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 under the action of the same Coulombic force that causes a given acceleration of an electron. We will use the relationship between force, mass, and acceleration.

### Step-by-Step Solution:

1. **Understand the relationship between force, mass, and acceleration:**
\[ F = ma \]
Here, \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.

2. **Express acceleration in terms of force and mass:**
\[ a = \frac{F}{m} \]
Since the force \( F \) is constant for both the electron and the proton, we can say that the acceleration is inversely proportional to the mass:
\[ a \propto \frac{1}{m} \]

3. **Set up the ratio of accelerations for the electron and proton:**
\[ \frac{a_e}{a_p} = \frac{m_p}{m_e} \]
Where \( a_e \) is the acceleration of the electron, \( a_p \) is the acceleration of the proton, \( m_p \) is the mass of the proton, and \( m_e \) is the mass of the electron.

4. **Insert known values for the masses:**
- Mass of electron \( m_e = 9.1 \times 10^{-31} \, \text{kg} \)
- Mass of proton \( m_p = 1.6 \times 10^{-27} \, \text{kg} \)

5. **Calculate the ratio of accelerations:**
\[ \frac{a_e}{a_p} = \frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}} \]

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

7. **Substitute the given value of electron's acceleration:**
\[ a_e = 2.5 \times 10^{22} \, \text{m/s}^2 \]
\[ a_p = 2.5 \times 10^{22} \cdot \frac{9.1 \times 10^{-31}}{1.6 \times 10^{-27}} \]

8. **Calculate the acceleration of the proton:**
\[ a_p = 2.5 \times 10^{22} \cdot \frac{9.1}{1.6} \times 10^{-31 + 27} \]
\[ a_p = 2.5 \times 10^{22} \cdot 5.6875 \times 10^{-4} \]
\[ a_p \approx 14.2 \times 10^{18} \, \text{m/s}^2 \]

9. **Convert to scientific notation:**
\[ a_p \approx 1.42 \times 10^{19} \, \text{m/s}^2 \]

10. **Round to the nearest significant figure:**
\[ a_p \approx 1.5 \times 10^{19} \, \text{m/s}^2 \]

### Final Answer:
The magnitude of the acceleration of a proton under the action of the same force is approximately \( 1.5 \times 10^{19} \, \text{m/s}^2 \).