The acceleration for electron and proton due to electrical force of their mutual attraction when they are 1 Å apart is

Options

A. `3.1xx10^(22)ms^(-2),1.3xx10^(19)ms^(-2)`
B. `3.3xx10^(28)ms^(-2),3.2xx10^(16)ms^(-2)`
C. `2.5xx10^(22)ms^(-2),1.4xx10^(19)ms^(-2)`
D. `2.5xx10^(18)ms^(-2),1.3xx10^(16)ms^(-2)`

Explanation

To find the acceleration of an electron and a proton due to the electrical force of their mutual attraction when they are 1 Å apart, we can follow these steps:

### Step 1: Determine the distance in meters
1 Å (angstrom) is equal to \( 1 \times 10^{-10} \) meters.

### Step 2: Calculate the electric force between the electron and proton
Using Coulomb's Law:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)
- \( q_1 = q_2 = e = 1.6 \times 10^{-19} \, \text{C} \) (magnitude of charge of electron and proton)
- \( r = 1 \times 10^{-10} \, \text{m} \)

Substituting the values:
\[ F = \frac{9 \times 10^9 \cdot (1.6 \times 10^{-19})^2}{(1 \times 10^{-10})^2} \]
Calculating this gives:
\[ F = \frac{9 \times 10^9 \cdot 2.56 \times 10^{-38}}{1 \times 10^{-20}} = 2.3 \times 10^{-8} \, \text{N} \]

### Step 3: Calculate the acceleration of the electron
Using Newton's second law:
\[ a = \frac{F}{m} \]
where \( m \) is the mass of the electron, \( m_e = 9 \times 10^{-31} \, \text{kg} \).

Substituting the values:
\[ a_e = \frac{2.3 \times 10^{-8}}{9 \times 10^{-31}} \approx 2.5 \times 10^{22} \, \text{m/s}^2 \]

### Step 4: Calculate the acceleration of the proton
Using the same formula:
\[ m_p = 1.67 \times 10^{-27} \, \text{kg} \, (\text{mass of proton}) \]
Substituting the values:
\[ a_p = \frac{2.3 \times 10^{-8}}{1.67 \times 10^{-27}} \approx 1.4 \times 10^{19} \, \text{m/s}^2 \]

### Final Result
- The acceleration of the electron is approximately \( 2.5 \times 10^{22} \, \text{m/s}^2 \).
- The acceleration of the proton is approximately \( 1.4 \times 10^{19} \, \text{m/s}^2 \).