The force between an `alpha`-particle and an electron separated by a distance of `1 Å` is -

Options

A. `2.3 xx 10^(-8)` N attractive
B. `2.3 xx 10^(-8)` N Repulsive
C. `4.6 xx 10^(-8)` N attractive
D. `4.6 xx 10^(-8)` N repulsive

Answer

To find the force between an alpha particle and an electron separated by a distance of 1 Å, we can use Coulomb's law, which 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 1: Identify the charges
An alpha particle consists of 2 protons and 2 neutrons, giving it a charge of \( +2e \). The charge of an electron is \( -e \).

\[ Q_1 = +2e \quad \text{(for the alpha particle)} \]
\[ Q_2 = -e \quad \text{(for the electron)} \]

The elementary charge \( e \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).

### Step 2: Calculate the magnitudes of the charges
Substituting the value of \( e \):

\[ Q_1 = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]
\[ Q_2 = -1.6 \times 10^{-19} \, \text{C} \]

### Step 3: Convert distance to meters
The distance \( r \) is given as \( 1 \, \text{Å} \):

\[ r = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \]

### Step 4: Substitute values into Coulomb's law
Now we can substitute the values into Coulomb's law:

\[ F = \frac{9 \times 10^9 \cdot |(3.2 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})|}{(1 \times 10^{-10})^2} \]

### Step 5: Calculate the force
Calculating the numerator:

\[ F = \frac{9 \times 10^9 \cdot (3.2 \times 10^{-19} \cdot 1.6 \times 10^{-19})}{(1 \times 10^{-10})^2} \]
\[ = \frac{9 \times 10^9 \cdot (5.12 \times 10^{-38})}{1 \times 10^{-20}} \]
\[ = 9 \times 10^9 \cdot 5.12 \times 10^{-18} \]
\[ = 4.608 \times 10^{-8} \, \text{N} \]

### Step 6: Determine the nature of the force
Since the alpha particle has a positive charge and the electron has a negative charge, the force will be attractive.

### Final Answer
The force between the alpha particle and the electron is:

\[ F \approx 4.6 \times 10^{-8} \, \text{N} \quad \text{(attractive)} \]

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