The acceleration of electron due to mutual attraction between electron and a proton when they are 1.6 Angstroms apart is

Options

A. `10^24 m/s^2`
B. `10^23 m/s^2`
C. `10^22 m/s^2`
D. `10^25 m/s^2`

Answer

To find the acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Angstroms apart, we can follow these steps:

### Step 1: Understand the Problem
We have an electron and a proton, which are oppositely charged particles. The force of attraction between them can be calculated using Coulomb's Law.

### Step 2: Write Down Coulomb's Law
Coulomb's Law states that the force \( F \) between two charges is given by:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges
- \( r \) is the distance between the charges

### Step 3: Identify the Charges and Distance
The charge of an electron \( q_e \) is \( -1.6 \times 10^{-19} \, \text{C} \) and the charge of a proton \( q_p \) is \( +1.6 \times 10^{-19} \, \text{C} \). The distance \( r \) is given as \( 1.6 \, \text{Å} = 1.6 \times 10^{-10} \, \text{m} \).

### Step 4: Calculate the Force
Substituting the values into Coulomb's Law:
\[ F = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{(1.6 \times 10^{-10})^2} \]

Calculating \( (1.6 \times 10^{-19})^2 \):
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]

Calculating \( (1.6 \times 10^{-10})^2 \):
\[ (1.6 \times 10^{-10})^2 = 2.56 \times 10^{-20} \]

Now substituting these values back into the force equation:
\[ F = \frac{(9 \times 10^9) \cdot (2.56 \times 10^{-38})}{2.56 \times 10^{-20}} = \frac{9 \times 10^9 \cdot 2.56 \times 10^{-38}}{2.56 \times 10^{-20}} = 9 \times 10^{-9} \, \text{N} \]

### Step 5: Calculate the Acceleration
Using Newton's second law, \( F = m \cdot a \), we can find the acceleration \( a \):
\[ a = \frac{F}{m} \]
where \( m \) is the mass of the electron, approximately \( 9.11 \times 10^{-31} \, \text{kg} \).

Substituting the values:
\[ a = \frac{9 \times 10^{-9}}{9.11 \times 10^{-31}} \approx 9.87 \times 10^{21} \, \text{m/s}^2 \]

### Final Answer
The acceleration of the electron due to mutual attraction with the proton when they are 1.6 Angstroms apart is approximately \( 9.87 \times 10^{21} \, \text{m/s}^2 \).

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