times 10^{-4} \]
\[ a_p \approx 14.21875 \times 10^{18} \, \text{m/s}^2 \]
Writing this in scientific notation:
\[ a_p \approx 1.42 \times 10^{19} \, \text{m/s}^2 \]

### Final Answer:
The magnitude of the acceleration of the proton under the action of the same force is approximately \(1.42 \times 10^{19} \, \text{m/s}^2\).