An `alpha`- particle is the nucleus of a helium atom. It has a mas `m=6.64xx10^-27` kg and as charge `1=+2e=3.2xx10^-19`c. compare the force of the electric repulsion between two `alpha`-particles with the force of gravitational attraction between them.

Explanation

To solve the problem of comparing the electric repulsion force between two alpha particles with the gravitational attraction force between them, we can follow these steps:

### Step 1: Calculate the Electric Force of Repulsion (Fe)
The electric force of repulsion between two charges can be calculated using Coulomb's law:

\[ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \]

Where:
- \( k \) is Coulomb's constant, \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 \) and \( q_2 \) are the charges of the alpha particles, which are both \( +2e \) (where \( e \approx 1.6 \times 10^{-19} \, \text{C} \)).
- \( r \) is the distance between the two particles.

Since both charges are the same, we can simplify this to:

\[ F_e = \frac{k \cdot q^2}{r^2} \]

Substituting \( q = 2e \):

\[ F_e = \frac{k \cdot (2e)^2}{r^2} = \frac{4k \cdot e^2}{r^2} \]

### Step 2: Calculate the Gravitational Force of Attraction (Fg)
The gravitational force between two masses can be calculated using Newton's law of gravitation:

\[ F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where:
- \( G \) is the gravitational constant, \( G \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( m_1 \) and \( m_2 \) are the masses of the alpha particles, which are both \( m = 6.64 \times 10^{-27} \, \text{kg} \).

Thus, we can simplify this to:

\[ F_g = \frac{G \cdot m^2}{r^2} \]

### Step 3: Find the Ratio of Electric Force to Gravitational Force
Now, we can find the ratio of the electric force to the gravitational force:

\[ \frac{F_e}{F_g} = \frac{\frac{4k \cdot e^2}{r^2}}{\frac{G \cdot m^2}{r^2}} = \frac{4k \cdot e^2}{G \cdot m^2} \]

### Step 4: Substitute Values
Substituting the known values:

- \( k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( e \approx 1.6 \times 10^{-19} \, \text{C} \)
- \( G \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( m = 6.64 \times 10^{-27} \, \text{kg} \)

Calculating \( e^2 \):

\[ e^2 \approx (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

Now substituting these values into the ratio:

\[ \frac{F_e}{F_g} = \frac{4 \cdot (9 \times 10^9) \cdot (2.56 \times 10^{-38})}{(6.67 \times 10^{-11}) \cdot (6.64 \times 10^{-27})^2} \]

Calculating \( (6.64 \times 10^{-27})^2 \):

\[ (6.64 \times 10^{-27})^2 \approx 4.40 \times 10^{-53} \, \text{kg}^2 \]

Now substituting this back into the ratio:

\[ \frac{F_e}{F_g} = \frac{4 \cdot (9 \times 10^9) \cdot (2.56 \times 10^{-38})}{(6.67 \times 10^{-11}) \cdot (4.40 \times 10^{-53})} \]

### Step 5: Final Calculation
Calculating the numerator:

\[ 4 \cdot (9 \times 10^9) \cdot (2.56 \times 10^{-38}) \approx 9.216 \times 10^{-28} \]

Calculating the denominator:

\[ (6.67 \times 10^{-11}) \cdot (4.40 \times 10^{-53}) \approx 2.9388 \times 10^{-63} \]

Now dividing:

\[ \frac{9.216 \times 10^{-28}}{2.9388 \times 10^{-63}} \approx 3.14 \times 10^{35} \]

### Conclusion
Thus, the ratio of the electric repulsion force to the gravitational attraction force between two alpha particles is approximately:

\[ \frac{F_e}{F_g} \approx 3.14 \times 10^{35} \]