Under the action of a given coulombic force the acceleration of an electron is `3.5 xx 10^(25) ms^(-2)`. 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. `1.9 xx 10^(22) 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 the same Coulombic force is applied to it as that which is applied to an electron. We will use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a), or \( F = ma \).

### Step-by-Step Solution:

1. **Identify the given values**:
- Acceleration of the electron (\( a_e \)) = \( 3.5 \times 10^{25} \, \text{m/s}^2 \)
- Mass of the electron (\( m_e \)) = \( 9.1 \times 10^{-31} \, \text{kg} \)
- Mass of the proton (\( m_p \)) = \( 1.7 \times 10^{-27} \, \text{kg} \)

2. **Apply Newton's second law**:
- For the electron, the force can be expressed as:
\[ F = m_e \cdot a_e \]
- For the proton, the force is:
\[ F = m_p \cdot a_p \]
- Since the force is the same for both the electron and the proton, we can set the two equations equal:
\[ m_e \cdot a_e = m_p \cdot a_p \]

3. **Rearranging the equation**:
- We can rearrange the equation to find the acceleration of the proton (\( a_p \)):
\[ a_p = \frac{m_e}{m_p} \cdot a_e \]

4. **Substituting the values**:
- Substitute the known values into the equation:
\[ a_p = \frac{9.1 \times 10^{-31}}{1.7 \times 10^{-27}} \cdot (3.5 \times 10^{25}) \]

5. **Calculating the mass ratio**:
- Calculate the mass ratio:
\[ \frac{9.1 \times 10^{-31}}{1.7 \times 10^{-27}} \approx 5.35 \times 10^{-4} \]

6. **Calculating the acceleration of the proton**:
- Now, substitute this ratio back into the equation for \( a_p \):
\[ a_p \approx (5.35 \times 10^{-4}) \cdot (3.5 \times 10^{25}) \approx 1.87 \times 10^{22} \, \text{m/s}^2 \]

7. **Final approximation**:
- We can approximate \( 1.87 \times 10^{22} \) to \( 1.9 \times 10^{22} \, \text{m/s}^2 \).

### Final Answer:
The magnitude of the acceleration of the proton under the action of the same force is approximately \( 1.9 \times 10^{22} \, \text{m/s}^2 \).

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